Chapter 12 Coordinate Geometry (Q & A)

This question asks us to find the distance between two given points in a Cartesian coordinate system. We will use the distance formula for this purpose.

Given:

The two points are $P_1(2, 3)$ and $P_2(5, 7)$.

Let $(x_1, y_1) = (2, 3)$.

Let $(x_2, y_2) = (5, 7)$.

To Find:

The distance between points $P_1$ and $P_2$.

Solution:

The distance $D$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian coordinate system is given by the distance formula:

Substitute the coordinates of the given points into the distance formula:

$x_1 = 2$, $y_1 = 3$

$x_2 = 5$, $y_2 = 7$

Now, plug these values into equation (i):

$D = \sqrt{(5 - 2)^2 + (7 - 3)^2}$

$D = \sqrt{(3)^2 + (4)^2}$

$D = \sqrt{9 + 16}$

$D = \sqrt{25}$

$D = 5$

Thus, the distance between the points $(2, 3)$ and $(5, 7)$ is 5 units.

Conclusion:

The calculated distance is 5. Comparing this with the given options:

• (A) 3
• (B) 4
• (C) 5
• (D) 7

The correct option is (C) 5.

This question asks for the standard formula for the slope of a straight line when two points on the line are known. The slope of a line represents its steepness or gradient.

Given:

Two arbitrary points on a line are $(x_1, y_1)$ and $(x_2, y_2)$.

To Find:

The formula for the slope of the line passing through these two points.

Solution:

The slope of a line, often denoted by $m$, is defined as the ratio of the change in the y-coordinate ($\Delta y$) to the change in the x-coordinate ($\Delta x$) between any two distinct points on the line.

Mathematically, this is expressed as:

$m = \frac{\text{Change in y}}{\text{Change in x}} = \frac{\Delta y}{\Delta x}$

Given the two points $(x_1, y_1)$ and $(x_2, y_2)$:

The change in y-coordinate is $y_2 - y_1$.

The change in x-coordinate is $x_2 - x_1$.

Therefore, the formula for the slope is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

It is crucial to note that this formula is valid only when $x_1 \neq x_2$. If $x_1 = x_2$, the line is a vertical line, and its slope is undefined (as it would involve division by zero).

Let's analyze the given options:

(A) $\frac{y_2 + y_1}{x_2 + x_1}$: This formula is incorrect. It represents the sum of coordinates, not their difference, which is essential for slope.

(B) $\frac{y_2 - y_1}{x_2 - x_1}$ (where $x_1 \neq x_2$): This is the correct and standard formula for the slope of a line passing through two points. The condition $x_1 \neq x_2$ correctly addresses the case of a vertical line.

(C) $\frac{x_2 - x_1}{y_2 - y_1}$ (where $y_1 \neq y_2$): This formula represents the reciprocal of the slope, or $\frac{\Delta x}{\Delta y}$. This is not the conventional slope formula. If $y_1 = y_2$, the line is horizontal, and its slope is 0, while this expression would be undefined.

(D) $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$: This is the distance formula between two points, not the slope formula.

Conclusion:

Based on the definition and standard formula for the slope of a line, the correct option is (B) $\frac{y_2 - y_1}{x_2 - x_1}$ (where $x_1 \neq x_2$).

This question asks for the slope of a line given its equation in a common algebraic form. We can determine the slope by comparing the given equation to the slope-intercept form of a linear equation.

Given:

The equation of the line is $y = 2x + 3$.

To Find:

The slope of the given line.

Solution:

The general form of a linear equation in slope-intercept form is:

$y = mx + c$

where:

• $m$ represents the slope of the line.
• $c$ represents the y-intercept (the point where the line crosses the y-axis, i.e., when $x=0$).

Given the equation of the line:

$y = 2x + 3$

By comparing this equation directly with the slope-intercept form ($y = mx + c$), we can identify the values of $m$ and $c$.

Here, the coefficient of $x$ is 2. Therefore, the slope $m = 2$.

The constant term is 3, which means the y-intercept $c = 3$.

So, the slope of the line $y = 2x + 3$ is 2.

Conclusion:

The calculated slope is 2. Comparing this with the given options:

• (A) 2
• (B) 3
• (C) -2
• (D) -3

The correct option is (A) 2.

This question asks us to identify the standard equation of a straight line in its slope-intercept form from the given options. Each option represents a different common form of a linear equation.

Given:

Four different forms of linear equations.

To Find:

The equation that represents the slope-intercept form of a straight line.

Solution:

Let's examine each option and determine what form of a linear equation it represents:

(A) $Ax + By + C = 0$

This is the general form or standard form of a linear equation, where A, B, and C are constants, and A and B are not both zero. This form is versatile but does not directly reveal the slope or y-intercept.

(B) $y - y_1 = m(x - x_1)$

This is the point-slope form of a linear equation. It is used when the slope $m$ of the line and a point $(x_1, y_1)$ on the line are known. It allows for easy construction of the equation if a point and slope are given.

(C) $y = mx + c$

This is the slope-intercept form of a linear equation. In this form:

• $m$ represents the slope of the line.
• $c$ (or $b$ in some notations) represents the y-intercept, which is the y-coordinate where the line crosses the y-axis (i.e., the point $(0, c)$).

This form is very useful as it directly provides the slope and y-intercept of the line, making it easy to graph.

(D) $\frac{x}{a} + \frac{y}{b} = 1$

This is the intercept form (or double-intercept form) of a linear equation. In this form:

• $a$ represents the x-intercept (the x-coordinate where the line crosses the x-axis, i.e., the point $(a, 0)$).
• $b$ represents the y-intercept (the y-coordinate where the line crosses the y-axis, i.e., the point $(0, b)$).

From the analysis, it is clear that option (C) correctly represents the slope-intercept form.

Conclusion:

The equation of a straight line in slope-intercept form is $y = mx + c$. Therefore, the correct option is (C) $y = mx + c$.

This question asks for the fundamental condition related to the slopes of two lines that are parallel. Understanding the geometric meaning of slope is key to answering this question.

Given:

Two lines are parallel.

To Find:

The relationship between their slopes.

Solution:

Let's consider two distinct lines, Line 1 and Line 2, with slopes $m_1$ and $m_2$ respectively.

By definition, parallel lines are lines in the same plane that are always the same distance apart and never intersect, no matter how far they are extended.

The slope of a line measures its steepness or gradient. It describes how much the line rises or falls vertically for a given horizontal change.

If two lines are parallel, it means they are heading in the exact same direction and have the same steepness. For them to have the same steepness, their slopes must be identical.

Therefore, if Line 1 is parallel to Line 2, then their slopes must be equal.

Let's evaluate the given options:

• (A) Equal: This aligns with our understanding that parallel lines have the same direction and steepness, thus their slopes are equal. This is the correct condition.
• (B) Negative reciprocals of each other: This means if $m_1$ is the slope of one line, the slope of the other line is $m_2 = -\frac{1}{m_1}$ (provided $m_1 \neq 0$). This is the condition for two lines to be perpendicular (intersecting at a $90^\circ$ angle), not parallel. For example, if $m_1 = 2$, then $m_2 = -\frac{1}{2}$.
• (C) Products equal to -1: This means $m_1 \times m_2 = -1$. This is mathematically equivalent to the slopes being negative reciprocals ($m_2 = -\frac{1}{m_1}$). Hence, this is also the condition for perpendicular lines (excluding cases where one line is horizontal and the other is vertical, where slopes are $0$ and undefined respectively).
• (D) Sums equal to 0: This means $m_1 + m_2 = 0$, or $m_1 = -m_2$. This implies that one slope is the negative of the other. For example, if $m_1 = 2$, then $m_2 = -2$. Lines with slopes of 2 and -2 are not parallel; they intersect. They would only be parallel if $m_1=m_2=0$, which is a specific case of equal slopes for horizontal lines.

Based on the definitions of parallel lines and slope, the only condition that ensures two lines are parallel is that their slopes are equal.

Conclusion:

Two lines are parallel if and only if their slopes are equal.

Therefore, the correct option is (A) Equal.

This question asks for the condition regarding the slopes of two lines that are perpendicular to each other. Perpendicular lines are fundamental in geometry and analytical geometry, forming right angles where they intersect.

Given:

Two lines are perpendicular.

To Find:

The relationship between the product of their slopes.

Solution:

Let's consider two lines, Line 1 and Line 2, with slopes $m_1$ and $m_2$ respectively.

By definition, perpendicular lines are lines that intersect at a right angle ($90^\circ$).

For two non-vertical lines to be perpendicular, the relationship between their slopes is that they are negative reciprocals of each other.

This means if $m_1$ is the slope of the first line, then the slope of the second perpendicular line $m_2$ is:

Multiplying both sides by $m_1$, we get the product of their slopes:

This condition holds true for all perpendicular lines except for special cases involving vertical and horizontal lines:

• A horizontal line has a slope of $0$ ($m_1 = 0$).
• A vertical line has an undefined slope ($m_2$ is undefined).

In this special case, a horizontal line is perpendicular to a vertical line. While their slopes do not technically satisfy the $m_1 m_2 = -1$ product (because one slope is undefined), the rule is generally applied to non-vertical lines. However, if we consider the general condition, for any pair of perpendicular lines, either their slopes are negative reciprocals (product is -1), or one is horizontal ($m=0$) and the other is vertical (slope undefined).

Let's evaluate the given options:

• (A) 0: If the product of slopes is 0, it means at least one of the slopes is 0. This occurs if one line is horizontal. However, this does not imply perpendicularity unless the other line is vertical (whose slope is undefined).
• (B) 1: If $m_1 m_2 = 1$, it means $m_2 = \frac{1}{m_1}$. This condition implies that the lines are reflections of each other across the line $y=x$, or more generally, they have inverse slopes. This does not correspond to perpendicularity.
• (C) -1: This is the correct condition for the product of the slopes of two perpendicular lines, excluding the case of a horizontal and a vertical line. However, in multiple-choice questions, this is the standard answer for the general condition.
• (D) Undefined: The product of slopes itself being undefined would mean one or both slopes are undefined, or one is zero and the other is undefined. While one of the slopes in a perpendicular pair can be undefined (for a vertical line), the product rule $m_1 m_2 = -1$ is for when both slopes are defined.

Therefore, the standard mathematical condition for two lines being perpendicular (assuming neither is vertical) is that the product of their slopes is -1.

Conclusion:

Two lines are perpendicular if the product of their slopes is -1.

Therefore, the correct option is (C) -1.

This question requires us to find the equation of a straight line when we are given a specific point it passes through and its slope. The most suitable form for this is the point-slope form of a linear equation.

Given:

A point on the line: $(x_1, y_1) = (2, 5)$

The slope of the line: $m = 3$

To Find:

The equation of the line.

Solution:

The point-slope form of a linear equation is given by:

Here, $x_1 = 2$, $y_1 = 5$, and $m = 3$.

Substitute these values into the point-slope formula (i):

$y - 5 = 3(x - 2)$

This equation is directly found among the given options.

Alternate Solution (Converting to Slope-Intercept Form):

We can further simplify the equation obtained in the previous step to the slope-intercept form ($y = mx + c$) to see if it matches any other options.

Start with the point-slope equation:

$y - 5 = 3(x - 2)$

Distribute the slope on the right side:

$y - 5 = 3x - 6$

Add 5 to both sides to isolate $y$:

$y = 3x - 6 + 5$

$y = 3x - 1$

This is the equation of the line in slope-intercept form. Both $y - 5 = 3(x - 2)$ and $y = 3x - 1$ represent the same straight line.

Conclusion:

By applying the point-slope formula directly, we get $y - 5 = 3(x - 2)$. This matches option (A).

Furthermore, simplifying option (A) yields $y = 3x - 1$, which matches option (D).

Both (A) and (D) are correct equations for the given line. However, option (A) is the direct form obtained from the point-slope formula, which is the most natural first step when given a point and a slope. In a multiple-choice setting where both are present, often the most direct application of the formula used for input is the intended answer, but it's important to recognize that they are equivalent.

Given the options, option (A) is the direct result of using the point-slope form with the provided data.

Therefore, the correct option is (A) $y - 5 = 3(x - 2)$.

This question is an Assertion-Reason type, where we need to evaluate the truthfulness of both the Assertion (A) and the Reason (R) and then determine if the Reason correctly explains the Assertion.

Assertion (A): The distance between $(0, 0)$ and $(x, y)$ is $\sqrt{x^2 + y^2}$.

To verify this, we use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is:

Let $(x_1, y_1) = (0, 0)$ (the origin) and $(x_2, y_2) = (x, y)$.

Substituting these values into the distance formula:

$D = \sqrt{(x - 0)^2 + (y - 0)^2}$

$D = \sqrt{x^2 + y^2}$

This matches the statement in Assertion (A).

Therefore, Assertion (A) is True.

Reason (R): This is a special case of the distance formula where one point is the origin.

As demonstrated in the verification of Assertion (A), the formula $\sqrt{x^2 + y^2}$ is indeed derived directly from the general distance formula by setting one of the points to the origin $(0, 0)$. This means it is a specific application or "special case" of the broader distance formula.

Therefore, Reason (R) is True.

Relationship between A and R:

Reason (R) explains *why* the formula in Assertion (A) is correct. It precisely describes the condition under which the general distance formula simplifies to the form given in Assertion (A). The fact that one point is the origin is exactly what makes the formula $D = \sqrt{x^2 + y^2}$ true for the distance between $(0,0)$ and $(x,y)$.

Therefore, R is the correct explanation of A.

Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Thus, the correct option is (A) Both A and R are true and R is the correct explanation of A.

This question is a case study that requires us to formulate a linear equation based on a real-world scenario of taxi fare calculation. The total fare comprises a fixed component and a variable component dependent on the distance traveled.

Given:

Fixed charge = $\textsf{₹}\,50$

Charge per kilometer = $\textsf{₹}\,15$

Distance traveled = $x$ kilometers

Total fare = $y$ Rupees

To Find:

The equation for the total fare $y$ as a function of the distance $x$.

Solution:

The total fare consists of two parts:

1. Fixed Charge: This is a one-time charge, regardless of the distance traveled. In this case, it is $\textsf{₹}\,50$.
2. Variable Charge: This charge depends on the distance traveled. Since the charge is $\textsf{₹}\,15$ per kilometer, for $x$ kilometers, the variable charge will be $15 \times x = 15x$.

To find the total fare $y$, we add the fixed charge and the variable charge:

We can rewrite this equation in the standard slope-intercept form ($y = mx + c$), where $m$ is the rate per unit and $c$ is the fixed amount:

Now, let's compare this derived equation with the given options:

• (A) $y = 50x + 15$: This implies a charge of $\textsf{₹}\,50$ per km and a fixed charge of $\textsf{₹}\,15$, which contradicts the problem statement.
• (B) $y = 15x + 50$: This equation correctly represents a charge of $\textsf{₹}\,15$ per km ($15x$) plus a fixed charge of $\textsf{₹}\,50$. This matches our derived equation.
• (C) $y = 65x$: This implies a charge of $\textsf{₹}\,65$ per km with no fixed charge. This is incorrect.
• (D) $y = 50 + 15/x$: This equation implies a fixed charge of $\textsf{₹}\,50$ and a variable charge that decreases as distance increases, which is illogical for a per-kilometer charge.

Therefore, the equation that correctly models the total fare is $y = 15x + 50$.

Conclusion:

The equation for the total fare $y$ as a function of the distance $x$ is $y = 15x + 50$.

The correct option is (B) $y = 15x + 50$.

This question is a continuation of the previous case study, asking us to calculate the total taxi fare for a specific distance using the fare equation derived earlier.

Given:

From the previous case study (Question 9), the fixed charge is $\textsf{₹}\,50$ and the charge per kilometer is $\textsf{₹}\,15$. The equation for the total fare $y$ as a function of distance $x$ (in kilometers) was established as:

For this specific question, the distance traveled is $10$ kilometers. So, $x = 10$.

To Find:

The total fare for a journey of 10 kilometers.

Solution:

To find the total fare for a journey of 10 kilometers, we need to substitute $x = 10$ into the fare equation (i):

$y = 15x + 50$

Substitute $x=10$:

$y = 15(10) + 50$

Perform the multiplication:

$y = 150 + 50$

Perform the addition:

$y = 200$

So, the total fare for a journey of 10 kilometers is $\textsf{₹}\,200$.

Conclusion:

The calculated fare for a 10 km journey is $\textsf{₹}\,200$. Let's compare this with the given options:

• (A) $\textsf{₹}\,150$
• (B) $\textsf{₹}\,200$
• (C) $\textsf{₹}\,650$
• (D) $\textsf{₹}\,500$

The correct option is (B) $\textsf{₹}\,200$.

This question asks for the standard algebraic equation of a circle given its center and radius. This formula is fundamental in coordinate geometry and is derived from the definition of a circle using the distance formula.

Given:

Center of the circle: $(h, k)$

Radius of the circle: $r$

To Find:

The equation of the circle.

Solution:

A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the center. The constant distance is called the radius.

Let $(x, y)$ be any arbitrary point on the circle.

Let the center of the circle be $C(h, k)$ and the radius be $r$.

According to the definition, the distance between any point $(x, y)$ on the circle and the center $(h, k)$ must be equal to the radius $r$.

We use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$:

In our case, $(x_1, y_1) = (h, k)$ and $(x_2, y_2) = (x, y)$, and $D = r$.

Substituting these values into the distance formula:

To remove the square root and obtain a more commonly used form of the equation, we square both sides of equation (i):

$r^2 = \left(\sqrt{(x - h)^2 + (y - k)^2}\right)^2$

This simplifies to:

This is the standard equation of a circle with center $(h, k)$ and radius $r$.

Let's examine the given options:

• (A) $(x-h)^2 + (y-k)^2 = r$: This is incorrect because the radius term on the right-hand side should be squared.
• (B) $(x-h)^2 + (y-k)^2 = r^2$: This perfectly matches the derived standard equation of a circle.
• (C) $x^2 + y^2 = r^2$: This is the equation of a circle centered at the origin $(0, 0)$ with radius $r$. It is a special case of the general equation (ii) where $h=0$ and $k=0$. While true for a specific center, it is not the general form for center $(h,k)$.
• (D) $(x+h)^2 + (y+k)^2 = r^2$: This equation would represent a circle with center $(-h, -k)$, not $(h, k)$, because the terms in the parentheses are $(x - (-h))$ and $(y - (-k))$.

Conclusion:

The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Therefore, the correct option is (B) $(x-h)^2 + (y-k)^2 = r^2$.

This question asks us to identify the center and radius of a circle given its equation. We will achieve this by comparing the given equation to the standard form of a circle's equation.

Given:

The equation of the circle is $(x-1)^2 + (y+2)^2 = 9$.

To Find:

The coordinates of the center $(h, k)$ and the value of the radius $r$ of the circle.

Solution:

The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is:

We are given the equation of the circle as:

To find $h$, $k$, and $r$, we compare equation (ii) with the standard form (i).

For the x-coordinate of the center (h):

Comparing $(x - h)^2$ with $(x - 1)^2$, we see that:

For the y-coordinate of the center (k):

Comparing $(y - k)^2$ with $(y + 2)^2$, we need to rewrite $(y + 2)^2$ as $(y - (-2))^2$.

So, we see that:

Therefore, the center of the circle is $(h, k) = (1, -2)$.

For the radius (r):

Comparing $r^2$ with $9$, we have:

$r^2 = 9$

To find $r$, we take the square root of both sides:

$r = \sqrt{9}$

Since the radius must be a positive value (distance), we take the positive square root:

Thus, the radius of the circle is 3 units.

Conclusion:

The center of the circle is $(1, -2)$ and the radius is 3.

Let's check the given options:

• (A) Center: $(1, 2)$, Radius: 3
• (B) Center: $(1, -2)$, Radius: 3
• (C) Center: $(-1, 2)$, Radius: 9
• (D) Center: $(-1, -2)$, Radius: 9

Our calculated center $(1, -2)$ and radius $3$ match option (B).

Therefore, the correct option is (B) Center: $(1, -2)$, Radius: 3.

This question asks us to determine the center and radius of a circle from its given equation. This involves recognizing a specific form of the circle's equation.

Given:

The equation of the circle is $x^2 + y^2 = 16$.

To Find:

The coordinates of the center $(h, k)$ and the value of the radius $r$ of the circle.

Solution:

The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is:

The given equation of the circle is:

We can rewrite equation (ii) to match the standard form (i) by noting that $x^2 = (x - 0)^2$ and $y^2 = (y - 0)^2$. Also, $16$ can be expressed as $4^2$.

So, equation (ii) can be written as:

Now, by comparing this form with the standard equation $(x - h)^2 + (y - k)^2 = r^2$:

For the x-coordinate of the center (h):

From $(x - 0)^2$, we get $h = 0$.

For the y-coordinate of the center (k):

From $(y - 0)^2$, we get $k = 0$.

Therefore, the center of the circle is $(h, k) = (0, 0)$. This indicates that the circle is centered at the origin.

For the radius (r):

From $r^2 = 4^2$, we get $r^2 = 16$.

Taking the positive square root (since radius is a length and must be positive):

Thus, the radius of the circle is 4 units.

Conclusion:

The center of the circle is $(0, 0)$ and the radius is 4.

Let's check the given options:

• (A) $(0, 0)$, 16 - Radius is incorrect.
• (B) $(0, 0)$, 4 - This matches our findings exactly.
• (C) $(16, 0)$, 4 - Center is incorrect.
• (D) $(0, 16)$, 16 - Both center and radius are incorrect.

Therefore, the correct option is (B) $(0, 0)$, 4.

This question asks us to find the equation of a circle given its center and a point it passes through. The fundamental step is to determine the radius of the circle, which is the distance from the center to any point on the circle.

Given:

Center of the circle: $(h, k) = (0, 0)$

A point on the circle: $(x_p, y_p) = (3, 4)$

To Find:

The equation of the circle.

Solution:

The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is:

Since the center is given as the origin, $(h, k) = (0, 0)$. Substituting these values into equation (i):

Next, we need to find the value of $r^2$. The radius $r$ is the distance from the center $(0, 0)$ to the point $(3, 4)$ which lies on the circle. We can use the distance formula for this:

Here, $D = r$, $(x_1, y_1) = (0, 0)$, and $(x_2, y_2) = (3, 4)$.

Substitute these values into the distance formula to find $r$:

$r = \sqrt{(3 - 0)^2 + (4 - 0)^2}$

$r = \sqrt{3^2 + 4^2}$

$r = \sqrt{9 + 16}$

$r = \sqrt{25}$

$r = 5$

Now, we need $r^2$ for the equation of the circle:

$r^2 = 5^2$

$r^2 = 25$

Finally, substitute the value of $r^2$ into equation (ii):

This is the required equation of the circle.

Conclusion:

The equation of the circle with center at $(0, 0)$ passing through the point $(3, 4)$ is $x^2 + y^2 = 25$.

Let's compare this with the given options:

• (A) $x^2 + y^2 = 5$: Incorrect, as $r^2$ should be 25, not 5.
• (B) $x^2 + y^2 = 25$: This matches our derived equation.
• (C) $(x-3)^2 + (y-4)^2 = 0$: This represents a single point $(3, 4)$ (a circle with radius 0), not the intended circle.
• (D) $x^2 + y^2 = \sqrt{5}$: Incorrect, as $r^2$ should be 25, not $\sqrt{5}$.

Therefore, the correct option is (B) $x^2 + y^2 = 25$.

This question requires us to evaluate an Assertion (A) and a Reason (R) related to the general equation of a circle. We need to determine if both statements are true and if Reason (R) provides a correct explanation for Assertion (A).

Assertion (A) Evaluation:

Assertion (A) states: "The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$."

In coordinate geometry, the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ is indeed recognized as the general form of the equation of a circle. This form arises from expanding the standard form $(x-h)^2 + (y-k)^2 = r^2$.

Therefore, Assertion (A) is True.

Reason (R) Evaluation:

Reason (R) states: "From this equation, the center is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$ (if $g^2 + f^2 - c \geq 0$)."

To verify this, we will convert the general form of the equation of a circle back to its standard form by completing the square.

Given the general equation:

$x^2 + y^2 + 2gx + 2fy + c = 0$

Group the x-terms and y-terms, and move the constant term to the right side:

$(x^2 + 2gx) + (y^2 + 2fy) = -c$

To complete the square for the x-terms, add $(g)^2$ to both sides. To complete the square for the y-terms, add $(f)^2$ to both sides:

$(x^2 + 2gx + g^2) + (y^2 + 2fy + f^2) = -c + g^2 + f^2$

Rewrite the grouped terms as perfect squares:

Now, compare this with the standard form of a circle equation: $(x - h)^2 + (y - k)^2 = r^2$.

From the comparison:

• $(x + g)^2$ can be written as $(x - (-g))^2$. So, the x-coordinate of the center $h = -g$.
• $(y + f)^2$ can be written as $(y - (-f))^2$. So, the y-coordinate of the center $k = -f$.
• The radius squared $r^2 = g^2 + f^2 - c$.

Thus, the center of the circle is $(-g, -f)$.

The radius $r = \sqrt{g^2 + f^2 - c}$.

For the equation to represent a real circle, the radius $r$ must be a real number, which means $r^2$ must be non-negative. Hence, the condition $g^2 + f^2 - c \geq 0$ is essential.

This matches the statement in Reason (R).

Therefore, Reason (R) is True.

Relationship between Assertion (A) and Reason (R):

Assertion (A) states the general equation of a circle. Reason (R) provides the formulas for calculating the center and radius *from* this general equation. Essentially, Reason (R) explains how to interpret and extract geometric properties from the algebraic form presented in Assertion (A). Therefore, Reason (R) is the correct explanation for Assertion (A).

Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

Thus, the correct option is (A) Both A and R are true and R is the correct explanation of A.

This question is a case study that asks us to find the radius of a circular park. We are given the coordinates of the park's center and a point on its boundary. The radius is defined as the distance from the center to any point on the boundary of the circle.

Given:

Center of the circular park: $(h, k) = (0, 0)$

A point on the boundary of the park: $(x_p, y_p) = (6, 8)$

To Find:

The radius $r$ of the park.

Solution:

The radius of the park is the distance between its center $(0, 0)$ and the point $(6, 8)$ on its boundary. We can use the distance formula for two points $(x_1, y_1)$ and $(x_2, y_2)$:

Let $(x_1, y_1) = (0, 0)$ (the center) and $(x_2, y_2) = (6, 8)$ (the point on the boundary).

The distance $D$ will represent the radius $r$ of the park.

Substitute the coordinates into the distance formula (i):

$r = \sqrt{(6 - 0)^2 + (8 - 0)^2}$

$r = \sqrt{(6)^2 + (8)^2}$

$r = \sqrt{36 + 64}$

$r = \sqrt{100}$

$r = 10$

Thus, the radius of the park is 10 units.

Conclusion:

The calculated radius is 10. Comparing this with the given options:

• (A) 6
• (B) 8
• (C) 10
• (D) $\sqrt{28}$

The correct option is (C) 10.

This question is a continuation of the case study from Question 16. We need to write the equation of the circular park given its center and the radius which was calculated in the previous question.

Given:

From Question 16:

• Center of the circular park: $(h, k) = (0, 0)$
• Radius of the park: $r = 10$ units (calculated as the distance between $(0,0)$ and $(6,8)$ in Q16).

To Find:

The equation representing the boundary of the circular park.

Solution:

The standard equation of a circle with center $(h, k)$ and radius $r$ is:

Substitute the given center $(h, k) = (0, 0)$ and the calculated radius $r = 10$ into this standard equation.

$(x - 0)^2 + (y - 0)^2 = (10)^2$

Simplify the equation:

$x^2 + y^2 = 100$

This is the equation representing the boundary of the circular park.

Conclusion:

The equation of the circular park is $x^2 + y^2 = 100$. Let's compare this with the given options:

• (A) $x^2 + y^2 = 10$: Incorrect, this would mean $r = \sqrt{10}$.
• (B) $x^2 + y^2 = 100$: This matches our derived equation exactly.
• (C) $(x-6)^2 + (y-8)^2 = 0$: This represents a single point at $(6,8)$ (a circle with radius 0), not the circular park.
• (D) $x^2 + y^2 = 1000$: Incorrect, this would mean $r = \sqrt{1000}$.

Therefore, the correct option is (B) $x^2 + y^2 = 100$.

This question asks us to evaluate the accuracy of the given definition of a parabola. This is a direct test of knowledge of conic sections definitions.

Given:

The statement: "A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed straight line (the directrix)."

To Determine:

Whether the given definition of a parabola is true or false, and why.

Solution:

Let's recall the standard geometric definitions of conic sections:

• A parabola is defined as the locus of points such that the distance from a fixed point (the focus) is equal to the distance from a fixed line (the directrix). That is, for any point P on the parabola, distance(P, Focus) = distance(P, Directrix). This is also known as having an eccentricity $e=1$.
• An ellipse is defined as the set of all points in a plane such that the sum of the distances from two fixed points (the foci) is constant. Alternatively, an ellipse can be defined as the locus of points such that the ratio of the distance from a fixed point (focus) to the distance from a fixed line (directrix) is a constant eccentricity $e < 1$.
• A hyperbola is defined as the set of all points in a plane such that the absolute difference of the distances from two fixed points (the foci) is constant. Alternatively, a hyperbola can be defined as the locus of points such that the ratio of the distance from a fixed point (focus) to the distance from a fixed line (directrix) is a constant eccentricity $e > 1$.

Comparing the given statement with these definitions, the statement precisely matches the geometric definition of a parabola.

Let's analyze the options:

• (A) True: This aligns perfectly with the accepted mathematical definition of a parabola.
• (B) False: The definition is correct, so this option is incorrect.
• (C) True, but only if the focus is at the origin: The definition of a parabola holds true regardless of the specific coordinates of the focus or the equation of the directrix. While the equation of the parabola changes with their positions, the fundamental geometric definition remains the same. So, this qualification is incorrect.
• (D) False, this definition is for an ellipse: As discussed, this definition is for a parabola, not an ellipse. The ellipse has a different definition involving two foci or a constant eccentricity less than 1 with a focus and directrix.

Conclusion:

The given definition accurately describes a parabola.

Therefore, the correct option is (A) True.

This question asks us to recall the standard form of the equation of a parabola given its vertex and focus. The position of the focus relative to the vertex determines the orientation of the parabola and, consequently, its equation.

Given:

Vertex of the parabola: $(0, 0)$

Focus of the parabola: $(a, 0)$, where $a > 0$

To Find:

The standard equation of this parabola.

Solution:

A parabola is defined as the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

When the vertex of a parabola is at the origin $(0,0)$, there are four standard forms of its equation, depending on its orientation:

1. Opens Right: If the focus is at $(a, 0)$ and $a > 0$, the directrix is the line $x = -a$. The equation is $y^2 = 4ax$.
2. Opens Left: If the focus is at $(-a, 0)$ and $a > 0$, the directrix is the line $x = a$. The equation is $y^2 = -4ax$.
3. Opens Upward: If the focus is at $(0, a)$ and $a > 0$, the directrix is the line $y = -a$. The equation is $x^2 = 4ay$.
4. Opens Downward: If the focus is at $(0, -a)$ and $a > 0$, the directrix is the line $y = a$. The equation is $x^2 = -4ay$.

In this question, we are given:

• Vertex at $(0, 0)$.
• Focus at $(a, 0)$, with $a > 0$.

Since the focus is on the positive x-axis and the vertex is at the origin, the parabola opens to the right.

Comparing this directly with the standard forms, the equation for a parabola with vertex at $(0,0)$ and focus at $(a,0)$ (where $a>0$) is:

Conclusion:

The standard equation of a parabola with vertex at $(0,0)$ and focus at $(a,0)$ where $a>0$ is $y^2 = 4ax$.

Let's check the given options:

• (A) $y^2 = 4ax$: This matches our derived equation.
• (B) $y^2 = -4ax$: This would be for a parabola opening left.
• (C) $x^2 = 4ay$: This would be for a parabola opening upward.
• (D) $x^2 = -4ay$: This would be for a parabola opening downward.

Therefore, the correct option is (A) $y^2 = 4ax$.

This question asks us to identify the focus of a parabola given its equation. The coordinates of the focus depend on the standard form of the parabola's equation and its orientation.

Given:

The equation of the parabola is $y^2 = 8x$.

To Find:

The coordinates of the focus of the parabola.

Solution:

We need to compare the given equation $y^2 = 8x$ with the standard forms of a parabola with its vertex at the origin $(0,0)$.

The four standard forms of a parabola with vertex at $(0,0)$ are:

• $y^2 = 4ax$ (opens right, focus at $(a,0)$)
• $y^2 = -4ax$ (opens left, focus at $(-a,0)$)
• $x^2 = 4ay$ (opens upward, focus at $(0,a)$)
• $x^2 = -4ay$ (opens downward, focus at $(0,-a)$)

The given equation is $y^2 = 8x$. This equation is of the form $y^2 = 4ax$.

By comparing the coefficients of $x$ in equation (i), we can find the value of $a$:

$4a = 8$

Divide both sides by 4:

$a = \frac{8}{4}$

For a parabola of the form $y^2 = 4ax$, with the vertex at $(0,0)$, the focus is located at $(a, 0)$.

Substitute the value of $a=2$ from equation (ii) into the coordinates of the focus $(a, 0)$:

Focus = $(2, 0)$

This parabola opens to the right, and its focus is at $(2,0)$.

Conclusion:

The coordinates of the focus of the parabola $y^2 = 8x$ are $(2, 0)$.

Let's check the given options:

• (A) $(2, 0)$: This matches our calculated focus.
• (B) $(0, 2)$: This would be for a parabola of the form $x^2 = 8y$.
• (C) $(-2, 0)$: This would be for a parabola of the form $y^2 = -8x$.
• (D) $(8, 0)$: This would imply $4a = 8$ and $a = 2$, but option (D) itself provides the coordinate as $(8,0)$, which is incorrect.

Therefore, the correct option is (A) $(2, 0)$.

This question asks us to determine the opening direction of a parabola given its equation. The form of the equation of a parabola dictates its orientation.

Given:

The equation of the parabola is $x^2 = 12y$.

To Determine:

The opening direction of the parabola.

Solution:

We need to compare the given equation with the standard forms of a parabola with its vertex at the origin $(0,0)$. The four basic standard forms for parabolas centered at the origin are:

• $y^2 = 4ax$: Represents a parabola opening to the right (focus at $(a,0)$, where $a>0$).

• $y^2 = -4ax$: Represents a parabola opening to the left (focus at $(-a,0)$, where $a>0$).

• $x^2 = 4ay$: Represents a parabola opening upward (focus at $(0,a)$, where $a>0$).

• $x^2 = -4ay$: Represents a parabola opening downward (focus at $(0,-a)$, where $a>0$).

The given equation is:

By observing equation (i), we notice the following characteristics:

• The $x$ term is squared ($x^2$). This indicates that the parabola opens either upward or downward. If the $y$ term were squared ($y^2$), it would open left or right.
• The coefficient of the linear term ($y$) is $12$, which is a positive value.

Comparing these characteristics with the standard forms:

The equation $x^2 = 12y$ matches the form $x^2 = 4ay$.

If we equate the coefficients of $y$:

$4a = 12$

$a = \frac{12}{4}$

$a = 3$

Since $a = 3$ is positive, and the $x$ term is squared, the parabola opens upward. The focus of this parabola would be at $(0, a) = (0, 3)$.

Conclusion:

The equation $x^2 = 12y$ represents a parabola opening Upwards.

Let's check the given options:

• (A) Upwards.
• (B) Downwards.
• (C) To the right.
• (D) To the left.

The correct option is (A) Upwards.

This question asks for the formula for the length of the latus rectum for a standard parabola. This is a standard property associated with conic sections.

Given:

The equation of the parabola is $y^2 = 4ax$.

To Find:

The length of the latus rectum.

Solution:

For a parabola, the latus rectum is a line segment that passes through the focus, is perpendicular to the axis of symmetry, and has its endpoints on the parabola.

Let's consider the standard parabola $y^2 = 4ax$:

• The vertex is at the origin $(0,0)$.
• The axis of symmetry is the x-axis ($y=0$).
• The focus is at $(a,0)$.

To find the length of the latus rectum, we need to find the coordinates of the endpoints of the latus rectum. Since the latus rectum passes through the focus $(a,0)$ and is perpendicular to the x-axis, the x-coordinate of its endpoints will be $a$.

Substitute $x = a$ into the equation of the parabola, $y^2 = 4ax$:

$y^2 = 4a(a)$

$y^2 = 4a^2$

Take the square root of both sides to find the y-coordinates:

$y = \pm\sqrt{4a^2}$

$y = \pm 2a$

So, the endpoints of the latus rectum are $L_1(a, 2a)$ and $L_2(a, -2a)$.

The length of the latus rectum is the distance between these two points. Since they have the same x-coordinate, the distance is simply the absolute difference of their y-coordinates:

In the context of the standard equation $y^2 = 4ax$, it's generally assumed that $a>0$ for the parabola to open to the right (in which case the length is $4a$). If $a<0$, the parabola opens to the left, and the length would still be $|4a|$. Therefore, the length is always positive $4a$.

Conclusion:

The length of the latus rectum of the parabola $y^2 = 4ax$ is $4a$.

Let's check the given options:

• (A) $a$
• (B) $2a$
• (C) $4a$
• (D) $a^2$

The correct option is (C) $4a$.

This question asks for the equation of the directrix of a parabola given its standard equation and the location of its vertex. The directrix is a key component in the definition of a parabola.

Given:

Equation of the parabola: $y^2 = 4ax$, with $a > 0$.

Vertex of the parabola: $(0, 0)$ (the origin).

To Find:

The equation of the directrix of this parabola.

Solution:

A parabola is defined as the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed straight line (the directrix).

For the standard parabola $y^2 = 4ax$ (where $a > 0$):

• The vertex is at the origin $(0,0)$.
• The axis of symmetry is the x-axis (since $y$ is squared, the parabola opens horizontally).
• The parabola opens to the right because $a > 0$.
• The focus is located at $(a, 0)$.

The directrix is a line perpendicular to the axis of symmetry and is located at the same distance from the vertex as the focus, but on the opposite side of the vertex.

Since the focus is at $(a, 0)$ (a distance of $a$ units to the right of the vertex), the directrix must be a vertical line located $a$ units to the left of the vertex.

Therefore, the equation of the directrix is $x = -a$.

Conclusion:

The equation of the directrix of the parabola $y^2 = 4ax$ is $x = -a$.

Let's check the given options:

• (A) $x = a$: This would be the directrix if the parabola were $y^2 = -4ax$ (opening left, focus at $(-a,0)$).
• (B) $x = -a$: This matches our derived equation for the directrix.
• (C) $y = a$: This would be the directrix for a parabola opening downwards (e.g., $x^2 = -4ay$).
• (D) $y = -a$: This would be the directrix for a parabola opening upwards (e.g., $x^2 = 4ay$).

Therefore, the correct option is (B) $x = -a$.

This is an Assertion-Reason type question. We need to evaluate the truthfulness of both the Assertion (A) and the Reason (R), and then determine if the Reason correctly explains the Assertion.

Assertion (A) Evaluation:

Assertion (A) states: "The path of a projectile under gravity (ignoring air resistance) is a parabola."

This statement is a fundamental principle in classical mechanics, specifically in the study of projectile motion. When an object is launched into the air and only experiences the force of gravity (neglecting air resistance), its trajectory forms a parabolic curve.

Therefore, Assertion (A) is True.

Reason (R) Evaluation:

Reason (R) states: "The equation describing the path is quadratic in one variable and linear in the other."

Let's derive the equation of the path of a projectile. Assume a projectile is launched from the origin $(0,0)$ with an initial velocity $u$ at an angle $\theta$ with the horizontal.

The initial velocity components are:

• Horizontal component: $u_x = u \cos\theta$
• Vertical component: $u_y = u \sin\theta$

Considering the motion under constant acceleration due to gravity $g$ (downwards) and no horizontal acceleration (ignoring air resistance):

Horizontal displacement ($x$) at time $t$:

Vertical displacement ($y$) at time $t$:

From equation (i), we can express $t$ in terms of $x$:

Now, substitute this expression for $t$ from equation (iii) into equation (ii):

$y = (u \sin\theta) \left(\frac{x}{u \cos\theta}\right) - \frac{1}{2}g \left(\frac{x}{u \cos\theta}\right)^2$

$y = \left(\frac{\sin\theta}{\cos\theta}\right) x - \frac{g}{2} \frac{x^2}{u^2 \cos^2\theta}$

Using the identity $\frac{\sin\theta}{\cos\theta} = \tan\theta$:

Equation (iv) is in the form $y = Ax - Bx^2$, where $A = \tan\theta$ and $B = \frac{g}{2u^2 \cos^2\theta}$ are constants (for a given launch velocity and angle). This equation is a quadratic equation in $x$ and a linear equation in $y$. Any equation of the form $y = ax^2 + bx + c$ (or $x = ay^2 + by + c$) represents a parabola.

Therefore, Reason (R) is True.

Relationship between A and R:

Assertion (A) states a fact about projectile motion (parabolic path). Reason (R) describes the mathematical form of the equation that governs this path. The fact that the equation is quadratic in one variable ($x^2$) and linear in the other ($y$) is precisely the algebraic characteristic that defines a parabola in coordinate geometry. Thus, Reason (R) correctly provides the mathematical basis for why the path of a projectile is parabolic.

Therefore, R is the correct explanation of A.

Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

This question is a case study about a parabolic satellite dish. We are given the location of its vertex and the position of its focus, and we need to determine a specific parameter 'a' from the standard equation of a parabola.

Given:

Vertex of the parabolic dish: $(0, 0)$ (at the origin).

Focus of the parabolic dish: 0.5 meters away from the vertex along the positive x-axis.

To Find:

The value of $a$ in the standard equation of the parabola.

Solution:

The definition of a parabola involves a vertex, a focus, and a directrix. For parabolas with their vertex at the origin $(0,0)$, there are four standard orientations:

1. If the parabola opens to the right, its focus is at $(a, 0)$ where $a > 0$. The equation is $y^2 = 4ax$.
2. If the parabola opens to the left, its focus is at $(-a, 0)$ where $a > 0$. The equation is $y^2 = -4ax$.
3. If the parabola opens upward, its focus is at $(0, a)$ where $a > 0$. The equation is $x^2 = 4ay$.
4. If the parabola opens downward, its focus is at $(0, -a)$ where $a > 0$. The equation is $x^2 = -4ay$.

In this problem, we are given:

• Vertex at $(0, 0)$.
• Focus is 0.5 meters away from the vertex along the positive x-axis.

This means the focus is located at the point $(0.5, 0)$.

Comparing this focus location with the standard forms, the focus $(a,0)$ corresponds to a parabola opening to the right. Therefore, the value of $a$ is the x-coordinate of the focus (when the vertex is at the origin and focus is on the x-axis, or y-coordinate if focus is on y-axis).

From this, we can directly identify the value of $a$:

The standard equation of this specific parabola would be $y^2 = 4(0.5)x$, which simplifies to $y^2 = 2x$.

Conclusion:

The value of $a$ in the standard equation of the parabola is 0.5.

Let's compare this with the given options:

• (A) 0.5
• (B) 1
• (C) 2
• (D) 4

The correct option is (A) 0.5.

This question is a continuation of the case study from Question 25. We need to formulate the equation of the parabolic satellite dish using the information about its vertex and focus, including the value of 'a' determined in the previous question.

Given:

From Question 25:

• Vertex of the parabolic dish: $(h, k) = (0, 0)$ (at the origin).
• Focus of the parabolic dish: 0.5 meters away from the vertex along the positive x-axis. This means the focus is at $(0.5, 0)$.
• The value of $a$ in the standard equation for this parabola is $0.5$.

To Find:

The equation of the parabola that describes the shape of the satellite dish.

Solution:

Since the vertex is at the origin $(0,0)$ and the focus is at $(a, 0)$ with $a = 0.5$ (a positive value on the x-axis), this parabola opens to the right.

The standard equation for a parabola with vertex at $(0,0)$ and opening to the right is:

We already determined from Question 25 that for this specific parabola, $a = 0.5$.

Substitute the value of $a = 0.5$ into equation (i):

$y^2 = 4(0.5)x$

$y^2 = 2x$

This is the equation describing the shape of the parabolic satellite dish.

Conclusion:

The equation of the parabola is $y^2 = 2x$. Let's compare this with the given options:

• (A) $y^2 = 0.5x$: Incorrect (this implies $4a = 0.5$, so $a=0.125$).
• (B) $y^2 = 2x$: This matches our derived equation.
• (C) $x^2 = 2y$: This would be a parabola opening upwards.
• (D) $x^2 = 0.5y$: This would be a parabola opening upwards with a different 'a' value.

Therefore, the correct option is (B) $y^2 = 2x$.

This question requires matching different standard forms of equations to the geometric shapes they represent. This is a fundamental concept in coordinate geometry, where algebraic equations describe geometric figures.

Given:

Three equation types and three conic section names.

To Match:

Match each equation type with its corresponding conic section.

Solution:

Let's analyze each equation type and identify the conic section it represents:

(i) $Ax + By + C = 0$ (where A, B not both zero)

This is the general form of a linear equation. A linear equation, when graphed, always forms a straight line. If $B \neq 0$, it can be rewritten as $y = (-\frac{A}{B})x - \frac{C}{B}$, which is the slope-intercept form $y = mx + c$. If $B = 0$ (and $A \neq 0$), it becomes $Ax + C = 0$, or $x = -\frac{C}{A}$, which is a vertical line.

Therefore, (i) matches with (c) Straight line.

(ii) $(x-h)^2 + (y-k)^2 = r^2$

This is the standard form of the equation of a circle. Here, $(h, k)$ represents the center of the circle, and $r$ represents its radius. This equation is derived from the distance formula, where every point $(x, y)$ on the circle is at a fixed distance $r$ from the center $(h, k)$.

Therefore, (ii) matches with (b) Circle.

(iii) $y^2 = 4ax$

This is one of the standard forms of the equation of a parabola with its vertex at the origin $(0,0)$ and its axis of symmetry along the x-axis. Since $y$ is squared and $x$ is linear, this type of equation represents a parabola. Specifically, for $a>0$, it opens to the right; for $a<0$, it opens to the left.

Therefore, (iii) matches with (a) Parabola.

Summary of Matches:

• (i) $Ax + By + C = 0 \implies$ (c) Straight line
• (ii) $(x-h)^2 + (y-k)^2 = r^2 \implies$ (b) Circle
• (iii) $y^2 = 4ax \implies$ (a) Parabola

Conclusion:

Based on our analysis, the correct matching is (i)-(c), (ii)-(b), (iii)-(a).

Let's check the given options:

• (A) (i)-(a), (ii)-(b), (iii)-(c) - Incorrect
• (B) (i)-(c), (ii)-(b), (iii)-(a) - This matches our findings.
• (C) (i)-(b), (ii)-(a), (iii)-(c) - Incorrect
• (D) (i)-(c), (ii)-(a), (iii)-(b) - Incorrect

Therefore, the correct option is (B) (i)-(c), (ii)-(b), (iii)-(a).

This question asks us to identify the correct section formula for finding the coordinates of a point that divides a line segment internally in a given ratio. This formula is a fundamental concept in coordinate geometry.

Given:

A line segment joining two points: $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$.

A point $P(x, y)$ that divides this line segment internally in the ratio $m:n$.

To Find:

The formula for the coordinates $(x, y)$ of point $P$.

Solution:

The Section Formula for internal division states that if a point $P(x, y)$ divides the line segment joining $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ internally in the ratio $m:n$, then its coordinates are given by:

Let's analyze each option provided:

• (A) $\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}\right)$: This is the formula for external division, not internal division. It is used when the point P lies outside the segment $P_1P_2$.
• (B) $\left(\frac{mx_1 + nx_2}{m+n}, \frac{my_1 + ny_2}{m+n}\right)$: This formula incorrectly swaps the $x_1, x_2$ and $y_1, y_2$ terms with $m$ and $n$. The ratio $m$ should be multiplied with the coordinates of the "far" point ($x_2, y_2$), and $n$ with the coordinates of the "near" point ($x_1, y_1$).
• (C) $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$: This exactly matches the standard section formula for internal division as presented in equation (i).
• (D) $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$: This is the midpoint formula, which is a special case of the section formula where the ratio is $1:1$ (i.e., $m=1, n=1$). While related, it's not the general section formula for any ratio $m:n$.

Based on the definitions of coordinate geometry formulas, option (C) correctly states the section formula for internal division.

Alternate Solution (Derivation Concept):

The section formula can be derived using similar triangles or vector methods. If point P divides $P_1P_2$ in ratio $m:n$, then $\vec{P_1P} = \frac{m}{m+n} \vec{P_1P_2}$.

Let $P_1 = (x_1, y_1)$, $P_2 = (x_2, y_2)$, and $P = (x, y)$.

Then, $(x - x_1, y - y_1) = \frac{m}{m+n} (x_2 - x_1, y_2 - y_1)$.

Equating components:

$x - x_1 = \frac{m(x_2 - x_1)}{m+n} \implies x(m+n) - x_1(m+n) = mx_2 - mx_1 \implies x(m+n) = mx_2 - mx_1 + mx_1 + nx_1 \implies x = \frac{mx_2 + nx_1}{m+n}$

$y - y_1 = \frac{m(y_2 - y_1)}{m+n} \implies y(m+n) - y_1(m+n) = my_2 - my_1 \implies y(m+n) = my_2 - my_1 + my_1 + ny_1 \implies y = \frac{my_2 + ny_1}{m+n}$

This derivation confirms the formula in option (C).

Conclusion:

The section formula for the coordinates of a point $(x, y)$ that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m:n$ is $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.

Therefore, the correct option is (C) $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.

This question asks us to find the midpoint of a line segment given the coordinates of its two endpoints. The midpoint formula is a special case of the section formula where the ratio of division is 1:1.

Given:

Two points defining the line segment: $P_1(4, -6)$ and $P_2(-2, 8)$.

Let $(x_1, y_1) = (4, -6)$.

Let $(x_2, y_2) = (-2, 8)$.

To Find:

The coordinates of the midpoint $(x_M, y_M)$ of the line segment.

Solution:

The midpoint formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

Now, substitute the given coordinates into the midpoint formula:

For the x-coordinate:

$x_M = \frac{4 + (-2)}{2}$

$x_M = \frac{4 - 2}{2}$

$x_M = \frac{2}{2}$

For the y-coordinate:

$y_M = \frac{-6 + 8}{2}$

$y_M = \frac{2}{2}$

Therefore, the coordinates of the midpoint are $(1, 1)$.

Conclusion:

The midpoint of the line segment joining $(4, -6)$ and $(-2, 8)$ is $(1, 1)$.

Let's check the given options:

• (A) $(2, 2)$
• (B) $(1, 1)$
• (C) $(1, 2)$
• (D) $(2, 1)$

The correct option is (B) $(1, 1)$.

This question relates the area of a triangle formed by three points to the geometric arrangement of those points. Understanding the implications of a zero area is key to solving this.

Given:

The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:

The condition given is that the Area $= 0$.

To Determine:

What geometric configuration the three points form if the area of the triangle they define is 0.

Solution:

The area of a triangle is a measure of the two-dimensional space enclosed by its three sides. For a triangle to exist and enclose any area, its three vertices must not lie on the same straight line.

If the area of a triangle is 0, it means that the "triangle" effectively collapses into a one-dimensional line segment. This can only happen if all three points lie on the same straight line.

Points that lie on the same straight line are called collinear points.

Let's consider the implications of Area $= 0$ from equation (i):

Since the area is $\frac{1}{2}$ times the absolute value of an expression, for the area to be zero, the expression inside the absolute value must be zero:

This condition (equation ii) is the algebraic criterion for three points to be collinear. When three points are collinear, they cannot form a triangle with a positive area; their "area" is zero.

Let's evaluate the given options:

• (A) Vertices of an equilateral triangle: An equilateral triangle has three equal sides and three $60^\circ$ angles. It always has a non-zero area.
• (B) Collinear: This means the three points lie on the same straight line. In this case, no actual triangle is formed, and its area is indeed 0. This is the correct condition.
• (C) Vertices of a right-angled triangle: A right-angled triangle has one angle equal to $90^\circ$. It always forms a triangle with a non-zero area.
• (D) Vertices of an isosceles triangle: An isosceles triangle has at least two equal sides. It always forms a triangle with a non-zero area.

Conclusion:

If the area of a triangle formed by three points is 0, it signifies that the three points are collinear.

Therefore, the correct option is (B) Collinear.

This question asks for the equation of a straight line given its slope and a point it passes through, specifically the origin. We will examine various forms of linear equations to determine the correct representation.

Given:

The line passes through the origin: $(x_1, y_1) = (0, 0)$.

The slope of the line: $m$.

To Find:

The equation of the line.

Solution:

We can use the point-slope form of a linear equation, which is generally applicable when a point $(x_1, y_1)$ on the line and its slope $m$ are known:

Substitute the given point $(0, 0)$ and slope $m$ into equation (i):

$y - 0 = m(x - 0)$

This simplifies to:

Now, let's analyze each of the given options:

• (A) $y = mx$: This is the simplified form derived directly from the point-slope form when the line passes through the origin. This is a correct equation for the line. This is also the slope-intercept form where the y-intercept $c = 0$.
• (B) $y - 0 = m(x - 0)$: This is the direct application of the point-slope formula with $(x_1, y_1) = (0,0)$. While it is unsimplified, it is mathematically identical to $y = mx$.
• (C) $y/x = m$: This form is obtained by dividing both sides of $y = mx$ by $x$. This form is valid only for $x \neq 0$. However, for a line passing through the origin, when $x=0$, $y=0$, which also satisfies $y=mx$. The expression $y/x = m$ implies the slope. It correctly represents the slope of any point on the line relative to the origin (except the origin itself). So, in the context of representing the relationship, it is fundamentally equivalent.

Since options (A), (B), and (C) are all equivalent mathematical representations of the same straight line passing through the origin with slope $m$ (with the domain consideration for option C being implicitly handled by the definition of slope, which is $\Delta y / \Delta x$), the most comprehensive answer is that all of them represent the same line.

Conclusion:

All the given equations, (A) $y = mx$, (B) $y - 0 = m(x - 0)$, and (C) $y/x = m$, correctly represent the same line passing through the origin $(0,0)$ with slope $m$.

Therefore, the correct option is (D) All of the above (representing the same line).

This question asks for the slope of a horizontal line. Understanding the geometric meaning of slope is crucial here.

Given:

A horizontal line.

To Find:

The slope of the horizontal line.

Solution:

The slope of a line measures its steepness or gradient. It is defined as the ratio of the change in the y-coordinate ($\Delta y$) to the change in the x-coordinate ($\Delta x$) between any two distinct points on the line:

A horizontal line is a straight line that runs parallel to the x-axis. This means that for any two points on a horizontal line, their y-coordinates will always be the same, while their x-coordinates will be different.

Let's consider two distinct points on a horizontal line, $P_1(x_1, y_0)$ and $P_2(x_2, y_0)$, where $y_0$ is a constant y-value, and $x_1 \neq x_2$.

Now, substitute these coordinates into the slope formula (i):

$m = \frac{y_0 - y_0}{x_2 - x_1}$

$m = \frac{0}{x_2 - x_1}$

Since $x_1 \neq x_2$, the denominator $(x_2 - x_1)$ will be a non-zero number. Dividing 0 by any non-zero number results in 0.

This result makes intuitive sense: a horizontal line has no vertical change (no steepness), so its slope is zero.

Conclusion:

The slope of a horizontal line is 0.

Let's check the given options:

• (A) 1: This is the slope of a line at $45^\circ$ to the x-axis (e.g., $y=x$).
• (B) -1: This is the slope of a line at $135^\circ$ to the x-axis (e.g., $y=-x$).
• (C) 0: This is the correct slope for a horizontal line.
• (D) Undefined: This is the slope of a vertical line, where $\Delta x = 0$, leading to division by zero in the slope formula.

Therefore, the correct option is (C) 0.

This question asks for the slope of a vertical line. Similar to horizontal lines, understanding the definition of slope and its implications for lines with no horizontal change is essential.

Given:

A vertical line.

To Find:

The slope of the vertical line.

Solution:

The slope of a line ($m$) is defined as the ratio of the change in the y-coordinate ($\Delta y$) to the change in the x-coordinate ($\Delta x$) between any two distinct points on the line:

A vertical line is a straight line that runs parallel to the y-axis. This means that for any two distinct points on a vertical line, their x-coordinates will always be the same, while their y-coordinates will be different.

Let's consider two distinct points on a vertical line, $P_1(x_0, y_1)$ and $P_2(x_0, y_2)$, where $x_0$ is a constant x-value, and $y_1 \neq y_2$.

Now, substitute these coordinates into the slope formula (i):

$m = \frac{y_2 - y_1}{x_0 - x_0}$

$m = \frac{y_2 - y_1}{0}$

Since $y_1 \neq y_2$, the numerator $(y_2 - y_1)$ will be a non-zero number. Division by zero is undefined in mathematics.

Therefore, the slope of a vertical line is undefined.

This also makes intuitive sense: a vertical line represents an infinite steepness. You cannot walk on it in the typical sense of traversing a slope, as there is no horizontal run for any vertical rise (or fall).

Conclusion:

The slope of a vertical line is Undefined.

Let's check the given options:

• (A) 0: This is the slope of a horizontal line.
• (B) 1: This is the slope of a line at $45^\circ$ to the x-axis.
• (C) -1: This is the slope of a line at $135^\circ$ to the x-axis.
• (D) Undefined: This is the correct slope for a vertical line.

Therefore, the correct option is (D) Undefined.

This question asks for the equation of a line that is parallel to the x-axis and passes through a specific point. Understanding the properties of lines parallel to the coordinate axes is crucial here.

Given:

The line is parallel to the x-axis.

The line passes through the point $(3, 5)$.

To Find:

The equation of this line.

Solution:

A line parallel to the x-axis is a horizontal line. A key characteristic of any horizontal line is that all points on it have the same y-coordinate.

The general equation of a horizontal line is $y = k$, where $k$ is a constant representing the y-coordinate through which the line passes.

We are given that the line passes through the point $(3, 5)$. This means that when $x = 3$, $y = 5$.

Since the line is horizontal, its y-coordinate must be constant throughout its length. For it to pass through $(3, 5)$, this constant y-coordinate must be 5.

Therefore, the equation of the line is:

Also, a horizontal line has a slope of 0. Using the slope-intercept form $y = mx + c$, with $m=0$ and passing through $(3,5)$:

$5 = (0)(3) + c$

$5 = 0 + c$

$c = 5$

So, $y = 0x + 5$, which simplifies to $y = 5$. This confirms our finding.

Conclusion:

The equation of the line parallel to the x-axis and passing through the point $(3, 5)$ is $y = 5$.

Let's check the given options:

• (A) $x = 3$: This is a vertical line passing through $(3,0)$, parallel to the y-axis.
• (B) $y = 5$: This matches our derived equation and is a horizontal line passing through $(0,5)$, $(3,5)$, etc.
• (C) $x = 5$: This is a vertical line passing through $(5,0)$, parallel to the y-axis.
• (D) $y = 3$: This is a horizontal line passing through $(0,3)$, not $(3,5)$.

Therefore, the correct option is (B) $y = 5$.

This question asks for the equation of a line that is perpendicular to the x-axis and passes through a given point. Understanding the orientation and properties of lines perpendicular to the coordinate axes is fundamental.

Given:

The line is perpendicular to the x-axis.

The line passes through the point $(-2, 4)$.

To Find:

The equation of this line.

Solution:

A line that is perpendicular to the x-axis is a vertical line. Vertical lines are characterized by having the same x-coordinate for every point on the line, regardless of the y-coordinate.

The general equation of a vertical line is $x = k$, where $k$ is a constant value representing the x-intercept or the x-coordinate through which the line passes.

We are given that the line passes through the point $(-2, 4)$. This means that the line must include the point where $x = -2$ and $y = 4$.

Since the line is a vertical line, its x-coordinate must be constant for all points on it. For it to pass through $(-2, 4)$, this constant x-coordinate must be $-2$.

Therefore, the equation of the line is:

This equation means that for any value of $y$, the x-coordinate is always -2, forming a vertical line passing through $x=-2$ on the x-axis.

Conclusion:

The equation of the line perpendicular to the x-axis and passing through the point $(-2, 4)$ is $x = -2$.

Let's check the given options:

• (A) $y = 4$: This is a horizontal line passing through $(0,4)$.
• (B) $x = -2$: This matches our derived equation and is a vertical line passing through $(-2,0)$, $(-2,4)$, etc.
• (C) $y = -2$: This is a horizontal line passing through $(0,-2)$.
• (D) $x = 4$: This is a vertical line passing through $(4,0)$.

Therefore, the correct option is (B) $x = -2$.

This question asks us to identify the form of a linear equation that directly provides the x-intercept and y-intercept. Understanding the structure and parameters of different linear equation forms is key here.

Given:

Four different forms of a straight line equation.

To Identify:

The form that directly reveals the x-intercept and y-intercept.

Solution:

Let's examine each option:

• (A) Slope-intercept form ($y=mx+c$):

  In this form:

  • $m$ is the slope.
  • $c$ is the y-intercept (the point where the line crosses the y-axis, i.e., $(0, c)$).

  This form directly gives the y-intercept but not the x-intercept. To find the x-intercept, you would set $y=0$ and solve for $x$ (i.e., $0 = mx+c \implies x = -c/m$).

• (B) Point-slope form ($y-y_1=m(x-x_1)$):

  In this form:

  • $m$ is the slope.
  • $(x_1, y_1)$ is a known point on the line.

  This form does not directly provide either intercept. You would need to set $x=0$ to find the y-intercept ($y = -my_1 + mx_1 + y_1$) and set $y=0$ to find the x-intercept ($x = x_1 - y_1/m$).

• (C) Intercept form ($\frac{x}{a} + \frac{y}{b} = 1$):

  In this form:

  • $a$ is the x-intercept (the x-coordinate where the line crosses the x-axis, i.e., $(a, 0)$). This is because if $y=0$, then $\frac{x}{a} = 1 \implies x=a$.
  • $b$ is the y-intercept (the y-coordinate where the line crosses the y-axis, i.e., $(0, b)$). This is because if $x=0$, then $\frac{y}{b} = 1 \implies y=b$.

  This form directly provides both the x-intercept and the y-intercept. This is precisely what the question asks for.

• (D) General form ($Ax+By+C=0$):

  In this form:

  • Neither the slope nor the intercepts are directly apparent.

  To find the x-intercept, set $y=0 \implies Ax+C=0 \implies x = -C/A$.

  To find the y-intercept, set $x=0 \implies By+C=0 \implies y = -C/B$.

  While the intercepts can be derived, they are not "directly" given as parameters in the equation itself.

From the analysis, the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ is specifically designed to highlight the x-intercept ($a$) and y-intercept ($b$).

Conclusion:

The Intercept form ($\frac{x}{a} + \frac{y}{b} = 1$) is useful for finding the x-intercept and y-intercept directly.

Therefore, the correct option is (C) Intercept form ($\frac{x}{a} + \frac{y}{b} = 1$).

This question asks for the equation of a straight line when two points on the line are given. The general approach involves two steps: first calculating the slope of the line, and then using either the point-slope form or the slope-intercept form to establish the equation.

Given:

Two points on the line: $P_1(x_1, y_1) = (1, 2)$ and $P_2(x_2, y_2) = (3, 6)$.

To Find:

The equation of the line passing through these two points.

Solution:

Step 1: Calculate the slope ($m$) of the line.

The formula for the slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

Substitute the given coordinates into the formula:

$m = \frac{6 - 2}{3 - 1}$

$m = \frac{4}{2}$

Step 2: Use the point-slope form to write the equation.

The point-slope form of a linear equation is given by:

We can use either of the given points and the calculated slope $m=2$. Let's use the first point $(x_1, y_1) = (1, 2)$:

$y - 2 = 2(x - 1)$

This equation matches option (A).

Alternate Solution (Using Slope-Intercept Form):

From Step 1, we know the slope $m=2$. The slope-intercept form of a linear equation is $y = mx + c$, where $c$ is the y-intercept.

Substitute $m=2$ and one of the points, say $(1, 2)$, into this form to find $c$:

$2 = 2(1) + c$

$2 = 2 + c$

$c = 0$

Now, substitute $m=2$ and $c=0$ back into the slope-intercept form:

$y = 2x + 0$

This equation matches option (D).

Conclusion:

We found two forms of the equation that represent the line:

• $y - 2 = 2(x - 1)$ (matches option A)
• $y = 2x$ (matches option D)

Both equations represent the exact same line. To confirm, let's expand option (A):

$y - 2 = 2x - 2$

$y = 2x - 2 + 2$

$y = 2x$

Since both (A) and (D) are algebraically equivalent and correctly represent the line, both are correct answers. In a multiple-choice setting, if both are listed, it implies both are valid. However, often one is considered more "direct" from a specific formula or a simplified form. Option (A) is a direct application of the point-slope formula with one of the given points.

Therefore, the most direct answer obtained using one of the points is (A) $y - 2 = 2(x - 1)$. Option (D) is also correct as it's the simplified slope-intercept form.

This question asks us to find the x-intercept of a given linear equation. The x-intercept is the point where the line crosses the x-axis.

Given:

The equation of the line is $2x + 3y = 6$.

To Find:

The x-intercept of the line.

Solution:

The x-intercept is the point where a line intersects the x-axis. At any point on the x-axis, the y-coordinate is always 0.

To find the x-intercept, we set $y = 0$ in the given equation and solve for $x$.

Given equation:

$2x + 3y = 6$

Substitute $y = 0$ into the equation:

$2x + 3(0) = 6$

Simplify the equation:

$2x + 0 = 6$

$2x = 6$

Divide both sides by 2 to solve for $x$:

$x = \frac{6}{2}$

$x = 3$

So, the x-intercept is at $(3, 0)$. The question asks for the value of the x-intercept, which is 3.

Conclusion:

The x-intercept of the line $2x + 3y = 6$ is 3.

Comparing this with the given options:

• (A) 3
• (B) 2
• (C) 6
• (D) 0

The correct option is (A) 3.

This question asks us to find the y-intercept of a given linear equation. The y-intercept is the point where the line crosses the y-axis.

Given:

The equation of the line is $2x + 3y = 6$.

To Find:

The y-intercept of the line.

Solution:

The y-intercept is the point where a line intersects the y-axis. At any point on the y-axis, the x-coordinate is always 0.

To find the y-intercept, we set $x = 0$ in the given equation and solve for $y$.

Given equation:

$2x + 3y = 6$

Substitute $x = 0$ into the equation:

$2(0) + 3y = 6$

Simplify the equation:

$0 + 3y = 6$

$3y = 6$

Divide both sides by 3 to solve for $y$:

$y = \frac{6}{3}$

$y = 2$

So, the y-intercept is at $(0, 2)$. The question asks for the value of the y-intercept, which is 2.

Conclusion:

The y-intercept of the line $2x + 3y = 6$ is 2.

Comparing this with the given options:

• (A) 3
• (B) 2
• (C) 6
• (D) 0

The correct option is (B) 2.

This question asks us to find the radius of a circle given its equation in the general form. To do this, we need to convert the general form into the standard form by completing the square.

Given:

The equation of the circle is $x^2 + y^2 - 4x + 6y - 12 = 0$.

To Find:

The radius $r$ of the circle.

Solution:

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.

The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

To convert the given general equation to the standard form, we use the method of completing the square for both the x-terms and the y-terms.

Given equation:

$x^2 + y^2 - 4x + 6y - 12 = 0$

Rearrange the terms by grouping x-terms, y-terms, and moving the constant to the right side:

$(x^2 - 4x) + (y^2 + 6y) = 12$

Now, complete the square for the x-terms and y-terms:

• For $(x^2 - 4x)$: Take half of the coefficient of $x$ (which is $-4/2 = -2$) and square it $(-2)^2 = 4$. Add this to both sides.
• For $(y^2 + 6y)$: Take half of the coefficient of $y$ (which is $6/2 = 3$) and square it $(3)^2 = 9$. Add this to both sides.

$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9$

Rewrite the trinomials as perfect squares:

$(x - 2)^2 + (y + 3)^2 = 25$

This equation is now in the standard form $(x - h)^2 + (y - k)^2 = r^2$.

By comparing, we can identify:

• $h = 2$
• $k = -3$ (since $y+3$ is $y - (-3)$)
• $r^2 = 25$

To find the radius $r$, take the square root of $r^2$:

$r = \sqrt{25}$

The center of the circle is $(2, -3)$ and its radius is 5.

Conclusion:

The radius of the circle $x^2 + y^2 - 4x + 6y - 12 = 0$ is 5.

Let's check the given options:

• (A) 5
• (B) $\sqrt{12}$
• (C) $\sqrt{13}$
• (D) 4

The correct option is (A) 5.

This question asks us to find the center of a circle given its equation in the general form. This involves converting the general form into the standard form by completing the square.

Given:

The equation of the circle is $x^2 + y^2 - 4x + 6y - 12 = 0$.

To Find:

The coordinates of the center $(h, k)$ of the circle.

Solution:

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. In this form, the center is $(-g, -f)$.

Alternatively, we can transform the given equation into the standard form $(x - h)^2 + (y - k)^2 = r^2$ by completing the square.

Given equation:

$x^2 + y^2 - 4x + 6y - 12 = 0$

Rearrange the terms by grouping x-terms, y-terms, and moving the constant to the right side:

$(x^2 - 4x) + (y^2 + 6y) = 12$

Now, complete the square for the x-terms and y-terms:

• For $(x^2 - 4x)$: Take half of the coefficient of $x$ (which is $-4/2 = -2$) and square it $(-2)^2 = 4$. Add this to both sides.
• For $(y^2 + 6y)$: Take half of the coefficient of $y$ (which is $6/2 = 3$) and square it $(3)^2 = 9$. Add this to both sides.

$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9$

Rewrite the trinomials as perfect squares:

This equation is now in the standard form $(x - h)^2 + (y - k)^2 = r^2$.

By comparing, we can identify:

• $(x - h)^2$ corresponds to $(x - 2)^2$, so $h = 2$.
• $(y - k)^2$ corresponds to $(y + 3)^2$, which is $(y - (-3))^2$, so $k = -3$.

Therefore, the center of the circle is $(h, k) = (2, -3)$.

Alternate Solution (Using General Form Parameters):

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.

The given equation is $x^2 + y^2 - 4x + 6y - 12 = 0$.

Comparing the coefficients:

• $2g = -4 \implies g = -2$
• $2f = 6 \implies f = 3$
• $c = -12$

The center of the circle is given by $(-g, -f)$.

Center $= (-(-2), -(3))$

Center $= (2, -3)$

This confirms the result obtained by completing the square.

Conclusion:

The center of the circle $x^2 + y^2 - 4x + 6y - 12 = 0$ is $(2, -3)$.

Let's check the given options:

• (A) $(2, -3)$
• (B) $(-2, 3)$
• (C) $(4, -6)$
• (D) $(-4, 6)$

The correct option is (A) $(2, -3)$.

This question asks about a specific property of the equation of a circle if it passes through the origin $(0,0)$. We need to analyze the standard and general forms of a circle's equation.

Given:

A circle passes through the origin $(0, 0)$.

To Determine:

Which statement correctly describes the equation or properties of such a circle.

Solution:

Let's consider the general equation of a circle, which is given by:

If a circle passes through a specific point, it means that the coordinates of that point must satisfy the circle's equation.

Since the circle passes through the origin $(0, 0)$, we can substitute $x=0$ and $y=0$ into equation (i):

$(0)^2 + (0)^2 + 2g(0) + 2f(0) + c = 0$

$0 + 0 + 0 + 0 + c = 0$

This means that for any circle passing through the origin, the constant term $c$ in its general equation must be 0. The general equation then simplifies to $x^2 + y^2 + 2gx + 2fy = 0$.

Now, let's evaluate each of the given options based on this finding:

• (A) Always $x^2 + y^2 = r^2$: This is the equation of a circle whose center is specifically at the origin $(0, 0)$. A circle can pass through the origin without being centered at the origin. For example, a circle with center $(1,0)$ and radius $1$ (equation $(x-1)^2 + y^2 = 1$) passes through the origin, but its equation expands to $x^2 - 2x + 1 + y^2 = 1 \implies x^2 + y^2 - 2x = 0$, which is not of the form $x^2+y^2=r^2$. So, this statement is false.
• (B) Always has $c=0$ in the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$: As derived above, if the circle passes through $(0,0)$, substituting these coordinates into the general equation forces $c$ to be 0. This statement is true.
• (C) Always has its center at the origin: This is incorrect. As shown in the example for option (A), a circle with center $(1,0)$ and radius $1$ passes through $(0,0)$ but its center is not the origin. So, this statement is false.
• (D) Cannot be determined without knowing the center: This is incorrect, as we were able to determine a property ($c=0$) of its equation simply by knowing it passes through the origin. So, this statement is false.

Conclusion:

If a circle passes through the origin $(0, 0)$, its general equation $x^2 + y^2 + 2gx + 2fy + c = 0$ must have the constant term $c = 0$.

Therefore, the correct option is (B) Always has $c=0$ in the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$.

This question asks for the condition that determines if a given point lies inside a circle, based on the circle's standard equation. This concept is derived directly from the definition of a circle and the distance formula.

Given:

Equation of a circle: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

A point: $(x_0, y_0)$.

To Determine:

The condition for the point $(x_0, y_0)$ to lie inside the circle.

Solution:

A circle is defined as the set of all points that are a fixed distance (radius $r$) from a fixed point (center $(h,k)$).

Let $D$ be the distance between the center of the circle $(h, k)$ and the given point $(x_0, y_0)$. We can calculate this distance using the distance formula:

By squaring both sides, we get $D^2 = (x_0 - h)^2 + (y_0 - k)^2$.

Now, let's consider the position of the point $(x_0, y_0)$ relative to the circle based on this distance $D$ compared to the radius $r$:

1. Point is ON the circle: If the point $(x_0, y_0)$ lies exactly on the boundary of the circle, its distance from the center must be equal to the radius.

   $D = r \implies D^2 = r^2$

   So, $(x_0 - h)^2 + (y_0 - k)^2 = r^2$.

2. Point is OUTSIDE the circle: If the point $(x_0, y_0)$ lies outside the circle, its distance from the center must be greater than the radius.

   $D > r \implies D^2 > r^2$

   So, $(x_0 - h)^2 + (y_0 - k)^2 > r^2$.

3. Point is INSIDE the circle: If the point $(x_0, y_0)$ lies inside the circle (but not on the boundary), its distance from the center must be less than the radius.

   $D < r \implies D^2 < r^2$

   So, $(x_0 - h)^2 + (y_0 - k)^2 < r^2$.

The question asks for the condition when the point lies inside the circle. Based on our analysis, this corresponds to the third case.

Let's check the given options:

• (A) $(x_0-h)^2 + (y_0-k)^2 > r^2$: This condition is for a point outside the circle.
• (B) $(x_0-h)^2 + (y_0-k)^2 < r^2$: This condition correctly represents a point inside the circle.
• (C) $(x_0-h)^2 + (y_0-k)^2 = r^2$: This condition is for a point on the circle.
• (D) $(x_0-h)^2 + (y_0-k)^2 \geq r^2$: This condition means the point is either on or outside the circle.

Conclusion:

A point $(x_0, y_0)$ lies inside the circle $(x-h)^2 + (y-k)^2 = r^2$ if the square of its distance from the center is less than the square of the radius.

Therefore, the correct option is (B) $(x_0-h)^2 + (y_0-k)^2 < r^2$.

This question asks for the standard equation of a parabola given its vertex and the position of its focus. The orientation of the parabola (which way it opens) is determined by the relative position of the focus and vertex.

Given:

Vertex of the parabola: $(0,0)$ (the origin).

Focus of the parabola: $(0, -a)$, where $a > 0$.

To Find:

The standard equation of this parabola.

Solution:

A parabola is defined by its vertex, focus, and directrix. When the vertex is at the origin $(0,0)$, there are four standard forms for its equation, each corresponding to a different opening direction:

1. Opens Right: Vertex $(0,0)$, Focus $(a, 0)$ (where $a > 0$).

   Equation: $y^2 = 4ax$.

2. Opens Left: Vertex $(0,0)$, Focus $(-a, 0)$ (where $a > 0$).

   Equation: $y^2 = -4ax$.

3. Opens Upward: Vertex $(0,0)$, Focus $(0, a)$ (where $a > 0$).

   Equation: $x^2 = 4ay$.

4. Opens Downward: Vertex $(0,0)$, Focus $(0, -a)$ (where $a > 0$).

   Equation: $x^2 = -4ay$.

In this problem, we are given the vertex at $(0,0)$ and the focus at $(0, -a)$, with $a>0$.

Since the focus is on the negative y-axis (because $a>0$, $-a$ is negative), the parabola must open downward.

Comparing this specific case to the list of standard forms, it matches the "Opens Downward" type.

Therefore, the equation of the parabola is:

Conclusion:

The standard equation of a parabola with vertex at $(0,0)$ and focus on the negative y-axis at $(0, -a)$ where $a>0$ is $x^2 = -4ay$.

Let's check the given options:

• (A) $y^2 = 4ax$: Opens right.
• (B) $y^2 = -4ax$: Opens left.
• (C) $x^2 = 4ay$: Opens upward.
• (D) $x^2 = -4ay$: This matches our derived equation (opens downward).

Therefore, the correct option is (D) $x^2 = -4ay$.

This question asks us to find the equation of the directrix for a given parabola. This requires recognizing the standard form of the parabola and identifying its key parameter 'a'.

Given:

The equation of the parabola is $x^2 = -16y$.

To Find:

The equation of the directrix of this parabola.

Solution:

We need to compare the given equation $x^2 = -16y$ with the standard forms of parabolas with vertices at the origin $(0,0)$.

The four standard forms for parabolas with vertex at $(0,0)$ and their corresponding directrices are:

• $y^2 = 4ax$: Opens right; Focus $(a,0)$; Directrix $x = -a$.
• $y^2 = -4ax$: Opens left; Focus $(-a,0)$; Directrix $x = a$.
• $x^2 = 4ay$: Opens upward; Focus $(0,a)$; Directrix $y = -a$.
• $x^2 = -4ay$: Opens downward; Focus $(0,-a)$; Directrix $y = a$.

The given equation is:

This equation matches the form $x^2 = -4ay$.

By comparing the coefficients of $y$ in equation (i) with $-4a$, we can find the value of $a$:

$-4a = -16$

Divide both sides by -4:

$a = \frac{-16}{-4}$

For a parabola of the form $x^2 = -4ay$, the parabola opens downward, and its directrix is a horizontal line given by $y = a$.

Substitute the value of $a=4$ into the directrix equation:

Conclusion:

The equation of the directrix of the parabola $x^2 = -16y$ is $y = 4$.

Let's check the given options:

• (A) $y = 4$: This matches our derived directrix equation.
• (B) $y = -4$: This would be the directrix for $x^2 = 4ay$.
• (C) $x = 4$: This would be a vertical line, not appropriate for a vertically opening parabola.
• (D) $x = -4$: This would also be a vertical line.

Therefore, the correct option is (A) $y = 4$.

This question asks for the length of the latus rectum of a given parabola. This requires identifying the standard form of the parabola and extracting the relevant parameter 'a'.

Given:

The equation of the parabola is $x^2 = 9y$.

To Find:

The length of the latus rectum of the parabola.

Solution:

The length of the latus rectum for any standard parabola (with vertex at the origin) is $|4a|$. The sign of $a$ indicates the direction of opening, but the length is always positive.

We need to compare the given equation $x^2 = 9y$ with one of the standard forms of a parabola with its vertex at the origin $(0,0)$.

The standard forms are:

• $y^2 = 4ax$ (opens right/left)
• $x^2 = 4ay$ (opens upward/downward)

The given equation $x^2 = 9y$ matches the form $x^2 = 4ay$. This means the parabola opens either upward or downward.

By comparing the coefficients of $y$ in equation (i), we can find the value of $a$:

$4a = 9$

Divide both sides by 4:

The length of the latus rectum is $|4a|$. Since $a = \frac{9}{4}$ is positive, $|4a| = 4a$.

Length of latus rectum $= 4 \times \left(\frac{9}{4}\right)$

Length of latus rectum $= \cancel{4} \times \frac{9}{\cancel{4}}$

Length of latus rectum $= 9$

Conclusion:

The length of the latus rectum of the parabola $x^2 = 9y$ is 9.

Let's check the given options:

• (A) 9
• (B) $9/4$
• (C) $9/2$
• (D) 18

The correct option is (A) 9.

This question asks for the equation of the axis of the parabola given its standard equation. The axis of a parabola is its line of symmetry.

Given:

The equation of the parabola is $y^2 = 4ax$.

The vertex is at the origin $(0,0)$.

To Find:

The equation of the axis of the parabola.

Solution:

The axis of a parabola is the straight line that passes through its vertex and its focus. It is the line about which the parabola is symmetric.

For the standard parabola $y^2 = 4ax$ (with $a \neq 0$):

• The vertex is at $(0,0)$.
• The focus is at $(a,0)$.

Since both the vertex $(0,0)$ and the focus $(a,0)$ lie on the x-axis, the x-axis must be the axis of symmetry for this parabola.

The equation of the x-axis is $y = 0$.

Alternatively, the variable that is not squared in the equation determines the axis of symmetry. In $y^2 = 4ax$, the $x$ term is not squared, meaning the parabola opens along the x-axis (either right or left). Therefore, the x-axis is its axis of symmetry.

Conclusion:

The axis of the parabola $y^2 = 4ax$ is the x-axis.

Let's check the given options:

• (A) x-axis: This is correct.
• (B) y-axis: This would be the axis for parabolas of the form $x^2 = 4ay$ or $x^2 = -4ay$.
• (C) $y=x$: This is a diagonal line.
• (D) $y=-x$: This is another diagonal line.

Therefore, the correct option is (A) x-axis.

This question asks us to identify the equation that represents a parabola opening to the left. This is a direct test of knowing the standard forms of parabola equations and their corresponding orientations.

Given:

Four standard forms of parabola equations, all assuming $a>0$.

To Determine:

Which equation represents a parabola opening to the left.

Solution:

For parabolas with their vertex at the origin $(0,0)$, their general shape and opening direction are determined by which variable is squared and the sign of the constant term associated with the linear variable. The parameter '$a$' is typically defined as a positive distance from the vertex to the focus.

Let's review the standard forms assuming $a > 0$:

1. $y^2 = 4ax$:

   • The $y$ term is squared, meaning the parabola opens horizontally (along the x-axis).
   • The coefficient $4a$ is positive (since $a>0$), so it opens in the positive x-direction.
   • This parabola opens to the right.

2. $y^2 = -4ax$:

   • The $y$ term is squared, meaning the parabola opens horizontally (along the x-axis).
   • The coefficient $-4a$ is negative (since $a>0$), so it opens in the negative x-direction.
   • This parabola opens to the left.

3. $x^2 = 4ay$:

   • The $x$ term is squared, meaning the parabola opens vertically (along the y-axis).
   • The coefficient $4a$ is positive (since $a>0$), so it opens in the positive y-direction.
   • This parabola opens upward.

4. $x^2 = -4ay$:

   • The $x$ term is squared, meaning the parabola opens vertically (along the y-axis).
   • The coefficient $-4a$ is negative (since $a>0$), so it opens in the negative y-direction.
   • This parabola opens downward.

The question specifically asks for the equation that represents a parabola opening to the left. Based on our analysis, this is $y^2 = -4ax$.

Conclusion:

The equation $y^2 = -4ax$ (with $a>0$) represents a parabola opening to the left.

Therefore, the correct option is (B) $y^2 = -4ax$, $a>0$.

This question provides a point that lies on a given line and asks for the slope of that line. If a point lies on a line, its coordinates must satisfy the line's equation.

Given:

The equation of the line is $y = mx + 1$.

A point on the line is $(2, 3)$.

Here, $x = 2$ and $y = 3$.

To Find:

The slope $m$ of the line.

Solution:

Since the point $(2, 3)$ lies on the line $y = mx + 1$, its coordinates must satisfy the equation of the line. This means we can substitute $x=2$ and $y=3$ into the equation and solve for $m$.

Given equation:

$y = mx + 1$

Substitute $y=3$ and $x=2$:

$3 = m(2) + 1$

$3 = 2m + 1$

Now, we need to solve this linear equation for $m$. Subtract 1 from both sides:

$3 - 1 = 2m$

$2 = 2m$

Divide both sides by 2:

$m = \frac{2}{2}$

So, the slope of the line is 1.

Conclusion:

The slope $m$ of the line is 1.

Let's check the given options:

• (A) 1
• (B) 2
• (C) 3
• (D) -1

The correct option is (A) 1.

This question asks about the slope of a line that is parallel to a given line. This is a direct application of the condition for parallel lines in coordinate geometry.

Given:

A line with the equation $y = 2x + 5$.

Another line is parallel to this given line.

To Find:

The slope of the line parallel to $y = 2x + 5$.

Solution:

The given line equation is $y = 2x + 5$. This is in the slope-intercept form, $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

By comparing $y = 2x + 5$ with $y = mx + c$, we can identify the slope of the given line:

Now, we recall the condition for parallel lines:

Two lines are parallel if and only if their slopes are equal.

Let the slope of the parallel line be $m_2$. According to the condition for parallel lines:

Since $m_1 = 2$, then:

Therefore, the slope of any line parallel to $y = 2x + 5$ is 2.

Conclusion:

The slope of a line parallel to $y = 2x + 5$ is 2.

Let's check the given options:

• (A) 2
• (B) 5
• (C) 1/2
• (D) -1/2

The correct option is (A) 2.

The correct option is (C) 1/3.

Explanation:

The given equation of the line is $y = -3x + 1$. This equation is in the slope-intercept form, $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

From the given equation, the slope of the line is $m_1 = -3$.

Two lines are perpendicular if the product of their slopes is -1. Let the slope of the perpendicular line be $m_2$.

The condition for perpendicular lines is:

Substitute the value of $m_1$ into Equation 1:

To find $m_2$, divide both sides by -3:

Therefore, the slope of a line perpendicular to $y = -3x + 1$ is $\frac{1}{3}$.

The correct option is (C) $y - 4 = -1/2 (x - 1)$.

Explanation:

First, we need to find the slope of the given line $2x - y = 5$. We can rewrite this equation in the slope-intercept form ($y = mx + c$):

$y = 2x - 5$

The slope of this line is $m_1 = 2$.

A line perpendicular to this line will have a slope $m_2$ such that $m_1 \times m_2 = -1$.

$2 \times m_2 = -1$

$m_2 = -\frac{1}{2}$

Now we need to find the equation of the line with slope $m_2 = -\frac{1}{2}$ that passes through the point $(1, 4)$. We can use the point-slope form of a linear equation, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point and $m$ is the slope.

Substituting the values:

$y - 4 = -\frac{1}{2}(x - 1)$

This matches option (C).

The correct option is (A) $\sqrt{8}$.

Explanation:

To find the length of the side AB, we use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is given by:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Here, the vertices of the triangle are A(1, 2) and B(3, 4).

Let $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 4)$.

The length of side AB is:

AB $= \sqrt{(3 - 1)^2 + (4 - 2)^2}$

AB $= \sqrt{(2)^2 + (2)^2}$

AB $= \sqrt{4 + 4}$

AB $= \sqrt{8}$

Therefore, the length of the side AB is $\sqrt{8}$.

The correct option is (A) $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Explanation:

The formula for the perpendicular distance of a point $(x_0, y_0)$ from a line given by the equation $Ax + By + C = 0$ is a standard result in coordinate geometry.

The formula is:

Where:

• $(x_0, y_0)$ are the coordinates of the point.
• $A$, $B$, and $C$ are the coefficients of the line equation $Ax + By + C = 0$.
• $|Ax_0 + By_0 + C|$ represents the absolute value of substituting the point's coordinates into the line's expression.
• $\sqrt{A^2 + B^2}$ is the magnitude of the normal vector to the line.

Option (B) is incorrect because the distance must be non-negative, hence the absolute value is required.

Option (C) is incorrect because the denominator should be the square root of $A^2 + B^2$, not $A^2 + B^2$ itself.

Option (D) is the distance of the origin $(0,0)$ from the line if $C \neq 0$, but it is not the general formula for any point $(x_0, y_0)$.

The correct option is (A) $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.

Explanation:

The formula for the distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is a standard result in coordinate geometry.

The formula is:

Where:

• $A$ and $B$ are the coefficients of $x$ and $y$ respectively, which must be the same for both parallel lines.
• $C_1$ and $C_2$ are the constant terms in the equations of the two parallel lines.
• $|C_1 - C_2|$ represents the absolute difference between the constant terms.
• $\sqrt{A^2 + B^2}$ is the magnitude of the normal vector to the lines.

This formula is derived by finding the distance from any point on one line to the other line.

Options (B), (C), and (D) are incorrect as they do not represent the correct formula for the distance between parallel lines.

The correct option is (B) $(-3, 5)$.

Explanation:

The standard equation of a circle with center $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

The given equation of the circle is:

$(x + 3)^2 + (y - 5)^2 = 20$

We can rewrite the given equation to match the standard form:

$(x - (-3))^2 + (y - 5)^2 = 20$

By comparing this with the standard equation, we can identify the center $(h, k)$:

$h = -3$

$k = 5$

Therefore, the center of the circle is $(-3, 5)$. The radius squared is 20, so the radius is $\sqrt{20}$.

The correct option is (A) $\sqrt{20}$.

Explanation:

The standard equation of a circle with center $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

The given equation of the circle is:

$(x + 3)^2 + (y - 5)^2 = 20$

Comparing this with the standard form, we can see that:

$r^2 = 20$

To find the radius $r$, we take the square root of both sides:

$r = \sqrt{20}$

Therefore, the radius of the circle is $\sqrt{20}$.

The correct option is (C) $x^2 = 20y$.

Explanation:

The vertex of the parabola is at the origin $(0,0)$.

The focus of the parabola is at $(0, 5)$.

Since the focus is on the y-axis and is above the vertex, the parabola opens upwards.

The standard equation of a parabola with vertex at the origin and opening upwards is:

$x^2 = 4py$

In this standard equation, the focus is at $(0, p)$.

Comparing the given focus $(0, 5)$ with the standard focus $(0, p)$, we have:

$p = 5$

Now, substitute the value of $p$ into the standard equation:

$x^2 = 4(5)y$

$x^2 = 20y$

This matches option (C).

The correct option is (B) $x = -4$.

Explanation:

The given equation of the parabola is $y^2 = 16x$.

This equation is in the standard form of a parabola with vertex at the origin and opening to the right:

$y^2 = 4px$

Comparing the given equation with the standard form, we have:

$4p = 16$

Dividing both sides by 4, we get:

$p = \frac{16}{4}$

$p = 4$

For a parabola of the form $y^2 = 4px$, the focus is at $(p, 0)$ and the equation of the directrix is $x = -p$.

Since $p = 4$, the equation of the directrix is:

$x = -4$

Therefore, the equation of the directrix is $x = -4$.

The correct option is (B) $(0, -6)$.

Explanation:

The given equation of the parabola is $x^2 = -24y$.

This equation is in the standard form of a parabola with vertex at the origin $(0,0)$ and opening downwards:

$x^2 = -4py$

In this standard form, the focus is located at $(0, -p)$.

Comparing the given equation $x^2 = -24y$ with the standard form $x^2 = -4py$, we can equate the coefficients of $y$:

$-4p = -24$

To find the value of $p$, divide both sides by -4:

$p = \frac{-24}{-4}$

$p = 6$

Now, using the coordinates of the focus $(0, -p)$, we substitute the value of $p$:

Focus $= (0, -6)$

Therefore, the focus of the parabola $x^2 = -24y$ is $(0, -6)$.

The correct option is (C) $x^2 + y^2 = 1$.

Explanation:

An equation of a straight line is a linear equation, meaning the highest power of the variables $x$ and $y$ is 1. The general form of a linear equation in two variables is $Ax + By + C = 0$, where $A$ and $B$ are not both zero. Let's analyze each option:

(A) $y = 5$

This can be rewritten as $0x + 1y - 5 = 0$. This is a linear equation where $A=0$, $B=1$, and $C=-5$. This represents a horizontal straight line.

(B) $x = -2$

This can be rewritten as $1x + 0y + 2 = 0$. This is a linear equation where $A=1$, $B=0$, and $C=2$. This represents a vertical straight line.

(C) $x^2 + y^2 = 1$

This equation involves terms with $x^2$ and $y^2$. This is the standard equation of a circle with center $(0,0)$ and radius 1. It is not a linear equation and therefore does not represent a straight line.

(D) $3x - 4y + 7 = 0$

This is a linear equation in the form $Ax + By + C = 0$, where $A=3$, $B=-4$, and $C=7$. This represents a straight line.

Therefore, the equation that is NOT an equation of a straight line is $x^2 + y^2 = 1$.

The correct option is (A) $(7/2, 0)$ or (D) $(3.5, 0)$, as they are equivalent.

Explanation:

A line intersects the x-axis at a point where the y-coordinate is 0.

Given the equation of the line: $2x + y = 7$.

To find the point of intersection with the x-axis, we set $y = 0$ in the equation:

$2x + 0 = 7$

$2x = 7$

Now, solve for $x$ by dividing both sides by 2:

$x = \frac{7}{2}$

So, the point where the line intersects the x-axis is $(\frac{7}{2}, 0)$.

Option (A) is $(\frac{7}{2}, 0)$.

Option (D) is $(3.5, 0)$, which is the decimal equivalent of $(\frac{7}{2}, 0)$.

Option (B) $(0, 7)$ is the y-intercept (where $x=0$).

Option (C) $(2, 3)$ is a point on the line, as $2(2) + 3 = 4 + 3 = 7$, but it is not on the x-axis.

The correct option is (B) $A = C \neq 0$ and $B = 0$.

Explanation:

The general equation of a second-degree curve in two variables is given by:

$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

For this equation to represent a circle, the following conditions must be met:

1. The coefficient of $x^2$ must be equal to the coefficient of $y^2$. That is, $A = C$.
2. The coefficient of the $xy$ term must be zero. That is, $B = 0$.
3. The coefficients $A$ and $C$ must be non-zero. If $A=C=0$, the equation would become linear ($Dx + Ey + F = 0$), representing a straight line, not a circle.

Therefore, the conditions for the general second-degree equation to represent a circle are $A = C \neq 0$ and $B = 0$.

Let's analyze the options:

• (A) $A = C$ and $B = 0$: This is almost correct, but it misses the condition that $A$ (and hence $C$) must be non-zero. If $A=C=0$, it's not a circle.
• (B) $A = C \neq 0$ and $B = 0$: This option includes all the necessary conditions.
• (C) $A \neq 0, C \neq 0, B=0$: This condition describes an ellipse or a hyperbola if $A \neq C$, or a circle if $A=C$. However, it doesn't explicitly state $A=C$.
• (D) $A=0, C=0, B=0$: This would result in $Dx + Ey + F = 0$, which is the equation of a straight line (or no locus if D, E, F are all zero).

Thus, option (B) is the most accurate and complete set of conditions for the general second-degree equation to represent a circle.

The correct option is (B) 1.

Explanation:

Eccentricity ($e$) is a measure of how much a conic section deviates from being circular. It is defined for different conic sections as follows:

• Circle: $e = 0$. A circle is the set of points equidistant from a central point, so it has no deviation from being "circular".
• Ellipse: $0 < e < 1$. An ellipse is a "squashed" circle. As $e$ approaches 0, the ellipse becomes more like a circle.
• Parabola: $e = 1$. A parabola is formed when the distance from any point on the curve to a fixed point (focus) is equal to the distance from that point to a fixed line (directrix). This equal distance property leads to an eccentricity of 1.
• Hyperbola: $e > 1$. A hyperbola is "more curved" than a parabola and diverges more rapidly.

Therefore, the eccentricity of a parabola is always equal to 1.

The correct option is (B) focus, axis.

Explanation:

The latus rectum of a parabola is defined as the chord passing through the focus of the parabola and perpendicular to the axis of symmetry.

Let's break down the definitions:

• Focus: A fixed point used in the definition of a parabola, ellipse, and hyperbola. For a parabola, all points on the parabola are equidistant from the focus and the directrix.
• Axis of symmetry: A line that divides a geometric figure into two congruent halves. For a parabola with vertex at the origin, the axis of symmetry is typically the y-axis (if the parabola is $x^2 = 4py$ or $x^2 = -4py$) or the x-axis (if the parabola is $y^2 = 4px$ or $y^2 = -4px$).
• Directrix: A fixed line used in the definition of a parabola, ellipse, and hyperbola. For a parabola, it is a line such that the distance from any point on the parabola to the focus is equal to its distance to the directrix.
• Vertex: The point where the axis of symmetry intersects the parabola.

The latus rectum is a special chord that passes through the focus and is perpendicular to the axis of symmetry. Its length is $4p$ for parabolas with vertex at the origin.

Therefore, the latus rectum of a parabola passes through the focus and is perpendicular to the axis.

The correct option is (D) All of the above (representing the same line).

Explanation:

A line parallel to $y = mx + c$ has slope $m$.

(A) $y - y_1 = m(x - x_1)$ is the point-slope form, which is correct.

(C) $y = mx + (y_1 - mx_1)$ is the slope-intercept form, derived from (A), and is also correct.

Since (A) and (C) are correct, and "All of the above" is an option, it implies (B) is also intended to be a correct representation of the same line.

The correct option is (A) 5.

Explanation:

The standard equation of a circle centered at the origin $(0,0)$ with radius $r$ is $x^2 + y^2 = r^2$.

The point $(-3, 4)$ lies on the circle, which means its coordinates must satisfy the circle's equation.

Substitute $x = -3$ and $y = 4$ into the equation:

$(-3)^2 + (4)^2 = r^2$

$9 + 16 = r^2$

$25 = r^2$

To find the radius $r$, take the square root of both sides:

$r = \sqrt{25}$

$r = 5$

Since the radius must be a positive value, $r=5$.

The correct option is (B) $y = -2$.

Explanation:

The given equation of the parabola is $x^2 = 8y$.

This equation is in the standard form of a parabola with its vertex at the origin $(0,0)$ and opening upwards:

$x^2 = 4py$

By comparing the given equation $x^2 = 8y$ with the standard form $x^2 = 4py$, we can find the value of $p$:

$4p = 8$

Divide both sides by 4:

$p = \frac{8}{4}$

$p = 2$

For a parabola of the form $x^2 = 4py$, the focus is at $(0, p)$ and the directrix is the line $y = -p$.

Since $p = 2$, the equation of the directrix is:

$y = -2$

The given line is $x - 2y = 4$.

We can rewrite this equation in the slope-intercept form ($y = mx + c$) to find its slope.

$x - 4 = 2y$

$y = \frac{1}{2}x - 2$

The slope of this line ($m_1$) is $\frac{1}{2}$.

A line perpendicular to this line will have a slope ($m_2$) that is the negative reciprocal of $m_1$.

So, $m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{1}{2}} = -2$.

The required line passes through the origin (0, 0).

The equation of a line passing through the origin with slope $m$ is $y = mx$.

Substituting the slope $m_2 = -2$, we get:

$y = -2x$

Comparing this with the given options, option (B) matches our result.

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$.

We can rewrite this equation by completing the square for the $x$ and $y$ terms.

$(x^2 + 2gx) + (y^2 + 2fy) = -c$

$(x^2 + 2gx + g^2) + (y^2 + 2fy + f^2) = -c + g^2 + f^2$

$(x + g)^2 + (y + f)^2 = g^2 + f^2 - c$

This is the standard equation of a circle with center $(-g, -f)$ and radius $r$, where $r^2 = g^2 + f^2 - c$.

For the equation to represent a point, the radius must be zero. This means $r^2 = 0$.

Therefore, $g^2 + f^2 - c = 0$.

If $g^2 + f^2 - c > 0$, the equation represents a real circle with a positive radius.

If $g^2 + f^2 - c < 0$, the equation represents an imaginary circle (no real locus).

Thus, the general equation of a circle represents a point if $g^2 + f^2 - c = 0$.

The given equation of the parabola is $x^2 = -4ay$.

This is the standard form of a parabola that opens downwards (if $a>0$) or upwards (if $a<0$), with its vertex at the origin (0,0).

The axis of symmetry for a parabola of the form $x^2 = 4py$ or $x^2 = -4py$ is the y-axis.

In this case, since the equation is $x^2 = -4ay$, the axis of symmetry is the y-axis, which has the equation $x=0$.

Therefore, the axis of symmetry of the parabola $x^2 = -4ay$ is the y-axis.

The equation of the parabola is given as $x^2 = -4ay$.

The problem states that the arch is 40 meters wide at the base and has a height of 10 meters at the center. Since the vertex is at the origin and the base is on the x-axis, this means the parabola opens downwards.

The width at the base being 40 meters means the parabola extends 20 meters to the left and 20 meters to the right of the y-axis. The height of 10 meters at the center means the lowest point on the base corresponds to a y-coordinate of -10 (since the vertex is at the origin and the arch goes downwards).

Therefore, the points $(-20, -10)$ and $(20, -10)$ lie on the parabola.

We can substitute one of these points into the equation $x^2 = -4ay$ to find the value of $a$. Let's use the point $(20, -10)$.

Here, $x = 20$ and $y = -10$.

Substituting these values:

$(20)^2 = -4a(-10)$

$400 = 40a$

To find $a$, divide both sides by 40:

$a = \frac{400}{40}$

$a = 10$

Thus, the value of $a$ is 10.

From the previous question (Q72), we determined that the equation of the parabola is of the form $x^2 = -4ay$, and we found that the value of $a$ is 10.

To find the equation of the parabolic arch, we substitute the value of $a$ back into the general form of the equation.

Equation: $x^2 = -4ay$

Substitute $a = 10$: $x^2 = -4(10)y$

$x^2 = -40y$

Therefore, the equation of the parabolic arch is $x^2 = -40y$.

Two lines are parallel if they have the same slope.

Let's find the slope of the given line $Ax + By + C = 0$.

Rearranging the equation to the slope-intercept form ($y = mx + c$):

$By = -Ax - C$

$y = -\frac{A}{B}x - \frac{C}{B}$

The slope of the given line is $m_1 = -\frac{A}{B}$.

Now let's examine the slopes of the given options:

(A) $Ax + By + k = 0$

$By = -Ax - k$

$y = -\frac{A}{B}x - \frac{k}{B}$

The slope is $m_A = -\frac{A}{B}$. This is the same as the slope of the given line.

(B) $Bx - Ay + k = 0$

$-Ay = -Bx - k$

$y = \frac{B}{A}x + \frac{k}{A}$

The slope is $m_B = \frac{B}{A}$. This is the slope of a perpendicular line (negative reciprocal).

(C) $Ax - By + k = 0$

$-By = -Ax - k$

$y = \frac{A}{B}x + \frac{k}{B}$

The slope is $m_C = \frac{A}{B}$.

(D) $-Ax - By + k = 0$

$-By = Ax - k$

$y = -\frac{A}{B}x + \frac{k}{B}$

The slope is $m_D = -\frac{A}{B}$. This is the same as the slope of the given line. Note that this equation is equivalent to $Ax + By - k = 0$, which is also parallel.

A line parallel to $Ax + By + C = 0$ must have the same slope. Both option (A) and option (D) satisfy this condition. However, option (A) explicitly states "$k \neq C$", which is a common way to ensure it's a distinct parallel line. Option (D) is also a valid form of a parallel line, as multiplying the equation by -1 does not change the line's orientation or slope. In the context of selecting *the* form, option (A) is the most direct representation of a parallel line where the constant term is changed.

The standard representation for a line parallel to $Ax + By + C = 0$ is $Ax + By + k = 0$.

The given line is $Ax + By + C = 0$.

The slope of this line ($m_1$) is found by rearranging it into the slope-intercept form ($y = mx + c$):

$By = -Ax - C$

$y = -\frac{A}{B}x - \frac{C}{B}$

So, $m_1 = -\frac{A}{B}$.

For a line to be perpendicular to the given line, its slope ($m_2$) must be the negative reciprocal of $m_1$.

$m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{A}{B}} = \frac{B}{A}$.

Now let's check the slopes of the given options:

(A) $Ax + By + k = 0 \implies y = -\frac{A}{B}x - \frac{k}{B}$. Slope is $-\frac{A}{B}$ (parallel).

(B) $Bx - Ay + k = 0 \implies -Ay = -Bx - k \implies y = \frac{B}{A}x + \frac{k}{A}$. Slope is $\frac{B}{A}$.

(C) $Ax - By + k = 0 \implies -By = -Ax - k \implies y = \frac{A}{B}x + \frac{k}{B}$. Slope is $\frac{A}{B}$.

(D) $Bx + Ay + k = 0 \implies Ay = -Bx - k \implies y = -\frac{B}{A}x - \frac{k}{A}$. Slope is $-\frac{B}{A}$.

Option (B) has a slope of $\frac{B}{A}$, which is the negative reciprocal of the slope of the given line ($-\frac{A}{B}$). Therefore, the line represented by option (B) is perpendicular to the given line.

The locus of a point that moves in a plane such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix) is the definition of a Parabola.

Let the fixed point be $F(a, 0)$ and the fixed line be $x = -a$. Let $P(x, y)$ be any point on the locus.

According to the definition, the distance from $P$ to $F$ is equal to the distance from $P$ to the line $x = -a$.

Distance $PF = \sqrt{(x-a)^2 + (y-0)^2}$

Distance from $P(x,y)$ to the line $x = -a$ is $|x - (-a)| = |x+a|$

Equating the distances:

Squaring both sides:

Simplifying the equation:

This is the standard equation of a parabola.

The correct option is (C) Parabola.

Let the endpoints of the diameter be $A = (1, 2)$ and $B = (3, 4)$.

The center of the circle is the midpoint of the diameter AB.

Midpoint $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

The radius $r$ of the circle is half the length of the diameter.

Length of diameter $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Radius $r = \frac{AB}{2} = \frac{\sqrt{8}}{2}$

So, $r^2 = \left(\frac{\sqrt{8}}{2}\right)^2 = \frac{8}{4} = 2$.

The equation of a circle with centre $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

Substituting the centre $(2, 3)$ and $r^2 = 2$:

Alternatively, we can use the property that if $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a diameter, the equation of the circle is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.

Here, $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 4)$.

Expanding this equation:

To convert this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, we complete the square:

Both methods yield the same result.

The correct option is (C) $(x-2)^2 + (y-3)^2 = 2$.

The formula for the distance of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by:

In this problem, the point is $(x_0, y_0) = (0, 0)$ and the line is $3x - 4y + 10 = 0$.

Here, $A = 3$, $B = -4$, and $C = 10$.

Substituting these values into the formula:

The distance of the point $(0, 0)$ from the line $3x - 4y + 10 = 0$ is 2.

The correct option is (B) 2.

The given equation of the parabola is $y^2 = -10x$.

This is in the standard form of a parabola $y^2 = 4ax$.

Comparing the given equation with the standard form:

Solving for $a$:

For a parabola of the form $y^2 = 4ax$, the focus is located at $(a, 0)$.

Therefore, the focus of the given parabola is $(-2.5, 0)$.

The correct option is (B) $(-2.5, 0)$.

The equation of a circle centered at the origin $(0,0)$ with radius $r$ is given by:

The equation of the tangent to this circle at a point $(x_1, y_1)$ on the circle is a standard result in coordinate geometry.

The general form of the tangent to the circle $x^2 + y^2 = r^2$ at the point $(x_1, y_1)$ is obtained by replacing one $x$ with $x_1$ and one $y$ with $y_1$ in the equation of the circle.

Therefore, the equation of the tangent is:

This equation holds true if the point $(x_1, y_1)$ lies on the circle, meaning $x_1^2 + y_1^2 = r^2$.

The correct option is (A) $x x_1 + y y_1 = r^2$.

Let's consider a standard parabola with its vertex at the origin $(0,0)$.

The standard form of such a parabola is $y^2 = 4ax$ (opening to the right).

The focus ($F$) of this parabola is at $(a, 0)$.

The directrix of this parabola is the vertical line $x = -a$.

The vertex ($V$) of this parabola is at $(0, 0)$.

The question asks for the midpoint of the segment joining the focus and the directrix.

To find this midpoint, we need a point on the directrix. The point on the directrix that is closest to the focus is the point where the axis of symmetry intersects the directrix. For $y^2 = 4ax$, the axis of symmetry is the x-axis ($y=0$).

So, the point on the directrix $x = -a$ that lies on the axis of symmetry is $(-a, 0)$.

Now we find the midpoint of the segment joining the focus $F(a, 0)$ and the point on the directrix $(-a, 0)$.

Midpoint $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

The midpoint is $(0, 0)$, which is the vertex of the parabola.

The correct option is (B) Vertex.

We need to find the equation of a line passing through two points: $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (5, 10)$.

The slope of the line ($m$) is given by the formula:

Substituting the given points:

Now we use the point-slope form of the equation of a line, which is $y - y_1 = m(x - x_1)$. We can use either of the given points. Let's use $(0, 0)$.

Alternatively, since the line passes through the origin $(0,0)$, its equation is of the form $y = mx$. We found the slope $m=2$, so the equation is $y = 2x$.

The correct option is (B) $y = 2x$.

To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian coordinate system, we use the distance formula:

Given the points A(-3, 5) and B(2, -7):

Let $(x_1, y_1) = (-3, 5)$ and $(x_2, y_2) = (2, -7)$.

Substitute these values into the distance formula:

First, calculate the differences in the x and y coordinates:

Now, square these differences:

Add the squared differences:

Finally, take the square root of the sum:

The distance between points A and B is 13 units.

To find the coordinates of a point that divides a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m:n$, we use the section formula:

Given the points P(4, -3) and Q(8, 5), and the ratio $m:n = 3:1$.

Let $(x_1, y_1) = (4, -3)$ and $(x_2, y_2) = (8, 5)$.

Here, $m = 3$ and $n = 1$.

Substitute these values into the section formula:

For the x-coordinate:

For the y-coordinate:

Therefore, the coordinates of the point that divides the line segment joining P(4, -3) and Q(8, 5) internally in the ratio 3:1 are (7, 3).

To find the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, we use the formula:

Given the vertices:

Vertex 1: $(x_1, y_1) = (1, 2)$

Vertex 2: $(x_2, y_2) = (-4, -3)$

Vertex 3: $(x_3, y_3) = (4, 1)$

Substitute these values into the formula:

Calculate the terms inside the absolute value:

Term 1: $1(-3 - 1) = 1(-4) = -4$

Term 2: $(-4)(1 - 2) = (-4)(-1) = 4$

Term 3: $4(2 - (-3)) = 4(2 + 3) = 4(5) = 20$

Now, sum these terms:

Now, apply the formula with the sum:

The area of the triangle is 10 square units.

To find the slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the slope formula:

Given the points (3, -2) and (-1, 4):

Let $(x_1, y_1) = (3, -2)$ and $(x_2, y_2) = (-1, 4)$.

Substitute these values into the slope formula:

Calculate the differences in the numerator and denominator:

Now, divide the numerator by the denominator:

Simplify the fraction:

The slope of the line passing through the given points is $-\frac{3}{2}$.

We need to find the equation of a line given a point $(-2, 3)$ and a slope $m = -4$.

We can use the point-slope form of the equation of a line, which is:

Here, $(x_1, y_1) = (-2, 3)$ and $m = -4$.

Substitute these values into the point-slope form:

Simplify the expression:

Distribute the -4 on the right side:

To express the equation in the slope-intercept form ($y = mx + c$), isolate $y$:

We can also express the equation in the general form ($Ax + By + C = 0$):

The equation of the line passing through the point (-2, 3) with slope -4 is $y = -4x - 5$ or $4x + y + 5 = 0$.

To find the equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$, we first find the slope ($m$) and then use the point-slope form of the equation of a line.

Given the points (1, -1) and (3, 5):

Let $(x_1, y_1) = (1, -1)$ and $(x_2, y_2) = (3, 5)$.

First, calculate the slope ($m$):

Now, we use the point-slope form of the equation of a line, which is $y - y_1 = m(x - x_1)$. We can use either of the given points. Let's use (1, -1).

Simplify the equation:

Distribute the 3 on the right side:

To express the equation in the slope-intercept form ($y = mx + c$), isolate $y$:

We can also express the equation in the general form ($Ax + By + C = 0$):

The equation of the line passing through the points (1, -1) and (3, 5) is $y = 3x - 4$ or $3x - y - 4 = 0$.

We are given the x-intercept and y-intercept of a line.

The x-intercept is the point where the line crosses the x-axis. If the x-intercept is 4, the point is $(4, 0)$.

The y-intercept is the point where the line crosses the y-axis. If the y-intercept is -3, the point is $(0, -3)$.

We can use these two points to find the equation of the line.

First, calculate the slope ($m$) using the two points $(x_1, y_1) = (4, 0)$ and $(x_2, y_2) = (0, -3)$:

Now we can use the slope-intercept form of the equation of a line, which is $y = mx + c$, where $c$ is the y-intercept.

We are given that the y-intercept is -3, so $c = -3$.

Substitute the slope $m = \frac{3}{4}$ and the y-intercept $c = -3$ into the slope-intercept form:

Alternatively, we can use the intercept form of the equation of a line, which is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.

Given x-intercept $a = 4$ and y-intercept $b = -3$:

To convert this to a more standard form, we can multiply by the least common multiple of the denominators (4 and -3), which is 12:

This can also be written as $3x - 4y - 12 = 0$.

Let's check if $y = \frac{3}{4}x - 3$ is equivalent to $3x - 4y = 12$:

Multiply $y = \frac{3}{4}x - 3$ by 4:

Rearrange to get $3x - 4y = 12$. The equations are consistent.

The equation of the line is $y = \frac{3}{4}x - 3$ or $3x - 4y = 12$.

The intercept form of the equation of a line is given by $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.

We are given the equation $3x + 4y - 12 = 0$.

To reduce this equation into the intercept form, we need to set the right-hand side of the equation to 1.

First, move the constant term to the right side:

Now, divide both sides of the equation by 12 to make the right side equal to 1:

Simplify the terms:

This is the intercept form of the equation of the line.

From the intercept form $\frac{x}{a} + \frac{y}{b} = 1$:

The x-intercept ($a$) is the denominator of the x term, which is 4.

The y-intercept ($b$) is the denominator of the y term, which is 3.

Therefore, the equation of the line in intercept form is $\frac{x}{4} + \frac{y}{3} = 1$, and the intercepts are:

x-intercept: 4

y-intercept: 3

To find the slope of the line given by the equation $2x - 3y + 5 = 0$, we can convert the equation into the slope-intercept form, which is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

Start with the given equation:

Our goal is to isolate $y$. First, move the terms involving $x$ and the constant term to the right side of the equation:

Now, divide both sides of the equation by -3 to solve for $y$:

Separate the terms on the right side:

Simplify the fractions:

This equation is now in the slope-intercept form $y = mx + c$. By comparing $y = \frac{2}{3}x + \frac{5}{3}$ with $y = mx + c$, we can identify the slope $m$ and the y-intercept $c$.

The slope of the line is $m = \frac{2}{3}$.

The y-intercept of the line is $c = \frac{5}{3}$.

The slope of the line $2x - 3y + 5 = 0$ is $\frac{2}{3}$.

We need to find the equation of a line that is parallel to the given line $2x + 5y - 1 = 0$ and passes through the point $(3, -2)$.

Two lines are parallel if they have the same slope.

First, let's find the slope of the given line $2x + 5y - 1 = 0$. We can rewrite this equation in the slope-intercept form ($y = mx + c$):

The slope of this line is $m = -\frac{2}{5}$.

Since the required line is parallel to this line, it will also have a slope of $m = -\frac{2}{5}$.

Now we have the slope of the new line and a point $(3, -2)$ through which it passes. We can use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute $m = -\frac{2}{5}$, $x_1 = 3$, and $y_1 = -2$:

Simplify the equation:

To eliminate the fraction, multiply both sides by 5:

Distribute on both sides:

Rearrange the terms to get the general form of the equation ($Ax + By + C = 0$):

Alternatively, we know that any line parallel to $Ax + By + C = 0$ has the form $Ax + By + K = 0$, where $K$ is a constant.

So, a line parallel to $2x + 5y - 1 = 0$ has the form $2x + 5y + K = 0$.

Since this line passes through the point $(3, -2)$, substitute $x = 3$ and $y = -2$ into this equation to find $K$:

Substitute $K = 4$ back into the parallel line equation:

The equation of the line parallel to $2x + 5y - 1 = 0$ and passing through the point (3, -2) is $2x + 5y + 4 = 0$.

We need to find the equation of a line that is perpendicular to the given line $x - 2y + 3 = 0$ and passes through the point $(1, 1)$.

First, let's find the slope of the given line $x - 2y + 3 = 0$. Rewrite it in the slope-intercept form ($y = mx + c$):

The slope of this line is $m_1 = \frac{1}{2}$.

For two lines to be perpendicular, the product of their slopes must be -1. Let the slope of the required line be $m_2$.

So, the slope of the required line is -2.

Now we have the slope $m_2 = -2$ and a point $(1, 1)$ through which the line passes. We use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute $m = -2$, $x_1 = 1$, and $y_1 = 1$:

Distribute the -2 on the right side:

Rearrange the terms to get the general form of the equation ($Ax + By + C = 0$):

Alternatively, we can note that a line perpendicular to $Ax + By + C = 0$ has the form $Bx - Ay + K = 0$, where $K$ is a constant.

For the given line $x - 2y + 3 = 0$ (where $A=1, B=-2$), the perpendicular line has the form $-2x - 1y + K = 0$, or $2x + y - K = 0$. Let's use $2x + y + K = 0$.

Since the line passes through the point $(1, 1)$, substitute $x=1$ and $y=1$ to find $K$:

Substitute $K = -3$ back into the equation $2x + y + K = 0$:

The equation of the line perpendicular to $x - 2y + 3 = 0$ and passing through the point (1, 1) is $2x + y - 3 = 0$.

To find the distance of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$, we use the formula:

In this problem, the point is $(x_0, y_0) = (2, 3)$ and the line is $4x + 3y - 12 = 0$.

Here, $A = 4$, $B = 3$, and $C = -12$.

Substitute these values into the formula:

Calculate the numerator:

Calculate the denominator:

Now, calculate the distance:

The distance of the point (2, 3) from the line $4x + 3y - 12 = 0$ is 1 unit.

To find the angle between two lines, we can use the formula involving their slopes.

The slope-intercept form of a line is $y = mx + c$, where $m$ is the slope.

The first line is $y = \sqrt{3}x + 5$. Its slope is $m_1 = \sqrt{3}$.

The second line is $y = -\frac{1}{\sqrt{3}}x - 2$. Its slope is $m_2 = -\frac{1}{\sqrt{3}}$.

The formula for the tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is:

Substitute the values of $m_1$ and $m_2$ into the formula:

Simplify the numerator:

Simplify the denominator:

Since the denominator is 0, the value of $\tan \theta$ is undefined.

When $\tan \theta$ is undefined, the angle $\theta$ is $90^\circ$ or $\frac{\pi}{2}$ radians.

This means the two lines are perpendicular to each other.

We can also observe this by checking the product of the slopes:

Since the product of the slopes is -1, the lines are perpendicular.

The angle between the lines is $90^\circ$.

The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Given the vertices of the triangle:

Vertex 1: $(x_1, y_1) = (0, 6)$

Vertex 2: $(x_2, y_2) = (8, 12)$

Vertex 3: $(x_3, y_3) = (8, 0)$

Substitute these coordinates into the centroid formula:

For the x-coordinate of the centroid:

For the y-coordinate of the centroid:

The coordinates of the centroid of the triangle are $\left(\frac{16}{3}, 6\right)$.

The standard equation of a circle with center $(h, k)$ and radius $r$ is given by:

In this problem, the center of the circle is $(h, k) = (2, -3)$ and the radius is $r = 4$.

Substitute these values into the standard equation:

Simplify the equation:

This is the equation of the circle in standard form.

If we need to expand it to the general form ($x^2 + y^2 + Dx + Ey + F = 0$):

Expand $(x - 2)^2$: $x^2 - 4x + 4$

Expand $(y + 3)^2$: $y^2 + 6y + 9$

Substitute these back into the equation:

Combine the terms:

The equation of the circle with center (2, -3) and radius 4 is $(x - 2)^2 + (y + 3)^2 = 16$.

To find the center and radius of the circle given by the equation $x^2 + y^2 - 6x + 4y - 3 = 0$, we need to convert it into the standard form of a circle's equation: $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

We will use the method of completing the square for the x-terms and y-terms.

First, group the x-terms and y-terms together:

Now, complete the square for the x-terms ($x^2 - 6x$). To do this, take half of the coefficient of $x$ (-6), square it $\left(\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9\right)$, and add it inside the parenthesis. We must also subtract this value outside the parenthesis to keep the equation balanced.

Complete the square for the y-terms ($y^2 + 4y$). Take half of the coefficient of $y$ (4), square it $\left(\left(\frac{4}{2}\right)^2 = (2)^2 = 4\right)$, and add it inside the parenthesis. Subtract this value outside the parenthesis.

Rewrite the terms in parenthesis as squared binomials:

Move the constant term to the right side of the equation:

Now, this equation is in the standard form $(x - h)^2 + (y - k)^2 = r^2$.

By comparing, we can identify the center and radius:

Center $(h, k)$: From $(x - 3)^2$, we have $h = 3$. From $(y + 2)^2$, which is $(y - (-2))^2$, we have $k = -2$. So, the center is $(3, -2)$.

Radius $r$: From $r^2 = 16$, we take the square root of both sides to find $r = \sqrt{16} = 4$. (Radius is always positive).

The center of the circle is $(3, -2)$ and the radius is 4.

Let the endpoints of the diameter be $A = (1, 2)$ and $B = (3, 4)$.

The center of the circle is the midpoint of the diameter AB.

Midpoint $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

The radius $r$ of the circle is half the length of the diameter.

Length of diameter $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Radius $r = \frac{AB}{2} = \frac{\sqrt{8}}{2}$

So, $r^2 = \left(\frac{\sqrt{8}}{2}\right)^2 = \frac{8}{4} = 2$.

The equation of a circle with centre $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

Substituting the centre $(2, 3)$ and $r^2 = 2$:

Alternatively, we can use the property that if $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a diameter, the equation of the circle is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.

Here, $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 4)$.

Expanding this equation:

To convert this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, we complete the square:

Both methods yield the same result.

The equation of the circle is $(x-2)^2 + (y-3)^2 = 2$.

The given equation of the parabola is $y^2 = 8x$.

This equation is in the standard form of a parabola that opens to the right: $y^2 = 4ax$.

The vertex of a parabola in the form $y^2 = 4ax$ is always at the origin, $(0, 0)$.

To find the focus, we first need to determine the value of $a$ by comparing the given equation with the standard form.

By comparing these two equations, we have:

Solving for $a$:

For a parabola in the form $y^2 = 4ax$, the focus is located at $(a, 0)$.

Substituting the value of $a = 2$, the focus is at $(2, 0)$.

Therefore:

Vertex: $(0, 0)$

Focus: $(2, 0)$

The given equation of the parabola is $x^2 = 12y$.

This equation is in the standard form of a parabola that opens upwards: $x^2 = 4ay$.

To find the equation of the directrix, we first need to determine the value of $a$ by comparing the given equation with the standard form.

By comparing these two equations, we have:

Solving for $a$:

For a parabola in the form $x^2 = 4ay$, the directrix is the horizontal line $y = -a$.

Substituting the value of $a = 3$, the equation of the directrix is $y = -3$.

The given equation of the parabola is $y^2 = 16x$.

This equation is in the standard form of a parabola that opens to the right: $y^2 = 4ax$.

The length of the latus rectum for a parabola in the form $y^2 = 4ax$ is given by $|4a|$.

To find the length of the latus rectum, we first need to determine the value of $a$ by comparing the given equation with the standard form.

By comparing these two equations, we have:

Solving for $a$:

The length of the latus rectum is $|4a|$.

Substituting the value of $a = 4$:

The length of the latus rectum of the parabola $y^2 = 16x$ is 16.

We are given the endpoints of a line segment, A(x, 4) and B(5, -2), and the midpoint P(2, y).

The midpoint formula states that the coordinates of the midpoint of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by:

In this case, $(x_1, y_1) = (x, 4)$, $(x_2, y_2) = (5, -2)$, and the midpoint is $(2, y)$.

Let's find the x-coordinate of the midpoint:

To solve for $x$, multiply both sides by 2:

Subtract 5 from both sides:

Now, let's find the y-coordinate of the midpoint:

Therefore, the values are $x = -1$ and $y = 1$.

To show that three points are collinear, we can demonstrate that the slope between the first two points is equal to the slope between the second two points (or any combination thereof). If the slopes are equal, it means the points lie on the same straight line.

Let the given points be A(1, 1), B(3, 4), and C(5, 7).

First, find the slope of the line segment AB (connecting points A and B).

The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

For points A(1, 1) and B(3, 4):

Next, find the slope of the line segment BC (connecting points B and C).

For points B(3, 4) and C(5, 7):

Since the slope of AB ($m_{AB}$) is equal to the slope of BC ($m_{BC}$), and point B is common to both segments, the points A, B, and C are collinear.

Alternatively, we can show that the area of the triangle formed by these three points is zero. The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

For points (1, 1), (3, 4), and (5, 7):

Since the area of the triangle formed by these points is 0, the points are collinear.

First, we need to find the slope of the line passing through the points (0, 0) and (5, 5).

Let $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (5, 5)$.

The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

The slope of the line passing through (0, 0) and (5, 5) is 1. This line is $y=x$.

Now, we need to find the slope of a line perpendicular to this line.

If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the original line be $m_1 = 1$. Let the slope of the perpendicular line be $m_2$.

The slope of the line perpendicular to the line passing through points (0, 0) and (5, 5) is -1.

The general form of the equation of a line is $ax + by + c = 0$.

To find the slope of this line, we can rewrite the equation in the slope-intercept form, which is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

Start with the given equation:

Our goal is to isolate $y$. First, move the terms involving $x$ and the constant term to the right side of the equation:

Now, divide both sides of the equation by $b$ (assuming $b \neq 0$) to solve for $y$:

Separate the terms on the right side:

This equation is now in the slope-intercept form $y = mx + c$. By comparing $y = -\frac{a}{b}x - \frac{c}{b}$ with $y = mx + c$, we can identify the slope $m$.

The slope of the line is $m = -\frac{a}{b}$.

Note: If $b=0$, the equation becomes $ax + c = 0$, or $x = -\frac{c}{a}$. This represents a vertical line, and its slope is undefined.

The slope of the line $ax + by + c = 0$ is $-\frac{a}{b}$, provided that $b \neq 0$.

The equation of the x-axis is y = 0.

This is because every point on the x-axis has a y-coordinate of zero. For example, points like (3, 0), (-5, 0), and (0, 0) all lie on the x-axis. The general form of any point on the x-axis can be represented as (x, 0), where 'x' can be any real number. Therefore, the defining characteristic of the x-axis is that the y-coordinate is always zero.

The equation of the y-axis is x = 0.

Similarly, every point on the y-axis has an x-coordinate of zero. Examples include points like (0, 7), (0, -2), and (0, 0). The general form of any point on the y-axis can be represented as (0, y), where 'y' can be any real number. This signifies that the x-coordinate is consistently zero for all points on the y-axis.

Let the first line be $L_1: x - y + 1 = 0$.

The slope of the first line, $m_1$, can be found by rewriting the equation in the form $y = mx + c$.

Rearranging $x - y + 1 = 0$, we get:

Comparing this with $y = m_1x + c_1$, we find that $m_1 = 1$.

Let the second line be $L_2: \sqrt{3}x + y - 2 = 0$.

The slope of the second line, $m_2$, can be found by rewriting the equation in the form $y = mx + c$.

Rearranging $\sqrt{3}x + y - 2 = 0$, we get:

Comparing this with $y = m_2x + c_2$, we find that $m_2 = -\sqrt{3}$.

The formula for the tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

Substituting the values of $m_1$ and $m_2$ into the formula:

To simplify the expression, we can multiply the numerator and denominator by the conjugate of the denominator:

We know that $\tan 75^\circ = 2 + \sqrt{3}$.

Therefore, the acute angle between the two lines is $75^\circ$.

We are given two parallel lines:

Line 1: $2x + 3y + 4 = 0$

Line 2: $2x + 3y - 6 = 0$

The general form of a linear equation is $Ax + By + C = 0$.

For Line 1, we have $A = 2$, $B = 3$, and $C_1 = 4$.

For Line 2, we have $A = 2$, $B = 3$, and $C_2 = -6$.

The formula for the distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is:

Substituting the values of A, B, $C_1$, and $C_2$ into the formula:

To rationalize the denominator, we multiply the numerator and the denominator by $\sqrt{13}$:

Therefore, the distance between the parallel lines is $\frac{10\sqrt{13}}{13}$ units.

The general equation of a circle with center $(h, k)$ and radius $r$ is given by:

In this problem, the center of the circle is given as $(h, k) = (-a, -b)$.

The circle passes through the origin (0, 0). The distance between the center of the circle and any point on the circle is equal to the radius ($r$).

Using the distance formula between the center $(-a, -b)$ and the point $(0, 0)$:

Now, we square the radius:

Now, substitute the center $(h, k) = (-a, -b)$ and $r^2 = a^2 + b^2$ into the general equation of the circle (i):

Expanding the terms:

Subtract $a^2 + b^2$ from both sides:

Rearranging the terms to the standard form:

This is the equation of the circle passing through the origin with center $(-a, -b)$.

The vertex of the parabola is at the origin, $V(0, 0)$.

The focus of the parabola is at $(0, -3)$.

Since the vertex is at the origin and the focus is on the y-axis (at $(0, -3)$), the parabola opens downwards.

The standard equation of a parabola with vertex at the origin and opening downwards is:

where 'a' is the distance from the vertex to the focus.

The focus of such a parabola is at $(0, -a)$.

We are given that the focus is at $(0, -3)$.

By comparing the general focus $(0, -a)$ with the given focus $(0, -3)$, we can equate the y-coordinates:

From this, we find that $a = 3$.

Now, substitute the value of $a = 3$ into the standard equation of the parabola (i):

Thus, the equation of the parabola with vertex at the origin and focus at (0, -3) is $x^2 = -12y$.

The equation of the circle is given by:

where $r$ is the radius of the circle. We assume $r > 0$ for a valid circle.

Domain:

To find the domain, we need to determine the possible values of $x$. We can rearrange the equation to solve for $x^2$:

For $x$ to be a real number, $x^2$ must be non-negative.

This implies $r^2 \geq y^2$, or $-r \leq y \leq r$.

Now, let's consider solving for $y^2$ from the original equation:

For $y$ to be a real number, $y^2$ must be non-negative.

This implies $r^2 \geq x^2$. Taking the square root of both sides, we get $|x| \leq r$, which means $-r \leq x \leq r$.

Therefore, the domain of the equation of the circle is all real numbers $x$ such that $-r \leq x \leq r$. In interval notation, the domain is $[-r, r]$.

Range:

To find the range, we need to determine the possible values of $y$. From equation (iv) above, we found that for $y$ to be a real number, $r^2 - x^2 \geq 0$, which leads to $-r \leq x \leq r$.

Similarly, from equation (ii), $x^2 = r^2 - y^2$, and for $x$ to be real, $r^2 - y^2 \geq 0$, which implies $r^2 \geq y^2$. Taking the square root of both sides, we get $|y| \leq r$, which means $-r \leq y \leq r$.

Therefore, the range of the equation of the circle is all real numbers $y$ such that $-r \leq y \leq r$. In interval notation, the range is $[-r, r]$.

Summary:

Domain: $[-r, r]$

Range: $[-r, r]$

Let the coordinates of point A be $(x_1, y_1) = (2, 1)$.

Let the coordinates of point B be $(x_2, y_2) = (5, 7)$.

The ratio in which the point divides the line segment externally is $m:n = 2:1$.

The formula for the coordinates of a point that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ externally in the ratio $m:n$ is given by:

Substitute the given values into the formula:

For the x-coordinate:

For the y-coordinate:

Therefore, the coordinates of the point that divides the line segment joining A(2, 1) and B(5, 7) externally in the ratio 2:1 are (8, 13).

We are looking for the equation of a line that passes through the point (4, 5).

The line is parallel to the y-axis.

Lines that are parallel to the y-axis are vertical lines.

The defining characteristic of a vertical line is that the x-coordinate of every point on the line is the same.

Since the line passes through the point (4, 5), the x-coordinate of every point on this line must be 4.

Therefore, the equation of the line passing through (4, 5) and parallel to the y-axis is:

This equation represents all points with an x-coordinate of 4, regardless of their y-coordinate. For example, (4, 0), (4, 5), (4, -2) all lie on this line.

The slope-intercept form of the equation of a line is given by:

where $m$ is the slope of the line and $c$ is the y-intercept.

In this problem, we are given:

Slope, $m = 2$.

y-intercept, $c = 5$.

Substitute these values into the slope-intercept form of the equation:

This is the equation of the line in slope-intercept form.

If we need to express it in the general form $Ax + By + C = 0$, we can rearrange the equation:

Thus, the equation of the line with slope 2 and y-intercept 5 is $y = 2x + 5$ or $2x - y + 5 = 0$.

We need to find the slopes of both lines to determine their relationship.

Let the first line be $L_1: 2x - y + 3 = 0$.

To find the slope of $L_1$, we can rewrite it in the slope-intercept form ($y = mx + c$):

The slope of the first line, $m_1$, is the coefficient of $x$, so $m_1 = 2$.

Let the second line be $L_2: 4x - 2y + 7 = 0$.

To find the slope of $L_2$, we rewrite it in the slope-intercept form:

The slope of the second line, $m_2$, is the coefficient of $x$, so $m_2 = 2$.

Now, let's compare the slopes:

We have $m_1 = 2$ and $m_2 = 2$.

Condition for parallel lines: Two lines are parallel if their slopes are equal ($m_1 = m_2$).

Condition for perpendicular lines: Two lines are perpendicular if the product of their slopes is -1 ($m_1 \times m_2 = -1$).

Since $m_1 = m_2 = 2$, the slopes are equal.

Therefore, the lines $2x - y + 3 = 0$ and $4x - 2y + 7 = 0$ are parallel.

The given point is $(x_0, y_0) = (-1, -3)$.

The given line is $y = 2x + 5$.

First, we need to rewrite the equation of the line in the general form $Ax + By + C = 0$.

Rearranging $y = 2x + 5$, we get:

From this equation, we have $A = 2$, $B = -1$, and $C = 5$.

The formula for the distance $d$ between a point $(x_0, y_0)$ and a line $Ax + By + C = 0$ is:

Substitute the values of $A$, $B$, $C$, $x_0$, and $y_0$ into the formula:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{5}$:

The distance between the point (-1, -3) and the line $y = 2x + 5$ is $\frac{6\sqrt{5}}{5}$ units.

The general equation of a circle with center $(h, k)$ and radius $r$ is:

We are given that the center of the circle is $(h, k) = (0, 0)$.

Substituting the center into the equation:

The circle passes through the point (1, 1). This means that the coordinates of this point must satisfy the equation of the circle.

Substitute $x = 1$ and $y = 1$ into equation (iii) to find the value of $r^2$:

Now, substitute the value of $r^2 = 2$ back into the equation of the circle (iii):

Therefore, the equation of the circle passing through the point (1, 1) with center at (0, 0) is $x^2 + y^2 = 2$.

The given equation of the parabola is $y^2 = -4ax$.

This is the standard form of a parabola with its vertex at the origin (0, 0) and its axis of symmetry along the x-axis.

For a parabola in the form $y^2 = -4ax$:

• The vertex is at the origin $(0, 0)$.
• The axis of symmetry is the x-axis ($y = 0$).
• The focus is located at $(-a, 0)$.

The directrix of a parabola is a line such that the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix.

For a parabola in the form $y^2 = -4ax$, the directrix is a vertical line located at a distance $a$ from the vertex on the opposite side of the focus.

Since the focus is at $(-a, 0)$ and the vertex is at $(0, 0)$, the directrix is to the right of the vertex.

The equation of the directrix is therefore $x = a$.

The equation of the directrix of the parabola $y^2 = -4ax$ is $x = a$.

The slope of a line is defined as the tangent of the angle it makes with the positive x-axis.

Let the angle be $\theta$. The slope $m$ is given by $m = \tan(\theta)$.

In this case, the line makes an angle of $45^\circ$ with the positive x-axis, so $\theta = 45^\circ$.

Therefore, the slope is:

We know the value of $\tan(45^\circ)$:

The slope of the line is 1.

Let the y-axis divide the line segment joining A(-4, 5) and B(3, -7) in the ratio $k:1$.

The y-axis is defined by the equation $x = 0$. This means any point lying on the y-axis has an x-coordinate of 0.

We will use the section formula to find the coordinates of the point of division. Let the point of division be P. The coordinates of P are given by:

P = $\left( \frac{k \cdot x_2 + 1 \cdot x_1}{k+1}, \frac{k \cdot y_2 + 1 \cdot y_1}{k+1} \right)$

Here, $(x_1, y_1) = A(-4, 5)$ and $(x_2, y_2) = B(3, -7)$.

So, the coordinates of P are:

Since the point P lies on the y-axis, its x-coordinate must be 0.

Therefore, we set the x-coordinate from P to 0:

For this fraction to be zero, the numerator must be zero (assuming the denominator is not zero):

Solve for $k$:

The ratio is $k:1$, which is $\frac{4}{3}:1$. To express this ratio with integers, we multiply both parts by 3:

Since $k$ is positive, the division is internal.

The y-axis divides the line segment joining A(-4, 5) and B(3, -7) in the ratio 4 : 3.

We are asked to find the equation of a line that passes through the point (5, -1) and has a slope of 0.

We can use the point-slope form of the equation of a line, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.

Given:

• Point $(x_1, y_1) = (5, -1)$
• Slope $m = 0$

Substitute these values into the point-slope form:

Simplify the equation:

Now, solve for y:

A line with a slope of 0 is a horizontal line. A horizontal line has an equation of the form $y = k$, where $k$ is the y-coordinate of every point on the line. In this case, the y-coordinate of the given point is -1, so the equation of the horizontal line is $y = -1$.

The equation of the line is $y = -1$.

The given equation of the line is $\frac{x}{3} + \frac{y}{4} = 1$.

To find the perpendicular distance of the origin (0, 0) from this line, we first need to convert the equation of the line into the general form $Ax + By + C = 0$.

Multiply the entire equation by the least common multiple of the denominators (3 and 4), which is 12:

Simplify the equation:

Now, rewrite it in the general form $Ax + By + C = 0$:

Here, $A = 4$, $B = 3$, and $C = -12$.

The perpendicular distance $d$ of a point $(x_0, y_0)$ from the line $Ax + By + C = 0$ is given by the formula:

We want to find the distance of the origin $(x_0, y_0) = (0, 0)$ from the line $4x + 3y - 12 = 0$.

Substitute the values into the formula:

Simplify the numerator and the denominator:

The perpendicular distance of the origin from the line $x/3 + y/4 = 1$ is $\frac{12}{5}$ units.

The standard equation of a circle with center $(h, k)$ and radius $r$ is given by:

We are given the center of the circle as $(h, k) = (-g, -f)$ and the radius as $r = \sqrt{g^2 + f^2 - c}$.

Substitute these values into the standard equation (i).

For the center $(h, k) = (-g, -f)$, we have:

$h = -g$

$k = -f$

So, $(x - h)^2 = (x - (-g))^2 = (x + g)^2$.

And $(y - k)^2 = (y - (-f))^2 = (y + f)^2$.

For the radius $r = \sqrt{g^2 + f^2 - c}$, we need to square it to find $r^2$:

$r^2 = (\sqrt{g^2 + f^2 - c})^2 = g^2 + f^2 - c$.

Now, substitute these into the standard equation (i):

Expand the terms on the left side:

Now, move all terms to one side to match the general form of the circle's equation ($x^2 + y^2 + 2gx + 2fy + c = 0$):

Simplify the equation by cancelling out $g^2$ and $f^2$ terms:

The equation of the circle is $(x + g)^2 + (y + f)^2 = g^2 + f^2 - c$, which can also be written as $x^2 + y^2 + 2gx + 2fy + c = 0$.

The given equation of the parabola is $x^2 = 4ay$.

This is the standard form of a parabola with its vertex at the origin (0, 0) and its axis of symmetry along the y-axis.

For a parabola in the form $x^2 = 4ay$:

• The vertex is at the origin $(0, 0)$.
• The axis of symmetry is the y-axis ($x = 0$).
• The directrix is the line $y = -a$.
• The focus is located at a distance of '$a$' units from the vertex along the axis of symmetry, in the direction opposite to the directrix.

Since the axis of symmetry is the y-axis and the parabola opens upwards if $a > 0$ or downwards if $a < 0$, the focus lies on the y-axis.

The focus is located at a distance $a$ from the vertex $(0, 0)$ along the y-axis. Therefore, the coordinates of the focus are $(0, a)$.

The focus of the parabola $x^2 = 4ay$ is $(0, a)$.

Let the points on the x-axis be denoted by $(x, 0)$, since the y-coordinate of any point on the x-axis is 0.

We are given a point P = (3, 4) and we need to find points on the x-axis that are 5 units away from P.

Let the point on the x-axis be Q = $(x, 0)$.

The distance between P and Q is given as 5 units. We use the distance formula:

Distance PQ = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Here, $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (x, 0)$.

So, 5 = $\sqrt{(x - 3)^2 + (0 - 4)^2}$

Square both sides of the equation to eliminate the square root:

Now, isolate the term $(x - 3)^2$:

Take the square root of both sides:

This gives us two possible values for $x$:

Case 1: $x - 3 = 3$

The first point on the x-axis is (6, 0).

Case 2: $x - 3 = -3$

The second point on the x-axis is (0, 0).

The points on the x-axis which are at a distance of 5 units from the point (3, 4) are (6, 0) and (0, 0).

First, we need to find the point of intersection of the two given lines: $x+y=5$ and $x-y=1$.

We can solve this system of linear equations. Adding the two equations will eliminate $y$:

Now, substitute the value of $x=3$ into either of the original equations to find $y$. Using $x+y=5$:

So, the point of intersection of the two lines is (3, 2).

Next, we need to find the slope of the line $3x - 2y + 1 = 0$. We can rewrite this equation in the slope-intercept form ($y = mx + c$) to find the slope ($m$).

The slope of this line is $m = \frac{3}{2}$.

The required line is parallel to $3x - 2y + 1 = 0$. Parallel lines have the same slope. Therefore, the slope of the required line is also $\frac{3}{2}$.

Now we have the slope of the required line ($m = \frac{3}{2}$) and a point it passes through (3, 2).

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

Multiply both sides by 2 to eliminate the fraction:

Expand both sides:

Rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

The equation of the line is $3x - 2y - 5 = 0$.

Let the equation of the circle be $(x-a)^2 + (y-b)^2 = r^2$.

Since the center of the circle is on the x-axis, the y-coordinate of the center is 0. So, let the center be $(a, 0)$.

The equation of the circle becomes $(x-a)^2 + (y-0)^2 = r^2$, which simplifies to $(x-a)^2 + y^2 = r^2$.

The circle passes through the point (0, 0). Substituting these coordinates into the equation:

This gives us $a^2 = r^2$.

The circle also passes through the point (6, 0). Substituting these coordinates into the equation:

This gives us $(6-a)^2 = r^2$.

Now we have two equations for $r^2$:

Equating (i) and (ii):

Taking the square root of both sides of (iii):

Case 1: $a = 6-a$

From (v), $a = 3$.

Substituting $a=3$ into equation (i):

Using the general equation of the circle $(x-a)^2 + y^2 = r^2$ with $a=3$ and $r^2=9$:

Expanding (vii):

Simplifying (viii):

Case 2: $a = -(6-a)$

From (x), $0 = -6$, which is impossible. Therefore, we only consider Case 1.

The equation of the circle is $x^2 + y^2 - 6x = 0$.

The given equation of the parabola is $x^2 = -8y$.

This is in the standard form of a parabola with its vertex at the origin and opening downwards, which is $x^2 = -4ay$.

Comparing the given equation with the standard form:

Equating the coefficients of y from both equations:

Solving for $a$ from equation (iii):

The length of the latus rectum of a parabola in the form $x^2 = -4ay$ is given by $4a$.

Substituting the value of $a$ we found:

The length of the latus rectum of the parabola $x^2 = -8y$ is 8.

The standard form of a parabola with its axis of symmetry parallel to the x-axis is given by:

In this form:

• The vertex of the parabola is at the point $(h, k)$.
• The axis of symmetry is the line $y = k$.
• If $a > 0$, the parabola opens to the right.
• If $a < 0$, the parabola opens to the left.

The vertex is the point where the parabola changes direction. In the given equation $(y-k)^2 = 4a(x-h)$, the term $(y-k)^2$ is always non-negative. For the equation to hold, $(x-h)$ must have the same sign as $4a$.

The vertex occurs when both squared terms in the shifted coordinate system are zero. If we consider a translation of coordinates such that $X = x-h$ and $Y = y-k$, the equation becomes $Y^2 = 4aX$.

In this translated system, the vertex is at $(X, Y) = (0, 0)$.

Substituting back the original variables:

Therefore, the vertex of the parabola $(y-k)^2 = 4a(x-h)$ is at the point $(h, k)$.

The vertex of the parabola $(y-k)^2 = 4a(x-h)$ is $(h, k)$.

The given equation of the circle is $2x^2 + 2y^2 - 8x + 12y - 24 = 0$.

To find the radius, we first need to convert this equation to the standard form of a circle's equation, which is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

Divide the entire equation by 2 to make the coefficients of $x^2$ and $y^2$ equal to 1:

Simplifying equation (i):

Now, rearrange the terms to group x-terms and y-terms:

Complete the square for the x-terms and y-terms. To complete the square for $x^2 - 4x$, we add $(\frac{-4}{2})^2 = (-2)^2 = 4$. To complete the square for $y^2 + 6y$, we add $(\frac{6}{2})^2 = (3)^2 = 9$.

Add these values to both sides of the equation:

Rewrite the expressions in parentheses as squared terms:

This equation is now in the standard form $(x-h)^2 + (y-k)^2 = r^2$, where $h=2$, $k=-3$, and $r^2=25$.

To find the radius $r$, take the square root of $r^2$:

The radius of the circle is 5.

First, let's find the slope of the segment joining the points (1, 2) and (3, 4).

The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Let $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 4)$.

The slope of the given segment ($m_1$) is:

We need to find the equation of a line that is perpendicular to this segment. If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the required perpendicular line be $m_2$. Then:

Substituting the value of $m_1$ from (iii):

Now we have the slope of the required line ($m_2 = -1$) and a point it passes through (5, 6).

We can use the point-slope form of the equation of a line, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is its slope.

Using the point (5, 6) and slope $m_2 = -1$:

Now, simplify the equation:

Rearrange the terms to get the equation in the general form ($Ax + By + C = 0$):

The equation of the line is $x + y - 11 = 0$.

Let the given points be A = (1, -1), B = (5, 2), and C = (9, 5).

To determine if these points form a right-angled triangle, we can calculate the lengths of the sides of the triangle and check if they satisfy the Pythagorean theorem ($a^2 + b^2 = c^2$), or we can calculate the slopes of the sides and check if any two sides are perpendicular (product of slopes is -1).

Method 1: Using Slopes

Let's calculate the slopes of the lines formed by these points.

The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Slope of AB ($m_{AB}$):

Slope of BC ($m_{BC}$):

Slope of AC ($m_{AC}$):

Since $m_{AB} = m_{BC} = m_{AC} = \frac{3}{4}$, all three points lie on the same straight line. This means the points are collinear and do not form a triangle at all, let alone a right-angled triangle.

Alternatively, we can check for perpendicularity by seeing if the product of any two slopes is -1.

$m_{AB} \times m_{BC} = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \neq -1$

$m_{AB} \times m_{AC} = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \neq -1$

$m_{BC} \times m_{AC} = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \neq -1$

Since none of the slopes have a product of -1, no two sides are perpendicular.

Method 2: Using Distances (Pythagorean Theorem)

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Length of AB ($AB$):

Length of BC ($BC$):

Length of AC ($AC$):

Now check if the Pythagorean theorem holds for any combination of sides:

$AB^2 + BC^2 = 5^2 + 5^2 = 25 + 25 = 50$

$AC^2 = 10^2 = 100$

Since $AB^2 + BC^2 \neq AC^2$ (50 $\neq$ 100), the triangle is not right-angled at B.

Let's check other combinations:

$AB^2 + AC^2 = 5^2 + 10^2 = 25 + 100 = 125$

$BC^2 = 5^2 = 25$

Since $AB^2 + AC^2 \neq BC^2$ (125 $\neq$ 25), the triangle is not right-angled at A.

$BC^2 + AC^2 = 5^2 + 10^2 = 25 + 100 = 125$

$AB^2 = 5^2 = 25$

Since $BC^2 + AC^2 \neq AB^2$ (125 $\neq$ 25), the triangle is not right-angled at C.

Conclusion from both methods:

The points (1, -1), (5, 2), and (9, 5) are collinear because the slopes between each pair of points are equal. Therefore, they do not form a triangle, and consequently, they cannot form a right-angled triangle.

The points (1, -1), (5, 2), and (9, 5) are collinear and do not form a right-angled triangle.

The vertex of the parabola is given as V = (2, 3).

The focus of the parabola is given as F = (2, 5).

Observe that the x-coordinates of the vertex and the focus are the same (both are 2). This indicates that the axis of symmetry of the parabola is a vertical line, $x = 2$.

For a parabola with a vertical axis of symmetry, the standard form of the equation is $(x-h)^2 = 4a(y-k)$, where $(h, k)$ is the vertex and $a$ is the distance from the vertex to the focus (and also from the vertex to the directrix).

From the given vertex (2, 3), we have $h = 2$ and $k = 3$.

The distance $a$ between the vertex (2, 3) and the focus (2, 5) is the difference in their y-coordinates, since the x-coordinates are the same:

Since the focus (2, 5) is above the vertex (2, 3), the parabola opens upwards. This means $a$ is positive, which is consistent with our calculation ($a=2$).

Now substitute the values of $h$, $k$, and $a$ into the standard equation $(x-h)^2 = 4a(y-k)$:

Simplify the equation:

Expanding the equation is optional, but if required:

The equation of the parabola is $(x - 2)^2 = 8(y - 3)$.

Let the equation of the line be in the intercept form, which is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.

The problem states that the line cuts off equal intercepts from the axes. This means that the magnitude of the x-intercept and the y-intercept are equal. So, $a = b$ or $a = -b$.

Case 1: Equal positive intercepts ($a = b$)

If the intercepts are equal and positive, then $a = b$. The equation of the line becomes:

Multiplying by $a$, we get:

The line passes through the point (2, 3). Substitute these coordinates into equation (ii):

Substituting the value of $a$ back into equation (ii), we get the equation of the line:

This can also be written as $x + y - 5 = 0$.

Case 2: Equal but opposite intercepts ($a = -b$)

If the intercepts are equal in magnitude but opposite in sign, then $b = -a$. The equation of the line becomes:

Multiplying by $a$, we get:

The line passes through the point (2, 3). Substitute these coordinates into equation (vii):

Substituting the value of $a$ back into equation (vii), we get the equation of the line:

This can also be written as $x - y + 1 = 0$.

The phrasing "equal intercepts" usually implies equal positive intercepts, but it's good practice to consider both cases if not specified.

The equations of the lines are $x + y - 5 = 0$ and $x - y + 1 = 0$.

Let the line $3x + 4y - 12 = 0$ divide the line segment joining A(-1, 3) and B(2, -5) in the ratio $k:1$.

Let the point of division be P. Using the section formula, the coordinates of P are:

P = $\left( \frac{k \cdot x_2 + 1 \cdot x_1}{k+1}, \frac{k \cdot y_2 + 1 \cdot y_1}{k+1} \right)$

Here, $(x_1, y_1) = (-1, 3)$ and $(x_2, y_2) = (2, -5)$.

So, the coordinates of P are:

Since this point P lies on the line $3x + 4y - 12 = 0$, its coordinates must satisfy the equation of the line.

Substitute the coordinates of P into the line equation:

To eliminate the denominator, multiply the entire equation by $(k+1)$:

Expand and simplify the equation:

Combine like terms:

Solve for $k$:

The ratio is $k:1$, which is $-\frac{3}{26}:1$. To express this ratio with positive integers, we can multiply both parts by 26:

The negative sign in the ratio indicates that the point of division lies outside the line segment AB, meaning the line divides the segment externally.

The line divides the line segment joining A(-1, 3) and B(2, -5) externally in the ratio 3 : 26.

The given circle equation is $x^2 + y^2 - 4x + 6y - 12 = 0$.

To find the center of this circle, we convert it to the standard form $(x-h)^2 + (y-k)^2 = r^2$.

Group the terms:

Complete the square for x and y terms:

Rewrite in standard form:

From this standard form, the center of the given circle is $(h, k) = (2, -3)$.

The required circle is concentric with this circle, which means it has the same center. So, the center of the required circle is also $(2, -3)$.

Let the equation of the required circle be $(x - h)^2 + (y - k)^2 = R^2$, where $(h, k) = (2, -3)$ and $R$ is the radius of the new circle.

The equation becomes:

The required circle passes through the point (-2, 1). We can substitute these coordinates into the equation (v) to find $R^2$.

Now, substitute the value of $R^2$ back into equation (v) to get the final equation of the circle:

Expanding this equation gives:

The equation of the circle is $(x - 2)^2 + (y + 3)^2 = 32$ or $x^2 + y^2 - 4x + 6y - 19 = 0$.

The given equation of the parabola is $y^2 - 4y - 4x + 8 = 0$.

To find the vertex, we need to rewrite this equation in the standard form of a parabola. Since the $y$ term is squared, the parabola's axis of symmetry is horizontal. The standard form for such a parabola is $(y-k)^2 = 4a(x-h)$, where $(h, k)$ is the vertex.

Let's rearrange the given equation to isolate the y-terms and the x-terms:

Now, complete the square for the y-terms. To do this, we take half of the coefficient of the y-term and square it. The coefficient of the y-term is -4. Half of -4 is -2, and $(-2)^2 = 4$.

Add 4 to both sides of the equation:

Rewrite the left side as a squared term and simplify the right side:

Now, factor out the coefficient of $x$ from the right side:

This equation is now in the standard form $(y-k)^2 = 4a(x-h)$.

Comparing $(y - 2)^2 = 4(x - 1)$ with $(y-k)^2 = 4a(x-h)$, we can identify the vertex $(h, k)$.

From $(y - 2)^2$, we have $k = 2$.

From $(x - 1)$, we have $h = 1$.

Therefore, the vertex of the parabola is $(h, k) = (1, 2)$.

The vertex of the parabola $y^2 - 4y - 4x + 8 = 0$ is (1, 2).

Let the vertices of the rhombus be A = (3, 0), B = (4, 5), C = (-1, 4), and D = (-2, -1).

The area of a rhombus can be found using the lengths of its diagonals. The formula for the area of a rhombus is $\frac{1}{2} d_1 d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.

The diagonals of the rhombus are AC and BD.

Calculate the length of diagonal AC:

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

For diagonal AC, let $(x_1, y_1) = A(3, 0)$ and $(x_2, y_2) = C(-1, 4)$.

We can simplify $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$.

Calculate the length of diagonal BD:

For diagonal BD, let $(x_1, y_1) = B(4, 5)$ and $(x_2, y_2) = D(-2, -1)$.

We can simplify $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$.

Calculate the area of the rhombus:

Area = $\frac{1}{2} \times d_{AC} \times d_{BD}$

The area of the rhombus is 24 square units.

The equation of a line can be found using the point-slope form: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is its slope.

We are given that the line passes through the point (1, 2). So, $(x_1, y_1) = (1, 2)$.

The line makes an angle of $120^\circ$ with the positive x-axis. The slope $m$ of a line is equal to the tangent of the angle it makes with the positive x-axis.

Therefore, $m = \tan(120^\circ)$.

We know that $\tan(120^\circ) = \tan(180^\circ - 60^\circ) = -\tan(60^\circ)$.

Now, substitute the point $(x_1, y_1) = (1, 2)$ and the slope $m = -\sqrt{3}$ into the point-slope form of the equation:

To express the equation in a standard form (like $Ax + By + C = 0$), we can rearrange it:

Move all terms to one side:

The equation of the line is $\sqrt{3}x + y - (2 + \sqrt{3}) = 0$.

Let the equation of the circle be in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

The circle passes through the origin (0, 0). Substituting these coordinates into the general equation:

This simplifies to $c = 0$.

So, the equation of the circle becomes $x^2 + y^2 + 2gx + 2fy = 0$.

The circle cuts off intercepts 3 and 4 from the positive axes. This means the points where the circle intersects the x-axis are (3, 0) and the point where it intersects the y-axis is (0, 4). Since it cuts off intercepts from the *positive* axes, and it passes through the origin, these are the only points on the axes that the circle intersects.

Since the circle passes through (3, 0), substitute these coordinates into the equation:

Simplifying equation (ii):

Solving for $g$:

Since the circle passes through (0, 4), substitute these coordinates into the equation:

Simplifying equation (vi):

Solving for $f$:

Now substitute the values of $g = -\frac{3}{2}$, $f = -2$, and $c = 0$ back into the general equation of the circle:

Simplifying:

Alternative Method (using intercept form properties):

A circle cutting off intercepts $p$ and $q$ from the positive axes and passing through the origin has the equation $x^2 + y^2 - px - qy = 0$.

Here, $p = 3$ (x-intercept) and $q = 4$ (y-intercept).

Substituting these values, we directly get:

The equation of the circle is $x^2 + y^2 - 3x - 4y = 0$.

The vertex of the parabola is at the origin, which is (0, 0).

The directrix of the parabola is given as $x = -5$.

For a parabola with vertex at the origin (0, 0), the standard form of the equation depends on whether the axis of symmetry is vertical or horizontal, and the direction it opens.

The directrix $x = -5$ is a vertical line. A parabola with a vertical directrix has a horizontal axis of symmetry.

The standard forms for a parabola with vertex at the origin are:

• If the axis of symmetry is the x-axis: $y^2 = 4ax$
• If the axis of symmetry is the y-axis: $x^2 = 4ay$

Since the directrix is $x = -5$ (a vertical line), the axis of symmetry is the x-axis (a horizontal line). Therefore, the form of the parabola's equation is $y^2 = 4ax$.

The directrix for $y^2 = 4ax$ is $x = -a$.

We are given that the directrix is $x = -5$.

Comparing the given directrix with the standard form, we have:

Equating (i) and (ii):

Solving for $a$:

Since the directrix is $x = -5$ and the vertex is at the origin, the parabola must open to the right (because the directrix is to the left of the vertex). This means $a$ must be positive, which is consistent with our finding $a=5$.

Now substitute the value of $a$ into the standard equation $y^2 = 4ax$:

The equation of the parabola is $y^2 = 20x$.

First, we need to find the slope of the line joining the points (3, 1) and (7, 4).

The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Let $(x_1, y_1) = (3, 1)$ and $(x_2, y_2) = (7, 4)$.

The slope of the line joining these points ($m_1$) is:

We need to find the equation of a line that is perpendicular to this line. If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the required perpendicular line be $m_2$. Then:

Substituting the value of $m_1$ from (ii):

Solving for $m_2$:

Now we have the slope of the required line ($m_2 = -\frac{4}{3}$) and a point it passes through (2, 2).

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

Multiply both sides by 3 to eliminate the fraction:

Expand both sides:

Rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

The equation of the line is $4x + 3y - 14 = 0$.

Given: The points A(1, 2), B(2, -3), C(-1, -5), and D(-2, 4).

To Prove: The points A, B, C, and D are the vertices of a parallelogram.

To Determine: If the parallelogram is a rectangle.

Solution:

To prove that the given points are the vertices of a parallelogram, we can show that the opposite sides are parallel. This can be done by calculating the slopes of the opposite sides.

1. Slope of AB:

2. Slope of BC:

3. Slope of CD:

4. Slope of DA:

Now, let's compare the slopes of opposite sides:

Slope of AB ($m_{AB}$) = -5

Slope of CD ($m_{CD}$) = -9

Since $m_{AB} \neq m_{CD}$, AB is not parallel to CD.

Let's recheck the coordinates or my understanding of opposite sides for a parallelogram ABCD. The opposite sides are AB and CD, and BC and DA.

Wait, I might have made a mistake in assuming the order of vertices. Let's check the slopes of AC and BD (diagonals) and also re-examine the order of vertices.

Let's consider the sides as AB and DC, and AD and BC.

1. Slope of AB:

2. Slope of DC:

Here also $m_{AB} \neq m_{DC}$. This means the order of vertices might not be ABCD in sequence.

Let's try the order AC and BD as sides, which is incorrect for a parallelogram. The correct way is to check if the midpoints of the diagonals coincide.

Midpoint of Diagonal AC:

Midpoint of Diagonal BD:

Since the midpoints of the diagonals AC and BD do not coincide ($ (0, -\frac{3}{2}) \neq (0, \frac{1}{2}) $), the given points do not form a parallelogram in the order ABCD or ADCB.

Let's re-evaluate the question and my interpretation. The question states "vertices of a parallelogram", implying that the order might not be consecutive.

Let's test the order ABDC.

Opposite sides would be AB and DC, and AD and BC.

We already calculated $m_{AB} = -5$ and $m_{DC} = -9$. Since they are not equal, ABDC is not a parallelogram.

Let's test the order ACBD.

Opposite sides would be AC and BD, and AB and CD.

Slope of AC ($m_{AC}$):

Slope of BD ($m_{BD}$):

Since $m_{AC} \neq m_{BD}$, ACBD is not a parallelogram.

Let's test the order ADBC.

Opposite sides would be AD and CB, and AC and DB.

Slope of AD ($m_{AD}$):

Slope of CB ($m_{CB}$):

Since $m_{AD} \neq m_{CB}$, ADBC is not a parallelogram.

There seems to be an error in my calculation or understanding of the problem statement. Let me re-calculate all slopes carefully.

A(1, 2), B(2, -3), C(-1, -5), D(-2, 4)

Slope of AB: $m_{AB} = \frac{-3 - 2}{2 - 1} = \frac{-5}{1} = -5$

Slope of BC: $m_{BC} = \frac{-5 - (-3)}{-1 - 2} = \frac{-2}{-3} = \frac{2}{3}$

Slope of CD: $m_{CD} = \frac{4 - (-5)}{-2 - (-1)} = \frac{9}{-1} = -9$

Slope of DA: $m_{DA} = \frac{2 - 4}{1 - (-2)} = \frac{-2}{3}$

Slope of AC: $m_{AC} = \frac{-5 - 2}{-1 - 1} = \frac{-7}{-2} = \frac{7}{2}$

Slope of BD: $m_{BD} = \frac{4 - (-3)}{-2 - 2} = \frac{7}{-4} = -\frac{7}{4}$

Let's reconsider the possibility of the question implying that *some* ordering of these points forms a parallelogram.

For ABCD to be a parallelogram, AB must be parallel to DC, and BC must be parallel to AD.

$m_{AB} = -5$. $m_{DC} = \frac{-5 - 4}{-1 - (-2)} = \frac{-9}{1} = -9$. $m_{AB} \neq m_{DC}$.

$m_{BC} = \frac{2}{3}$. $m_{AD} = \frac{4 - 2}{-2 - 1} = \frac{2}{-3} = -\frac{2}{3}$. $m_{BC} \neq m_{AD}$.

Let's check the lengths of the sides as well, as another property of a parallelogram is that opposite sides are equal in length.

Length of AB:

Length of BC:

Length of CD:

Length of DA:

From the lengths, we see that $BC = DA = \sqrt{13}$.

However, $AB = \sqrt{26}$ and $CD = \sqrt{82}$. Since $AB \neq CD$, these points do not form a parallelogram with sides AB and CD.

Let's re-check the slopes calculation one more time. It's possible I made a calculation error.

A(1, 2), B(2, -3), C(-1, -5), D(-2, 4)

Slope of AB: $m_{AB} = \frac{-3 - 2}{2 - 1} = \frac{-5}{1} = -5$

Slope of DC: $m_{DC} = \frac{-5 - 4}{-1 - (-2)} = \frac{-9}{-1+2} = \frac{-9}{1} = -9$

Slope of BC: $m_{BC} = \frac{-5 - (-3)}{-1 - 2} = \frac{-2}{-3} = \frac{2}{3}$

Slope of AD: $m_{AD} = \frac{4 - 2}{-2 - 1} = \frac{2}{-3} = -\frac{2}{3}$

It seems there's a misstatement in the problem or my approach. Let me try using the property that diagonals bisect each other. We already calculated the midpoints.

Midpoint of AC = $(0, -\frac{3}{2})$

Midpoint of BD = $(0, \frac{1}{2})$

Since the midpoints are not the same, ABCD is not a parallelogram.

Let's consider another ordering of points, for instance, ACBD.

Midpoint of AB: $M_{AB} = \left(\frac{1+2}{2}, \frac{2+(-3)}{2}\right) = \left(\frac{3}{2}, -\frac{1}{2}\right)$

Midpoint of CD: $M_{CD} = \left(\frac{-1+(-2)}{2}, \frac{-5+4}{2}\right) = \left(-\frac{3}{2}, -\frac{1}{2}\right)$

Midpoints of AB and CD do not coincide. So ACBD is not a parallelogram.

Let's consider ADBC.

Midpoint of AC: $M_{AC} = (0, -\frac{3}{2})$

Midpoint of DB: $M_{DB} = \left(\frac{-2+2}{2}, \frac{4+(-3)}{2}\right) = \left(\frac{0}{2}, \frac{1}{2}\right) = (0, \frac{1}{2})$

These are the same midpoints as for ABCD diagonals. My previous midpoint calculation was correct.

Let's check the problem again. It's possible there's a typo in the coordinates provided in the question.

However, if we assume the question is correct and there must be a parallelogram, let's re-examine slopes. My slope calculations for AB and AD seem correct. Let me recalculate the slopes for DC and BC.

A(1, 2), B(2, -3), C(-1, -5), D(-2, 4)

Slope of AB: $m_{AB} = \frac{-3 - 2}{2 - 1} = -5$

Slope of AD: $m_{AD} = \frac{4 - 2}{-2 - 1} = \frac{2}{-3} = -\frac{2}{3}$

For ADBC to be a parallelogram, AD must be parallel to CB, and AC must be parallel to DB.

Slope of CB: $m_{CB} = \frac{-3 - (-5)}{2 - (-1)} = \frac{2}{3}$

Slope of AC: $m_{AC} = \frac{-5 - 2}{-1 - 1} = \frac{-7}{-2} = \frac{7}{2}$

Slope of DB: $m_{DB} = \frac{-3 - 4}{2 - (-2)} = \frac{-7}{4} = -\frac{7}{4}$

It appears that with the given coordinates, these points do not form a parallelogram in any standard sequential order.

Let's assume there might be a typo in the question and proceed with a hypothetical case where it *is* a parallelogram, to show how to check for a rectangle.

Hypothetical Scenario: If ABCD were a parallelogram

To check if it's a rectangle, we need to see if adjacent sides are perpendicular. This means the product of their slopes should be -1.

Let's check the slopes we calculated:

$m_{AB} = -5$

$m_{BC} = \frac{2}{3}$

Product of slopes of adjacent sides AB and BC:

Since $m_{AB} \times m_{BC} \neq -1$, the sides AB and BC are not perpendicular. Therefore, even if it were a parallelogram, it would not be a rectangle based on these slopes.

Justification for not being a rectangle (based on the slope product rule):

For a parallelogram to be a rectangle, its adjacent sides must be perpendicular. This condition is met if the product of the slopes of adjacent sides is -1.

Let's consider the product of slopes of AB and BC:

As $-\frac{10}{3} \neq -1$, the sides AB and BC are not perpendicular. Therefore, the parallelogram (if it were one) would not be a rectangle.

Conclusion based on the given points:

Upon recalculating slopes and midpoints, it appears that the given points A(1, 2), B(2, -3), C(-1, -5), and D(-2, 4) do not form a parallelogram in any sequential order. This suggests a potential error in the coordinates provided in the question.

However, if we were to assume it were a parallelogram, the check for a rectangle would involve verifying if adjacent sides are perpendicular. Based on the calculated slopes of adjacent sides AB and BC ($m_{AB} = -5$ and $m_{BC} = \frac{2}{3}$), their product is $-\frac{10}{3}$, which is not -1. Thus, it would not be a rectangle.

Given: The midpoints of the sides of a triangle are P(1, 5), Q(2, 3), and R(3, 4).

To Find: The coordinates of the vertices of the triangle.

Solution:

Let the vertices of the triangle be A($x_1$, $y_1$), B($x_2$, $y_2$), and C($x_3$, $y_3$).

Let P(1, 5) be the midpoint of side BC.

Let Q(2, 3) be the midpoint of side AC.

Let R(3, 4) be the midpoint of side AB.

Using the midpoint formula, which states that the midpoint of a segment with endpoints ($x_a$, $y_a$) and ($x_b$, $y_b$) is $\left(\frac{x_a + x_b}{2}, \frac{y_a + y_b}{2}\right)$, we can set up the following equations:

1. Midpoint P(1, 5) of BC:

2. Midpoint Q(2, 3) of AC:

3. Midpoint R(3, 4) of AB:

Now we have a system of linear equations:

For x-coordinates:

$x_2 + x_3 = 2$ … (ii)

$x_1 + x_3 = 4$ … (vi)

$x_1 + x_2 = 6$ … (x)

Add equations (ii), (vi), and (x):

Subtract equation (ii) from (xv):

Subtract equation (vi) from (xv):

Subtract equation (x) from (xv):

Similarly, for y-coordinates:

$y_2 + y_3 = 10$ … (iv)

$y_1 + y_3 = 6$ … (viii)

$y_1 + y_2 = 8$ … (xii)

Add equations (iv), (viii), and (xii):

Subtract equation (iv) from (xxi):

Subtract equation (viii) from (xxi):

Subtract equation (xii) from (xxi):

Therefore, the coordinates of the vertices of the triangle are:

Vertex A ($x_1$, $y_1$) = (4, 2)

Vertex B ($x_2$, $y_2$) = (2, 6)

Vertex C ($x_3$, $y_3$) = (0, 4)

Verification:

Midpoint of BC = $\left(\frac{2+0}{2}, \frac{6+4}{2}\right) = \left(\frac{2}{2}, \frac{10}{2}\right) = (1, 5)$, which matches P.

Midpoint of AC = $\left(\frac{4+0}{2}, \frac{2+4}{2}\right) = \left(\frac{4}{2}, \frac{6}{2}\right) = (2, 3)$, which matches Q.

Midpoint of AB = $\left(\frac{4+2}{2}, \frac{2+6}{2}\right) = \left(\frac{6}{2}, \frac{8}{2}\right) = (3, 4)$, which matches R.

The calculated vertices are correct.

Given:

1. The straight line passes through the point (3, 4).

2. The sum of the intercepts on the axes is 14.

To Find: The equation of the straight line.

Solution:

The equation of a straight line in intercept form is given by:

where 'a' is the x-intercept and 'b' is the y-intercept.

From the given information, the sum of the intercepts is 14:

From equation (ii), we can express b in terms of a:

The line passes through the point (3, 4). Substituting x=3 and y=4 into the intercept form of the equation (i):

Now, substitute the expression for b from equation (iii) into equation (iv):

To solve for 'a', we find a common denominator:

Cross-multiply:

Rearrange the terms to form a quadratic equation:

Factor the quadratic equation:

This gives two possible values for 'a':

Case 1: $a - 6 = 0 \implies a = 6$

Case 2: $a - 7 = 0 \implies a = 7$

Now, we find the corresponding values of 'b' using equation (iii) ($b = 14 - a$):

Case 1: If $a = 6$

$b = 14 - 6 = 8$

The equation of the line is $\frac{x}{6} + \frac{y}{8} = 1$.

Multiplying by the LCM of 6 and 8 (which is 24) to clear the denominators:

Case 2: If $a = 7$

$b = 14 - 7 = 7$

The equation of the line is $\frac{x}{7} + \frac{y}{7} = 1$.

Multiplying by 7 to clear the denominators:

We can verify that both lines pass through (3, 4):

For $4x + 3y = 24$: $4(3) + 3(4) = 12 + 12 = 24$. This is correct.

For $x + y = 7$: $3 + 4 = 7$. This is also correct.

Therefore, there are two possible equations for the straight line.

The equations of the straight lines are:

$4x + 3y = 24$

$x + y = 7$

Given:

1. The lines pass through the point (2, 3).

2. The angle between the required lines and the line $x - 2y + 5 = 0$ is $45^\circ$.

To Find: The equations of the two lines.

Solution:

The equation of the given line is $x - 2y + 5 = 0$.

We can rewrite this in the slope-intercept form ($y = mx + c$) to find its slope.

The slope of the given line ($m_1$) is $\frac{1}{2}$.

Let the slope of the required lines be $m$. The angle ($\theta$) between two lines with slopes $m_1$ and $m$ is given by the formula:

We are given that $\theta = 45^\circ$ and $m_1 = \frac{1}{2}$. We know that $\tan 45^\circ = 1$.

Substituting these values into the formula:

This gives us two possibilities:

Possibility 1: $\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = 1$

Multiply by 2 to clear fractions:

Rearrange to solve for m:

Possibility 2: $\frac{m - \frac{1}{2}}{1 + \frac{m}{2}} = -1$

Multiply by 2 to clear fractions:

Rearrange to solve for m:

So, the slopes of the two required lines are $m=3$ and $m=-\frac{1}{3}$.

Now we find the equations of these lines using the point-slope form of a line, $y - y_1 = m(x - x_1)$, where ($x_1, y_1$) = (2, 3).

Equation of the first line (with $m = 3$):

Equation of the second line (with $m = -\frac{1}{3}$):

Multiply by 3:

The equations of the lines are:

$3x - y - 3 = 0$

$x + 3y - 11 = 0$

Given: Two parallel lines

Line 1: $3x - 4y + 5 = 0$

Line 2: $6x - 8y + 15 = 0$

To Find: The distance between these two lines.

Solution:

First, we need to ensure that both lines are in the same form, i.e., the coefficients of x and y are the same (or opposite). We can rewrite the second line by dividing it by 2.

Line 1: $3x - 4y + 5 = 0$

Line 2: $\frac{6x - 8y + 15}{2} = \frac{0}{2}$

Now we have two parallel lines in the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, where:

For Line 1: $A = 3$, $B = -4$, $C_1 = 5$

For Line 2 (rewritten): $A = 3$, $B = -4$, $C_2 = \frac{15}{2}$

The formula for the distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is:

Substitute the values of A, B, $C_1$, and $C_2$ into the formula:

Calculate the numerator:

Calculate the denominator:

Now, calculate the distance:

The distance between the lines is $\frac{1}{2}$.

Given: The circle passes through the points A(2, -2), B(3, 4), and C(-1, -4).

To Find: The equation of the circle.

Solution:

The general equation of a circle is given by:

Since the circle passes through the given points, these points must satisfy the equation of the circle.

1. Substituting point A (2, -2):

2. Substituting point B (3, 4):

3. Substituting point C (-1, -4):

Now we have a system of three linear equations with three unknowns (g, f, and c):

1) $4g - 4f + c + 8 = 0$ (from iv)

2) $6g + 8f + c + 25 = 0$ (from vii)

3) $-2g - 8f + c + 17 = 0$ (from x)

Subtract equation (1) from equation (2):

Subtract equation (3) from equation (2):

Divide equation (xiv) by 8:

Now we have a system of two equations with two unknowns (g and f):

1') $2g + 12f + 17 = 0$ (from xii)

2') $g + 2f + 1 = 0$ (from xv)

From equation (2'), we can express g as $g = -2f - 1$. Substitute this into equation (1'):

Now substitute the value of f back into the expression for g:

Now substitute the values of g and f into equation (iv) to find c:

$4g - 4f + c + 8 = 0$

Now substitute the values of g, f, and c back into the general equation of the circle (i):

$x^2 + y^2 + 2gx + 2fy + c = 0$

To remove fractions, multiply the entire equation by 4:

The equation of the circle is $4x^2 + 4y^2 + 22x - 15y - 106 = 0$.

Given:

1. The center of the circle lies on the line $x + 2y = 4$.

2. The circle passes through the points A(1, 2) and B(4, -1).

To Find: The equation of the circle.

Solution:

Let the center of the circle be $(h, k)$. Since the center lies on the line $x + 2y = 4$, it must satisfy this equation:

The general equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Since the circle passes through points A(1, 2) and B(4, -1), the distance from the center $(h, k)$ to each of these points is equal to the radius ($r$).

1. Distance from center (h, k) to A(1, 2):

2. Distance from center (h, k) to B(4, -1):

Since both expressions are equal to $r^2$, we can equate them:

Expand both sides:

Cancel out $h^2$, $k^2$, and simplify:

Rearrange the terms to form a linear equation in h and k:

Divide by 6:

Now we have a system of two linear equations with two unknowns (h and k):

1) $h + 2k = 4$ (from i)

2) $h - k = 2$ (from ix)

Subtract equation (2) from equation (1):

Substitute the value of k back into equation (ix) to find h:

So, the center of the circle is $(h, k) = (\frac{8}{3}, \frac{2}{3})$.

Now, we need to find the radius squared ($r^2$) by substituting the center coordinates into either equation (ii) or (iii). Let's use equation (ii):

Now, substitute the center $(h, k) = (\frac{8}{3}, \frac{2}{3})$ and $r^2 = \frac{41}{9}$ into the standard equation of a circle:

$(x - h)^2 + (y - k)^2 = r^2$

Expand the equation:

Multiply by 3 to clear the fractions:

The equation of the circle is $3x^2 + 3y^2 - 16x - 4y + 9 = 0$.

Given:

1. Focus of the parabola: F(2, 0)

2. Directrix of the parabola: $x = -2$

To Find:

a) The equation of the parabola.

b) The length of the latus rectum.

c) The equation of the axis of the parabola.

Solution:

A parabola is defined as the set of all points that are equidistant from the focus and the directrix.

Let P(x, y) be any point on the parabola.

The distance from P to the focus F(2, 0) is PF.

The distance from P to the directrix $x = -2$ is the perpendicular distance from P to the line $x = -2$. This distance is $|x - (-2)| = |x + 2|$.

By the definition of a parabola, PF = Distance from P to the directrix.

Using the distance formula for PF:

Equating PF and the distance to the directrix:

Squaring both sides to eliminate the square root:

Expand the squared terms:

Cancel out $x^2$ and 4 from both sides:

Rearrange the terms to get the equation of the parabola:

This is the equation of the parabola.

a) Equation of the parabola: $y^2 = 8x$

To find the length of the latus rectum and the equation of the axis, we can compare the obtained equation with the standard forms.

The standard form of a parabola with vertex at the origin and opening to the right is $y^2 = 4ax$.

Comparing $y^2 = 8x$ with $y^2 = 4ax$, we get $4a = 8$, which means $a = 2$.

The focus of such a parabola is at $(a, 0)$, and the directrix is $x = -a$.

Given focus is (2, 0) and directrix is $x = -2$. This matches the standard form with $a = 2$, and the vertex is at the origin (0, 0).

b) Length of the latus rectum:

The length of the latus rectum for a parabola of the form $y^2 = 4ax$ is $4a$.

c) Equation of the axis:

The axis of symmetry for a parabola of the form $y^2 = 4ax$ is the x-axis, which has the equation $y = 0$.

Summary:

Equation of the parabola: $y^2 = 8x$

Length of the latus rectum: 8 units

Equation of the axis: $y = 0$

Given:

1. Vertex of the parabola is at (0, 0).

2. The parabola passes through the point (5, 2).

3. The parabola is symmetric with respect to the y-axis.

To Find:

a) The equation of the parabola.

b) The coordinates of its focus.

c) The equation of the directrix.

Solution:

Since the vertex is at the origin (0, 0) and the parabola is symmetric with respect to the y-axis, its standard equation is of the form:

where 'a' is a constant that determines the position of the focus and the directrix.

The parabola passes through the point (5, 2). We can substitute these coordinates into the standard equation to find the value of 'a'.

Substitute $x = 5$ and $y = 2$ into equation (i):

Solve for 'a':

a) Equation of the parabola:

Substitute the value of 'a' back into the standard equation (i):

b) Coordinates of the focus:

For a parabola of the form $x^2 = 4ay$, the focus is located at $(0, a)$.

Using the value $a = \frac{25}{8}$, the coordinates of the focus are:

c) Equation of the directrix:

For a parabola of the form $x^2 = 4ay$, the equation of the directrix is $y = -a$.

Using the value $a = \frac{25}{8}$, the equation of the directrix is:

Summary:

Equation of the parabola: $x^2 = \frac{25}{2} y$

Coordinates of the focus: $(0, \frac{25}{8})$

Equation of the directrix: $y = -\frac{25}{8}$

Given: Vertices of a triangle ABC are A(2, 1), B(-2, 3), and C(4, -1).

To Find: The equation of the median from vertex A.

Solution:

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

The median from vertex A will connect vertex A to the midpoint of the side BC.

First, find the midpoint of the side BC. Let the midpoint be M.

The coordinates of B are ($x_B$, $y_B$) = (-2, 3).

The coordinates of C are ($x_C$, $y_C$) = (4, -1).

The midpoint formula is $M = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right)$.

So, the midpoint of BC is M(1, 1).

Now, we need to find the equation of the line passing through vertex A(2, 1) and the midpoint M(1, 1).

We can use the two-point form of the equation of a straight line: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$.

Let $(x_1, y_1) = A(2, 1)$ and $(x_2, y_2) = M(1, 1)$.

The slope of the median AM is:

Since the slope is 0, the median AM is a horizontal line.

The equation of a horizontal line passing through a point $(x_1, y_1)$ is $y = y_1$.

Here, the point is A(2, 1) (or M(1, 1)), and the y-coordinate is 1.

Therefore, the equation of the median from vertex A is $y = 1$.

Alternatively, using the point-slope form with point A(2, 1) and slope 0:

The equation of the median from vertex A is $y = 1$.

Given:

1. The point P is (-1, 3).

2. The line L is $3x - 4y - 16 = 0$.

To Find:

a) The coordinates of the foot of the perpendicular from P to L.

b) The perpendicular distance from P to L.

Solution:

Let the given line be $Ax + By + C = 0$, so $A=3$, $B=-4$, $C=-16$.

Let the given point be P($x_1$, $y_1$), so $x_1=-1$, $y_1=3$.

a) Coordinates of the foot of the perpendicular:

Let the foot of the perpendicular from P to the line L be Q($x_2$, $y_2$).

The formula for the coordinates of the foot of the perpendicular is:

First, calculate the value of the term $-\frac{Ax_1 + By_1 + C}{A^2 + B^2}$:

$Ax_1 + By_1 + C = 3(-1) - 4(3) - 16 = -3 - 12 - 16 = -31$.

$A^2 + B^2 = 3^2 + (-4)^2 = 9 + 16 = 25$.

So, $-\frac{Ax_1 + By_1 + C}{A^2 + B^2} = -\frac{-31}{25} = \frac{31}{25}$.

Now, use the formula from (i) to find $x_2$ and $y_2$:

For $x_2$:

For $y_2$:

So, the coordinates of the foot of the perpendicular are $(\frac{68}{25}, -\frac{49}{25})$.

b) Perpendicular distance:

The perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by the formula:

We have already calculated the numerator and denominator in part (a):

$Ax_1 + By_1 + C = -31$

$A^2 + B^2 = 25$

Substituting these values:

The coordinates of the foot of the perpendicular are $(\frac{68}{25}, -\frac{49}{25})$.

The perpendicular distance is $\frac{31}{5}$.

To show that three lines are concurrent, we need to find the point of intersection of any two lines and then check if this point lies on the third line.

Given lines:

Line 1: $x - 2y + 3 = 0$ …(i)

Line 2: $2x - 3y + 4 = 0$ …(ii)

Line 3: $3x - 4y + 5 = 0$ …(iii)

To find the point of concurrency.

Solution:

First, let's find the point of intersection of Line 1 and Line 2. We can use the method of substitution or elimination.

Method of Elimination:

Multiply equation (i) by 2:

This gives:

Now, subtract equation (ii) from equation (iv):

$-4y + 3y + 6 - 4 = 0$

$ -y + 2 = 0$

Substitute the value of $y = 2$ into equation (i):

$x - 2(2) + 3 = 0$

$x - 4 + 3 = 0$

$x - 1 = 0$

So, the point of intersection of Line 1 and Line 2 is (1, 2).

Now, we need to check if this point (1, 2) lies on Line 3 ($3x - 4y + 5 = 0$). Substitute $x = 1$ and $y = 2$ into equation (iii):

$3(1) - 4(2) + 5$

$3 - 8 + 5$

$-5 + 5 = 0$

Since substituting the point (1, 2) into the equation of Line 3 results in 0, the point (1, 2) lies on Line 3.

Conclusion:

Since the point of intersection of the first two lines also lies on the third line, the three given lines are concurrent.

The point of concurrency is (1, 2).

To show that three lines are concurrent, we need to find the point of intersection of any two lines and then check if this point lies on the third line.

Given lines:

Line 1: $x - 2y + 3 = 0$ …(i)

Line 2: $2x - 3y + 4 = 0$ …(ii)

Line 3: $3x - 4y + 5 = 0$ …(iii)

To find the point of concurrency.

Solution:

First, let's find the point of intersection of Line 1 and Line 2. We can use the method of substitution or elimination.

Method of Elimination:

Multiply equation (i) by 2:

This gives:

Now, subtract equation (ii) from equation (iv):

$-4y + 3y + 6 - 4 = 0$

$ -y + 2 = 0$

Substitute the value of $y = 2$ into equation (i):

$x - 2(2) + 3 = 0$

$x - 4 + 3 = 0$

$x - 1 = 0$

So, the point of intersection of Line 1 and Line 2 is (1, 2).

Now, we need to check if this point (1, 2) lies on Line 3 ($3x - 4y + 5 = 0$). Substitute $x = 1$ and $y = 2$ into equation (iii):

$3(1) - 4(2) + 5$

$3 - 8 + 5$

$-5 + 5 = 0$

Since substituting the point (1, 2) into the equation of Line 3 results in 0, the point (1, 2) lies on Line 3.

Conclusion:

Since the point of intersection of the first two lines also lies on the third line, the three given lines are concurrent.

The point of concurrency is (1, 2).

Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

Given:

1. The circle passes through the points (4, 1) and (6, 5).

2. The center $(h, k)$ of the circle lies on the line $4x + y = 16$.

To find: The equation of the circle.

Solution:

Since the points (4, 1) and (6, 5) lie on the circle, they must satisfy the equation of the circle. Let the center of the circle be $(h, k)$. The distance from the center to any point on the circle is the radius ($r$). Therefore, the distance from $(h, k)$ to (4, 1) is equal to the distance from $(h, k)$ to (6, 5).

Distance from $(h, k)$ to (4, 1): $r^2 = (h-4)^2 + (k-1)^2$ …(i)

Distance from $(h, k)$ to (6, 5): $r^2 = (h-6)^2 + (k-5)^2$ …(ii)

Equating (i) and (ii):

Expanding both sides:

$h^2 - 8h + 16 + k^2 - 2k + 1 = h^2 - 12h + 36 + k^2 - 10k + 25$

$-8h - 2k + 17 = -12h - 10k + 61$

Rearranging the terms to group $h$ and $k$:

$-8h + 12h - 2k + 10k = 61 - 17$

$4h + 8k = 44$

Dividing by 4:

We are also given that the center $(h, k)$ lies on the line $4x + y = 16$. So, $(h, k)$ must satisfy this equation:

Now we have a system of two linear equations with two variables $h$ and $k$:

1) $h + 2k = 11$

2) $4h + k = 16$

From equation (iv), we can express $k$ in terms of $h$: $k = 16 - 4h$.

Substitute this expression for $k$ into equation (iii):

$h + 2(16 - 4h) = 11$

$h + 32 - 8h = 11$

$-7h = 11 - 32$

$-7h = -21$

Now, substitute the value of $h = 3$ back into the expression for $k$:

$k = 16 - 4(3)$

$k = 16 - 12$

So, the center of the circle is $(h, k) = (3, 4)$.

Now, we need to find the radius ($r$). We can use either equation (i) or (ii). Let's use equation (i) with the center (3, 4) and the point (4, 1):

$r^2 = (3-4)^2 + (4-1)^2$

$r^2 = (-1)^2 + (3)^2$

$r^2 = 1 + 9$

The equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values of $h$, $k$, and $r^2$:

$(x-3)^2 + (y-4)^2 = 10$

Expanding this equation:

$x^2 - 6x + 9 + y^2 - 8y + 16 = 10$

$x^2 + y^2 - 6x - 8y + 25 - 10 = 0$

The equation of the circle is $\boldsymbol{(x-3)^2 + (y-4)^2 = 10}$ or $\boldsymbol{x^2 + y^2 - 6x - 8y + 15 = 0}$.

To find the vertex, focus, directrix, and length of the latus rectum of the parabola, we need to convert the given equation into its standard form.

Given equation of the parabola:

$y^2 + 4y + 4x + 8 = 0$

To find:

1. Vertex

2. Focus

3. Directrix

4. Length of the latus rectum

Solution:

We will rearrange the given equation to complete the square for the terms involving $y$.

$y^2 + 4y = -4x - 8$

To complete the square for $y^2 + 4y$, we need to add $(\frac{4}{2})^2 = 2^2 = 4$ to both sides of the equation:

$y^2 + 4y + 4 = -4x - 8 + 4$

$(y+2)^2 = -4x - 4$

$(y+2)^2 = -4(x+1)$

This is the standard form of a parabola that opens to the left: $(y-k)^2 = -4a(x-h)$, where $(h, k)$ is the vertex and $a$ is the distance from the vertex to the focus and from the vertex to the directrix.

Comparing $(y+2)^2 = -4(x+1)$ with $(y-k)^2 = -4a(x-h)$, we get:

$k = -2$

$h = -1$

$-4a = -4 \implies a = 1$

1. Vertex:

The vertex is at $(h, k)$.

Vertex = (-1, -2)

2. Focus:

For a parabola of the form $(y-k)^2 = -4a(x-h)$, the focus is located at $(h-a, k)$.

Focus = $(-1 - 1, -2) = (-2, -2)$

Focus = (-2, -2)

3. Directrix:

For a parabola of the form $(y-k)^2 = -4a(x-h)$, the directrix is the vertical line $x = h-a$.

Directrix: $x = -1 - 1$

Directrix: $x = -2$

4. Length of the latus rectum:

The length of the latus rectum for a parabola in this standard form is $4a$.