Chapter 10 Straight Lines (Additional Questions)

Given:

Two points $A(x_1, y_1)$ and $B(x_2, y_2)$ are given as $A(3, 4)$ and $B(-1, 1)$.

To Find:

The distance between the points $A(3, 4)$ and $B(-1, 1)$.

Solution:

We use the distance formula to find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a Cartesian coordinate system.

The distance $d$ is given by the formula:

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

Here, $x_1 = 3$, $y_1 = 4$, $x_2 = -1$, and $y_2 = 1$.

Substitute the values into the distance formula:

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

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

Calculate the squares:

$(-4)^2 = 16$

$(-3)^2 = 9$

Substitute these values back into the formula:

$d = \sqrt{16 + 9}$

Add the numbers under the square root:

$d = \sqrt{25}$

Calculate the square root:

$d = 5$

The distance between the points $(3, 4)$ and $(-1, 1)$ is 5 units.

Comparing the calculated distance with the given options:

(A) 4

(B) 5

(C) 6

(D) 7

The calculated distance is 5, which matches option (B).

Final Answer:

The distance between the points $(3, 4)$ and $(-1, 1)$ is 5.

The correct option is (B) 5.

Given:

The endpoints of the line segment are $A(x_1, y_1) = (1, -2)$ and $B(x_2, y_2) = (4, 7)$.

The line segment is divided internally in the ratio $m:n = 1:2$.

To Find:

The coordinates $(x, y)$ of the point that divides the line segment AB internally in the ratio $1:2$.

Solution:

We use the section formula for internal division.

If a point $P(x, y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m:n$, then the coordinates of $P$ are given by:

$x = \frac{mx_2 + nx_1}{m+n}$

$y = \frac{my_2 + ny_1}{m+n}$

In this problem, we have:

$x_1 = 1$, $y_1 = -2$

$x_2 = 4$, $y_2 = 7$

$m = 1$, $n = 2$

Substitute these values into the section formula for $x$:

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

$x = \frac{4 + 2}{3}$

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

$x = 2$

Now substitute the values into the section formula for $y$:

$y = \frac{(1)(7) + (2)(-2)}{1+2}$

$y = \frac{7 - 4}{3}$

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

$y = 1$

The coordinates of the point are $(2, 1)$.

Comparing the calculated coordinates with the given options:

(A) $(2, 1)$

(B) $(2, 3)$

(C) $(3, 1)$

(D) $(3, 3)$

The calculated coordinates are $(2, 1)$, which matches option (A).

Final Answer:

The coordinates of the point that divides the line segment joining $(1, -2)$ and $(4, 7)$ internally in the ratio $1:2$ are $(2, 1)$.

The correct option is (A) $(2, 1)$.

Given:

Two points on the line are $A(x_1, y_1) = (2, 3)$ and $B(x_2, y_2) = (4, 7)$.

To Find:

The slope of the line passing through the points $(2, 3)$ and $(4, 7)$.

Solution:

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

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

Substitute the coordinates of the given points into the formula:

$x_1 = 2$, $y_1 = 3$

$x_2 = 4$, $y_2 = 7$

$m = \frac{7 - 3}{4 - 2}$

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

$m = 2$

The slope of the line passing through the points $(2, 3)$ and $(4, 7)$ is 2.

Comparing the calculated slope with the given options:

(A) 1

(B) 2

(C) 3

(D) 4

The calculated slope is 2, which matches option (B).

Final Answer:

The slope of the line is 2.

The correct option is (B) 2.

Given:

Slope of the line, $m = 2$.

Y-intercept of the line, $c = 5$.

To Find:

The equation of the line.

Solution:

The equation of a line with slope $m$ and y-intercept $c$ is given by the slope-intercept form:

$y = mx + c$

Substitute the given values of $m$ and $c$ into this equation.

$m = 2$

$c = 5$

So, the equation of the line is:

$y = (2)x + (5)$

$y = 2x + 5$

This is the required equation of the line.

Comparing the derived equation with the given options:

(A) $y = 2x + 5$

(B) $y = 5x + 2$

(C) $2x + y = 5$ (This can be written as $y = -2x + 5$)

(D) $x + 2y = 5$ (This can be written as $2y = -x + 5 \implies y = -\frac{1}{2}x + \frac{5}{2}$)

The derived equation $y = 2x + 5$ matches option (A).

Final Answer:

The equation of the line is $y = 2x + 5$.

The correct option is (A) $y = 2x + 5$.

Given:

A point on the line $(x_0, y_0)$.

The slope of the line is $m$.

To Find:

The equation of the line.

Solution:

The standard form for the equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is the point-slope form, given by:

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

In this problem, the given point is $(x_0, y_0)$, which corresponds to $(x_1, y_1)$ in the general formula.

Substitute $x_1 = x_0$ and $y_1 = y_0$ into the point-slope formula:

$y - y_0 = m(x - x_0)$

This is the equation of the line passing through $(x_0, y_0)$ with slope $m$.

Comparing the derived equation with the given options:

(A) $y - y_0 = m(x - x_0)$

(B) $y - x_0 = m(x - y_0)$

(C) $x - x_0 = m(y - y_0)$

(D) $x - y_0 = m(y - x_0)$

The derived equation $y - y_0 = m(x - x_0)$ matches option (A).

Final Answer:

The equation of a line passing through $(x_0, y_0)$ with slope $m$ is $y - y_0 = m(x - x_0)$.

The correct option is (A) $y - y_0 = m(x - x_0)$.

Given:

The equation of the line is $3x + 4y = 12$.

To Find:

The slope of the given line.

Solution:

We need to find the slope of the line given by the equation $3x + 4y = 12$.

The equation of a line can be written in the slope-intercept form, which is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

We can rearrange the given equation to put it in this form.

Start with the given equation:

$3x + 4y = 12$

Subtract $3x$ from both sides of the equation to isolate the term with $y$:

$4y = 12 - 3x$

It is usually written with the $x$ term first:

$4y = -3x + 12$

Now, divide both sides of the equation by 4 to solve for $y$:

$y = \frac{-3x + 12}{4}$

Separate the terms on the right side:

$y = \frac{-3}{4}x + \frac{12}{4}$

Simplify the second term:

$y = -\frac{3}{4}x + 3$

This equation is now in the slope-intercept form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

Comparing $y = -\frac{3}{4}x + 3$ with $y = mx + c$, we can see that the slope $m$ is $-\frac{3}{4}$ and the y-intercept $c$ is 3.

Therefore, the slope of the line $3x + 4y = 12$ is $-\frac{3}{4}$.

Comparing the calculated slope with the given options:

(A) $3/4$

(B) $-3/4$

(C) $4/3$

(D) $-4/3$

The calculated slope is $-3/4$, which matches option (B).

Final Answer:

The slope of the line $3x + 4y = 12$ is $-\frac{3}{4}$.

The correct option is (B) $-3/4$.

Given:

Assertion (A): The angle between two parallel lines is $0^\circ$ or $\pi$ radians.

Reason (R): Parallel lines have the same slope.

Evaluation:

Let's evaluate the truthfulness of Assertion (A) and Reason (R).

Assertion (A): Parallel lines are lines that never intersect. If two distinct lines are parallel, they have the same direction. The angle between two lines running in the same direction is $0^\circ$. If we consider the angle measured from one line to the other in a different orientation, it could be $180^\circ$, which is $\pi$ radians. If the lines are coincident, the angle between them is also considered $0^\circ$. So, the statement that the angle between two parallel lines is $0^\circ$ or $\pi$ radians is True.

Reason (R): In coordinate geometry, two non-vertical lines are parallel if and only if they have the same slope. Vertical lines have undefined slopes but are parallel to each other. When discussing slopes as a condition for parallelism, it is generally understood for non-vertical lines. For example, if line $L_1$ has slope $m_1$ and line $L_2$ has slope $m_2$, they are parallel if and only if $m_1 = m_2$ (assuming they are not vertical). If they are both vertical, their slopes are undefined, but they are parallel. The statement "Parallel lines have the same slope" is a fundamental property used to identify parallel lines. Thus, Reason (R) is True.

Now, let's evaluate if Reason (R) is the correct explanation for Assertion (A).

Consider two lines with slopes $m_1$ and $m_2$. The angle $\theta$ between these lines is given by the formula:

$\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$

If the lines are parallel, then according to Reason (R), they have the same slope, so $m_1 = m_2$.

Substituting $m_1 = m_2$ into the angle formula:

$\tan \theta = \left|\frac{m_1 - m_1}{1 + m_1 m_1}\right| = \left|\frac{0}{1 + m_1^2}\right|$

Assuming $1 + m_1^2 \neq 0$ (which is always true for real $m_1$), we get:

$\tan \theta = 0$

If $\tan \theta = 0$, then the angle $\theta$ must be $0^\circ$ or $180^\circ$ ($\pi$ radians).

This means that having the same slope directly implies that the angle between the lines is $0^\circ$ or $\pi$ radians. Thus, Reason (R) provides the correct mathematical explanation for why Assertion (A) is true for non-vertical parallel lines.

For vertical parallel lines, the concept of slope isn't directly used in the formula this way, but the geometric fact remains that they are parallel and the angle is $0^\circ$. The property of having the same slope (or both being vertical with undefined slope) is the characteristic that defines parallel lines and leads to the $0^\circ$ or $\pi$ angle between them.

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

Comparing our conclusion with the given options:

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

(B) Both A and R are true but R is not the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.

Our conclusion matches option (A).

Final Answer:

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.

Given:

The line passes through the point $(2, 5)$.

The line is parallel to the x-axis.

To Find:

The equation of the line.

Solution:

A line that is parallel to the x-axis is a horizontal line.

The slope of a horizontal line is always 0.

Also, every point on a horizontal line has the same y-coordinate.

Since the line passes through the point $(2, 5)$, the y-coordinate of every point on this line must be the same as the y-coordinate of $(2, 5)$, which is 5.

Therefore, the equation of the line is $y = 5$.

Alternatively, we can use the point-slope form of the equation of a line, which is $y - y_0 = m(x - x_0)$, where $(x_0, y_0)$ is a point on the line and $m$ is the slope.

The given point is $(x_0, y_0) = (2, 5)$.

Since the line is parallel to the x-axis, its slope is $m = 0$.

Substitute these values into the point-slope form:

$y - 5 = 0(x - 2)$

$y - 5 = 0$

$y = 5$

The equation of the line is $y = 5$.

Comparing the derived equation with the given options:

(A) $x = 2$ (This is a vertical line)

(B) $y = 5$ (This is a horizontal line)

(C) $x = 5$ (This is a vertical line)

(D) $y = 2$ (This is a horizontal line passing through a different y-coordinate)

The derived equation $y = 5$ matches option (B).

Final Answer:

The equation of the line is $y = 5$.

The correct option is (B) $y = 5$.

Given:

Four equations representing forms of a line and four names of these forms.

To Match:

Match each equation with its corresponding form.

Solution:

Let's identify each equation form:

(i) $y = mx + c$: This form shows the slope ($m$) and the y-intercept ($c$). It is the Slope-intercept form.

(ii) $y - y_1 = m(x - x_1)$: This form uses the slope ($m$) and a specific point $(x_1, y_1)$ on the line. It is the Point-slope form.

(iii) $\frac{x}{a} + \frac{y}{b} = 1$: This form shows the x-intercept ($a$) and the y-intercept ($b$). It is the Intercept form.

(iv) $x \cos \alpha + y \sin \alpha = p$: This form relates the equation of the line to the perpendicular distance ($p$) from the origin to the line and the angle ($\alpha$) that the normal (perpendicular) from the origin to the line makes with the positive x-axis. It is the Normal form (also called the perpendicular form).

Based on the identifications, the correct matching is:

(i) matches (b) - Slope-intercept form

(ii) matches (c) - Point-slope form

(iii) matches (a) - Intercept form

(iv) matches (d) - Normal form

So the required matching is (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d).

Comparing the required matching with the given options:

(A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d) - Incorrect.

(B) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d) - Correct.

(C) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d) - Incorrect.

(D) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a) - Incorrect.

Final Answer:

The correct matching is (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d).

The correct option is (B) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d).

Given:

The equations of two lines are $L_1: y = \sqrt{3}x + 5$ and $L_2: y = x + 2$.

To Find:

The angle between the lines $L_1$ and $L_2$.

Solution:

The equations of the lines are given in the slope-intercept form $y = mx + c$, where $m$ is the slope.

For line $L_1: y = \sqrt{3}x + 5$, the slope is $m_1 = \sqrt{3}$.

For line $L_2: y = x + 2$, the slope is $m_2 = 1$.

Let $\theta$ be the angle between the two lines. The tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$

Substitute the values of $m_1$ and $m_2$ into the formula:

$m_1 = \sqrt{3}$

$m_2 = 1$

$\tan \theta = \left|\frac{1 - \sqrt{3}}{1 + (\sqrt{3})(1)}\right|$

$\tan \theta = \left|\frac{1 - \sqrt{3}}{1 + \sqrt{3}}\right|$

To simplify, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $1 - \sqrt{3}$:

$\tan \theta = \left|\frac{(1 - \sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}\right|$

$\tan \theta = \left|\frac{1^2 - 2(1)(\sqrt{3}) + (\sqrt{3})^2}{1^2 - (\sqrt{3})^2}\right|$

$\tan \theta = \left|\frac{1 - 2\sqrt{3} + 3}{1 - 3}\right|$

$\tan \theta = \left|\frac{4 - 2\sqrt{3}}{-2}\right|$

$\tan \theta = \left|\frac{-2( \sqrt{3} - 2)}{-2}\right|$

$\tan \theta = \left|\sqrt{3} - 2\right|$

Since $\sqrt{3} \approx 1.732$, $\sqrt{3} - 2$ is a negative value. The absolute value makes it positive:

$\tan \theta = -(\sqrt{3} - 2) = 2 - \sqrt{3}$

We know that $\tan(15^\circ) = 2 - \sqrt{3}$.

Therefore, $\theta = 15^\circ$.

Alternatively, we can find the angles of inclination of the lines.

For $L_1$, $m_1 = \sqrt{3}$. If $\alpha_1$ is the angle of inclination, $\tan \alpha_1 = \sqrt{3}$. This means $\alpha_1 = 60^\circ$.

For $L_2$, $m_2 = 1$. If $\alpha_2$ is the angle of inclination, $\tan \alpha_2 = 1$. This means $\alpha_2 = 45^\circ$.

The angle between the two lines is the absolute difference between their angles of inclination:

$\theta = |\alpha_1 - \alpha_2| = |60^\circ - 45^\circ| = |15^\circ| = 15^\circ$

Or the obtuse angle is $180^\circ - 15^\circ = 165^\circ$. When asked for the angle between lines, the acute angle is usually implied.

So the angle between the lines is $15^\circ$.

Comparing the calculated angle with the given options:

(A) $15^\circ$

(B) $30^\circ$

(C) $45^\circ$

(D) $60^\circ$

The calculated angle is $15^\circ$, which matches option (A).

Final Answer:

The angle between the lines is $15^\circ$.

The correct option is (A) $15^\circ$.

Given:

Two points on the line are $A(x_1, y_1) = (-1, 3)$ and $B(x_2, y_2) = (4, -2)$.

To Find:

The equation of the line passing through the points $(-1, 3)$ and $(4, -2)$.

Solution:

We can find the equation of the line using the two-point form, which is:

$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

Substitute the coordinates of the given points into the formula:

$x_1 = -1$, $y_1 = 3$

$x_2 = 4$, $y_2 = -2$

$\frac{y - 3}{x - (-1)} = \frac{-2 - 3}{4 - (-1)}$

Simplify the denominators:

$\frac{y - 3}{x + 1} = \frac{-5}{4 + 1}$

$\frac{y - 3}{x + 1} = \frac{-5}{5}$

Simplify the fraction on the right side:

$\frac{y - 3}{x + 1} = -1$

Multiply both sides by $(x + 1)$:

$y - 3 = -1(x + 1)$

$y - 3 = -x - 1$

Move all terms to one side to get the equation in the form $Ax + By + C = 0$:

$x + y - 3 + 1 = 0$

$x + y - 2 = 0$

This is the equation of the line passing through the given points.

Comparing the derived equation with the given options:

(A) $x+y-2=0$

(B) $x-y+4=0$

(C) $x+y+2=0$

(D) $x-y-4=0$

The derived equation $x + y - 2 = 0$ matches option (A).

Final Answer:

The equation of the line passing through $(-1, 3)$ and $(4, -2)$ is $x + y - 2 = 0$.

The correct option is (A) $x+y-2=0$.

Given:

The vertices of the triangle are $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$.

To Find:

The coordinates of the centroid of the triangle.

Solution:

The centroid of a triangle is the point of intersection of its medians. A median joins a vertex to the midpoint of the opposite side.

The coordinates of the centroid $(x, y)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are given by the average of the x-coordinates and the average of the y-coordinates of the vertices.

The formula for the coordinates of the centroid is:

$x = \frac{x_1 + x_2 + x_3}{3}$

$y = \frac{y_1 + y_2 + y_3}{3}$

So, the coordinates of the centroid are $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.

Comparing the formula with the given options:

(A) $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$

(B) $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ (This is the midpoint formula for two points)

(C) $(\frac{x_1+x_2+x_3}{2}, \frac{y_1+y_2+y_3}{2})$ (Incorrect denominator)

(D) $(\frac{x_1+x_2}{3}, \frac{y_1+y_2}{3})$ (Incorrect inclusion of third vertex and incorrect formula)

The derived formula matches option (A).

Final Answer:

The centroid of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.

The correct option is (A) $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.

Given:

The point is $(x_1, y_1) = (3, -5)$.

The equation of the line is $3x - 4y - 26 = 0$.

To Find:

The distance of the point $(3, -5)$ from the line $3x - 4y - 26 = 0$.

Solution:

The equation of the line is in the form $Ax + By + C = 0$, where $A = 3$, $B = -4$, and $C = -26$.

The coordinates of the point are $(x_1, y_1) = (3, -5)$.

The distance $d$ of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Substitute the given values into the formula:

$d = \frac{|(3)(3) + (-4)(-5) + (-26)|}{\sqrt{(3)^2 + (-4)^2}}$

Calculate the terms in the numerator and the denominator:

Numerator: $|9 + 20 - 26| = |29 - 26| = |3| = 3$

Denominator: $\sqrt{9 + 16} = \sqrt{25} = 5$

So, the distance is:

$d = \frac{3}{5}$

$d = 0.6$

The calculated distance is 0.6.

Upon comparing the calculated distance with the given options (A) 1, (B) 2, (C) 3, (D) 4, we notice that the calculated distance 0.6 does not match any of the options.

There might be a typo in the question or the provided options.

Alternate Solution (Assuming a typo in the point's x-coordinate):

Let's assume the point was $(-3, -5)$ instead of $(3, -5)$, as this is a common type of error in problem statements.

The point is $(x_1, y_1) = (-3, -5)$.

The equation of the line is $3x - 4y - 26 = 0$, so $A=3, B=-4, C=-26$.

Using the distance formula:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Substitute the alternate point coordinates and line coefficients:

$d = \frac{|(3)(-3) + (-4)(-5) + (-26)|}{\sqrt{(3)^2 + (-4)^2}}$

Calculate the terms in the numerator and the denominator:

Numerator: $|-9 + 20 - 26| = |11 - 26| = |-15| = 15$

Denominator: $\sqrt{9 + 16} = \sqrt{25} = 5$

So, the distance is:

$d = \frac{15}{5}$

$d = 3$

If the point was $(-3, -5)$, the distance from the line $3x - 4y - 26 = 0$ would be 3.

Comparing the distance calculated with the alternate point $(-3, -5)$ with the given options:

(A) 1

(B) 2

(C) 3

(D) 4

The calculated distance of 3 matches option (C).

Final Answer (Based on assumed typo in the point):

Assuming the intended point was $(-3, -5)$, the distance is 3.

Based on the provided options, it is highly probable that the point in the question was intended to be $(-3, -5)$ instead of $(3, -5)$. With the given point $(3, -5)$, the distance is $0.6$, which is not among the options.

If we consider the possibility of a typo in the question leading to one of the options being correct, the most likely intended answer is 3, which corresponds to the point $(-3, -5)$.

Therefore, based on the probable intended question:

The correct option is (C) 3.

Given:

The equation of line $L_1$ is $2x + 3y = 6$.

To Find:

The slope of the line $L_1$.

Solution:

We are given the equation of the line $L_1$ as $2x + 3y = 6$.

To find the slope of the line, we can rearrange 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:

$2x + 3y = 6$

Subtract $2x$ from both sides of the equation to isolate the term containing $y$:

$3y = 6 - 2x$

Rewrite the right side with the $x$ term first:

$3y = -2x + 6$

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

$y = \frac{-2x + 6}{3}$

Separate the terms on the right side:

$y = \frac{-2}{3}x + \frac{6}{3}$

Simplify the second term:

$y = -\frac{2}{3}x + 2$

This equation is now in the slope-intercept form $y = mx + c$.

Comparing $y = -\frac{2}{3}x + 2$ with $y = mx + c$, we can identify the slope $m$ and the y-intercept $c$.

The slope $m$ of the line $L_1$ is $-\frac{2}{3}$.

Comparing the calculated slope with the given options:

(A) $2/3$

(B) $-2/3$

(C) $3/2$

(D) $-3/2$

The calculated slope is $-\frac{2}{3}$, which matches option (B).

Final Answer:

The slope of the line $L_1$ is $-\frac{2}{3}$.

The correct option is (B) $-2/3$.

Given:

The equation of line $L_2$ is $4x + 6y = 24$.

To Find:

The slope of the line $L_2$.

Solution:

We are given the equation of the line $L_2$ as $4x + 6y = 24$.

To find the slope of the line, we can rearrange 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:

$4x + 6y = 24$

Subtract $4x$ from both sides of the equation to isolate the term containing $y$:

$6y = 24 - 4x$

Rewrite the right side with the $x$ term first:

$6y = -4x + 24$

Divide both sides by 6 to solve for $y$:

$y = \frac{-4x + 24}{6}$

Separate the terms on the right side:

$y = \frac{-4}{6}x + \frac{24}{6}$

Simplify the fractions:

$y = -\frac{2}{3}x + 4$

This equation is now in the slope-intercept form $y = mx + c$.

Comparing $y = -\frac{2}{3}x + 4$ with $y = mx + c$, we can identify the slope $m$ and the y-intercept $c$.

The slope $m$ of the line $L_2$ is $-\frac{2}{3}$.

Comparing the calculated slope with the given options:

(A) $4/6 = 2/3$

(B) $-4/6 = -2/3$

(C) $6/4 = 3/2$

(D) $-6/4 = -3/2$

The calculated slope is $-\frac{2}{3}$, which matches option (B).

Final Answer:

The slope of the line $L_2$ is $-\frac{2}{3}$.

The correct option is (B) $-4/6 = -2/3$.

Given:

The equations of the two lines are $L_1: 2x + 3y = 6$ and $L_2: 4x + 6y = 24$.

To Find:

The relationship between the lines $L_1$ and $L_2$.

Solution:

To determine the relationship between the two lines, we can compare their slopes and y-intercepts.

We first convert the equations into the slope-intercept form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

For line $L_1: 2x + 3y = 6$

$3y = -2x + 6$

$y = -\frac{2}{3}x + \frac{6}{3}$

$y = -\frac{2}{3}x + 2$

The slope of $L_1$ is $m_1 = -\frac{2}{3}$ and the y-intercept is $c_1 = 2$.

For line $L_2: 4x + 6y = 24$

$6y = -4x + 24$

$y = -\frac{4}{6}x + \frac{24}{6}$

$y = -\frac{2}{3}x + 4$

The slope of $L_2$ is $m_2 = -\frac{2}{3}$ and the y-intercept is $c_2 = 4$.

We observe that the slopes are equal ($m_1 = m_2 = -\frac{2}{3}$).

Lines with equal slopes are either parallel or are the same line.

To check if they are the same line, we compare their y-intercepts. The y-intercept of $L_1$ is $c_1 = 2$, and the y-intercept of $L_2$ is $c_2 = 4$.

Since the y-intercepts are different ($c_1 \neq c_2$), the lines are distinct.

Therefore, the lines $L_1$ and $L_2$ are parallel and distinct.

Comparing the derived relationship with the given options:

(A) They are perpendicular. (Perpendicular lines have slopes whose product is -1, unless one is horizontal and the other vertical. Here, $(-\frac{2}{3}) \times (-\frac{2}{3}) = \frac{4}{9} \neq -1$. So, not perpendicular.)

(B) They are parallel and distinct. (Slopes are equal, y-intercepts are different. This matches our finding.)

(C) They are the same line. (Slopes are equal, but y-intercepts are different. So, not the same line.)

(D) They intersect at a single point. (Lines intersect at a single point if their slopes are different. Here, the slopes are equal. So, they do not intersect at a single point unless they are the same line, which they are not.)

Our finding matches option (B).

Final Answer:

The lines $L_1$ and $L_2$ are parallel and distinct.

The correct option is (B) They are parallel and distinct.

Given:

The coordinates of the point are $P(x_1, y_1) = (1, 2)$.

The equation of the line $L_1$ is $2x + 3y - 6 = 0$.

To Find:

The distance of the point $P(1, 2)$ from the line $L_1: 2x + 3y - 6 = 0$.

Solution:

The equation of the line $L_1$ is given in the general form $Ax + By + C = 0$, where $A = 2$, $B = 3$, and $C = -6$.

The coordinates of the point are $(x_1, y_1) = (1, 2)$.

The distance $d$ of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Substitute the values of $A$, $B$, $C$, $x_1$, and $y_1$ into the formula:

$d = \frac{|(2)(1) + (3)(2) + (-6)|}{\sqrt{(2)^2 + (3)^2}}$

Calculate the expression inside the absolute value in the numerator:

$2(1) + 3(2) - 6 = 2 + 6 - 6 = 8 - 6 = 2$

The numerator is $|2| = 2$.

Calculate the expression under the square root in the denominator:

$(2)^2 + (3)^2 = 4 + 9 = 13$

The denominator is $\sqrt{13}$.

So, the distance is:

$d = \frac{2}{\sqrt{13}}$

The distance of the point $P(1, 2)$ from the line $L_1$ is $\frac{2}{\sqrt{13}}$.

Comparing the calculated distance with the given options:

(A) 0

(B) $\frac{|2(1) + 3(2) - 6|}{\sqrt{2^2 + 3^2}} = \frac{|2+6-6|}{\sqrt{4+9}} = \frac{2}{\sqrt{13}}$

(C) $\frac{2}{\sqrt{13}}$

(D) $\frac{2}{13}$

The calculated distance $\frac{2}{\sqrt{13}}$ matches option (C).

Final Answer:

The distance of the point $P(1, 2)$ from the line $L_1$ is $\frac{2}{\sqrt{13}}$.

The correct option is (C) $\frac{2}{\sqrt{13}}$.

Given:

The vertices of the triangle are $A(x_1, y_1) = (2, 3)$, $B(x_2, y_2) = (-1, 0)$, and $C(x_3, y_3) = (2, -4)$.

To Find:

The area of the triangle formed by the given points.

Solution:

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

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates of the given vertices into this formula:

$x_1 = 2$, $y_1 = 3$

$x_2 = -1$, $y_2 = 0$

$x_3 = 2$, $y_3 = -4$

Area $= \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$

Perform the calculations inside the absolute value:

Area $= \frac{1}{2} |2(0 + 4) + (-1)(-7) + 2(3)|$

Area $= \frac{1}{2} |2(4) + 7 + 6|$

Area $= \frac{1}{2} |8 + 7 + 6|$

Area $= \frac{1}{2} |21|$

Area $= \frac{21}{2}$

Area $= 10.5$ square units.

The calculated area of the triangle is 10.5.

Comparing the calculated area with the given options:

(A) 7

(B) 14

(C) 21

(D) 28

The calculated area, 10.5, does not match any of the provided options.

However, the value 21 appears in the intermediate calculation before dividing by 2 (i.e., the absolute value of the determinant). In some multiple-choice questions where there might be a typo, the value before the final division is sometimes listed as an option.

Based on the standard formula for the area of a triangle, the correct area is 10.5. Since 10.5 is not among the options, there appears to be an error in the question or the provided options.

If we were forced to choose the closest or most likely intended answer based on the intermediate calculation, option (C) 21 is the absolute value of the determinant calculated in the area formula.

Final Answer:

The calculated area of the triangle is 10.5 square units.

Based on the provided options, none of them match the calculated area. However, option (C) is 21, which is twice the calculated area or the absolute value of the determinant used in the calculation. Assuming a potential error in the question's options, we note the discrepancy.

Based on the most likely intended question (where the determinant value is presented as an option), the answer would be 21. However, mathematically, the area is 10.5.

Given the options, and the presence of 21 as a calculated intermediate value, we select option (C) as the likely intended answer, while acknowledging it is not the correct area per the standard formula.

The correct option (based on likely intended answer among choices) is (C) 21.

Given:

Two points $A(a, 0)$ and $B(-a, 0)$.

A point $P(x, y)$ such that its distance from $A$ is equal to its distance from $B$.

$PA = PB$

To Find:

The locus of the point $P(x, y)$ that satisfies the given condition.

Solution:

Let the coordinates of the point be $P(x, y)$.

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

The distance $PA$ between $P(x, y)$ and $A(a, 0)$ is:

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

The distance $PB$ between $P(x, y)$ and $B(-a, 0)$ is:

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

The given condition is $PA = PB$.

So, $\sqrt{(x - a)^2 + y^2} = \sqrt{(x + a)^2 + y^2}$

Square both sides of the equation to eliminate the square roots:

$(x - a)^2 + y^2 = (x + a)^2 + y^2$

Expand the squared terms:

$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2 + y^2$

Subtract $x^2$, $a^2$, and $y^2$ from both sides of the equation:

$-2ax = 2ax$

Move the $2ax$ term from the right side to the left side:

$-2ax - 2ax = 0$

Combine the terms:

$-4ax = 0$

Assuming $a \neq 0$ (if $a=0$, the two points are the same, and any point $(x,y)$ is equidistant from itself, leading to the entire plane), we can divide both sides by $-4a$:

$x = 0$

The equation $x = 0$ represents a vertical line, which is the y-axis.

The locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining those two points.

The midpoint of the segment joining $(a, 0)$ and $(-a, 0)$ is $(\frac{a + (-a)}{2}, \frac{0 + 0}{2}) = (\frac{0}{2}, \frac{0}{2}) = (0, 0)$.

The segment joining $(a, 0)$ and $(-a, 0)$ lies on the x-axis (a horizontal line). The perpendicular bisector of a horizontal segment is a vertical line passing through its midpoint.

The vertical line passing through the origin $(0, 0)$ is the y-axis, whose equation is $x = 0$.

Comparing the derived locus with the given options:

(A) A circle

(B) A parabola

(C) A straight line (the y-axis, $x=0$)

(D) A single point

The derived equation $x = 0$ represents a straight line (the y-axis), which matches option (C).

Final Answer:

The locus of the point is the straight line $x = 0$ (the y-axis).

The correct option is (C) A straight line (the y-axis, $x=0$).

Given:

The line passes through the origin, which is the point $(0, 0)$.

The slope of the line is $m = 3$.

To Find:

The equation of the line.

Solution:

We can use the slope-intercept form of the equation of a line, which is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

We are given the slope $m = 3$.

The line passes through the origin $(0, 0)$. This means when $x = 0$, the value of $y$ is $0$. We can use this point to find the y-intercept $c$.

Substitute the coordinates of the origin $(0, 0)$ and the slope $m = 3$ into the slope-intercept form:

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

$0 = 0 + c$

$c = 0$

So, the y-intercept is 0.

Now, substitute the slope $m = 3$ and the y-intercept $c = 0$ back into the slope-intercept form $y = mx + c$:

$y = (3)x + (0)$

$y = 3x$

This is the equation of the line.

We can also write this as $y = 3x + 0$, which is an equivalent form.

Alternatively, 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.

The given point is $(x_1, y_1) = (0, 0)$ (the origin), and the slope is $m = 3$.

Substitute these values into the point-slope form:

$y - 0 = 3(x - 0)$

$y = 3x$

This is the equation of the line.

Comparing the derived equation with the given options:

(A) $y = 3x$ - This matches our derived equation.

(B) $x = 3y$ - Dividing by 3 gives $y = \frac{1}{3}x$, which has a slope of $\frac{1}{3}$, not 3. So, this is incorrect.

(C) $y = 3x + 0$ - This is exactly the same as $y = 3x$. This also matches our derived equation.

(D) Both (A) and (C) - Since both (A) and (C) represent the same correct equation, this option is the most appropriate answer.

Final Answer:

The equation of the line passing through the origin and having a slope of 3 is $y = 3x$ or $y = 3x + 0$.

The correct option is (D) Both (A) and (C).

Given:

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

To Find:

The equation of the line passing through these two points.

Solution:

There are standard formulas for the equation of a line passing through two given points $(x_1, y_1)$ and $(x_2, y_2)$.

One common form is derived from the point-slope form $y - y_1 = m(x - x_1)$, where $m$ is the slope.

The slope of the line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$, provided $x_1 \neq x_2$.

Substituting this slope into the point-slope form using the point $(x_1, y_1)$, we get:

$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

This equation is valid for $x_1 \neq x_2$. If $x_1 = x_2$, the line is vertical and its equation is $x = x_1$. In this case, the denominator $x_2 - x_1$ is zero, and the formula is not directly applicable in this division form, but the original relationship from equating slopes $\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$ implies $x - x_1 = 0$ when $x_2 - x_1 = 0$ and $y_2 - y_1 \neq 0$.

Another standard form is obtained by rearranging the equation from equating slopes $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$ to the symmetric form:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

This form is valid provided $x_1 \neq x_2$ and $y_1 \neq y_2$. Special cases for horizontal ($y_1=y_2$) or vertical ($x_1=x_2$) lines need to be considered if denominators are zero.

Let's examine the given options:

(A) $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$: This matches the standard two-point form derived from the point-slope form.

(B) $y - x_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - y_1)$: Incorrectly mixes coordinates.

(C) $x - x_1 = \frac{x_2 - x_1}{y_2 - y_1}(y - y_1)$: This is also a valid form, derived from equating slopes $\frac{x-x_1}{y-y_1} = \frac{x_2-x_1}{y_2-y_1}$, or by rearranging option (A). It is valid when $y_1 \neq y_2$.

(D) $\frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}$: This is another standard two-point form, often referred to as the symmetric form.

Options (A), (C), and (D) are all mathematically correct representations of the line equation, with conditions on the denominators being non-zero. However, in a multiple-choice question expecting a single answer, option (A) is a very direct result of substituting the slope into the fundamental point-slope form. Option (D) is also very standard. Option (C) is a less frequently cited primary form, though equivalent.

Given that Question 5 explicitly asks for and provides the point-slope form $y - y_0 = m(x - x_0)$, it is likely that the intended answer here is the direct application of substituting the two-point slope formula into the point-slope form, which is option (A).

Comparing our derived equation form with the given options:

The equation $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$ matches option (A).

Final Answer:

The equation of the line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is commonly written as $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.

The correct option is (A) $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.

Given:

A point $(x_1, y_1)$.

A line with the equation $Ax + By + C = 0$.

The foot of the perpendicular from $(x_1, y_1)$ to the line is $(x_2, y_2)$.

To Find:

The formula that gives the coordinates $(x_2, y_2)$ of the foot of the perpendicular.

Solution:

The formula for the coordinates $(x_2, y_2)$ of the foot of the perpendicular from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is a standard result in coordinate geometry.

The formula is given by:

$\frac{x_2 - x_1}{A} = \frac{y_2 - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$

This formula equates three ratios. The first ratio involves the difference in x-coordinates scaled by the coefficient of $x$ in the line equation. The second ratio involves the difference in y-coordinates scaled by the coefficient of $y$ in the line equation. Both of these ratios are equal to a third ratio, which is the negative of the value obtained by substituting the point $(x_1, y_1)$ into the line equation ($Ax_1 + By_1 + C$) divided by the sum of the squares of the coefficients of $x$ and $y$ ($A^2 + B^2$).

By setting each of the first two ratios equal to the third ratio, we can solve for $x_2$ and $y_2$ in terms of $x_1$, $y_1$, $A$, $B$, and $C$.

For example, to find $x_2$:

$\frac{x_2 - x_1}{A} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$

$x_2 - x_1 = -A \left(\frac{Ax_1 + By_1 + C}{A^2 + B^2}\right)$

$x_2 = x_1 - \frac{A(Ax_1 + By_1 + C)}{A^2 + B^2}$

Similarly, for $y_2$:

$\frac{y_2 - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$

$y_2 - y_1 = -B \left(\frac{Ax_1 + By_1 + C}{A^2 + B^2}\right)$

$y_2 = y_1 - \frac{B(Ax_1 + By_1 + C)}{A^2 + B^2}$

The question asks for the formula relating $(x_1, y_1)$, $(x_2, y_2)$, and the line coefficients, which is the chain of equalities.

Comparing the standard formula with the given options:

(A) $\frac{x_2 - x_1}{A} = \frac{y_2 - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$ - This matches the standard formula.

(B) $\frac{x_2 - x_1}{A} = \frac{y_2 - y_1}{B} = \frac{Ax_1 + By_1 + C}{A^2 + B^2}$ - Incorrect sign on the third term.

(C) $\frac{x_2 + x_1}{A} = \frac{y_2 + y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$ - Incorrect terms in the numerators of the first two ratios.

(D) $\frac{x_2 + x_1}{A} = \frac{y_2 + y_1}{B} = \frac{Ax_1 + By_1 + C}{A^2 + B^2}$ - Incorrect terms and sign.

The correct option is (A).

Final Answer:

The coordinates $(x_2, y_2)$ of the foot of the perpendicular from $(x_1, y_1)$ to the line $Ax + By + C = 0$ are given by the relation $\frac{x_2 - x_1}{A} = \frac{y_2 - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$.

The correct option is (A) $\frac{x_2 - x_1}{A} = \frac{y_2 - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}$.

Given:

The x-intercept of the line is $a = 3$.

The y-intercept of the line is $b = 4$.

To Find:

The equation of the line with the given intercepts.

Solution:

The equation of a line with x-intercept $a$ and y-intercept $b$ is given by the intercept form:

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

Substitute the given values of the intercepts into this formula:

$a = 3$

$b = 4$

So, the equation of the line is:

$\frac{x}{3} + \frac{y}{4} = 1$

To convert this equation to the standard form $Ax + By + C = 0$ or $Ax + By = C$, we can find a common denominator for the fractions on the left side.

The least common multiple (LCM) of 3 and 4 is 12.

Multiply every term in the equation by 12:

$12 \left(\frac{x}{3}\right) + 12 \left(\frac{y}{4}\right) = 12(1)$

Simplify the terms:

$\frac{12x}{3} + \frac{12y}{4} = 12$

$4x + 3y = 12$

This is the equation of the line in the form $Ax + By = C$.

Comparing the derived equation with the given options:

(A) $4x + 3y = 12$ - This matches our derived equation.

(B) $3x + 4y = 12$ - The coefficients of $x$ and $y$ are swapped compared to our equation.

(C) $4x + 3y = 1$ - The constant term is incorrect.

(D) $3x + 4y = 1$ - The coefficients and the constant term are incorrect.

The derived equation $4x + 3y = 12$ matches option (A).

Final Answer:

The equation of the line with x-intercept 3 and y-intercept 4 is $4x + 3y = 12$.

The correct option is (A) $4x + 3y = 12$.

Given:

The angle of inclination of the line with the positive x-axis is $\theta = 135^\circ$.

To Find:

The slope of the line.

Solution:

The slope $m$ of a line that makes an angle $\theta$ with the positive x-axis is given by the formula:

$m = \tan \theta$

Substitute the given angle $\theta = 135^\circ$ into the formula:

$m = \tan(135^\circ)$

To evaluate $\tan(135^\circ)$, we can use the property $\tan(180^\circ - \alpha) = -\tan(\alpha)$.

Here, $135^\circ = 180^\circ - 45^\circ$.

So, $m = \tan(180^\circ - 45^\circ) = -\tan(45^\circ)$

We know that $\tan(45^\circ) = 1$.

Therefore, $m = -1$.

The slope of the line making an angle of $135^\circ$ with the positive x-axis is -1.

Comparing the calculated slope with the given options:

(A) 1

(B) -1

(C) $\sqrt{3}$

(D) $-\sqrt{3}$

The calculated slope is -1, which matches option (B).

Final Answer:

The slope of the line is -1.

The correct option is (B) -1.

Given:

A point $P(x, y)$ moves such that its distance from the fixed point $O(0, 0)$ is always equal to 5.

Distance $PO = 5$.

To Find:

The locus of the point $P(x, y)$ that satisfies the given condition.

Solution:

Let the coordinates of the moving point be $P(x, y)$. The fixed point is the origin $O(0, 0)$.

The distance between $P(x, y)$ and $O(0, 0)$ is given by the distance formula:

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

According to the given condition, the distance $PO$ is always 5.

So, we have:

$PO = 5$

$\sqrt{x^2 + y^2} = 5$

To eliminate the square root, square both sides of the equation:

$({\sqrt{x^2 + y^2}})^2 = 5^2$

$x^2 + y^2 = 25$

This is the equation of the locus of the point $P(x, y)$.

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

In our case, $r^2 = 25$, so the radius is $r = \sqrt{25} = 5$ (since radius must be positive).

Thus, the locus of the point is a circle centered at the origin with a radius of 5.

Comparing the type of locus with the given options:

(A) line

(B) circle

(C) parabola

(D) point

The derived locus is a circle, which matches option (B).

Final Answer:

The locus of a point that moves such that its distance from the point $(0, 0)$ is always 5 is a circle.

The correct option is (B) circle.

Given:

The equation of a line is $Ax + By + C = 0$, where A, B, and C are constants.

To Find:

The general form of the equation of a line parallel to the given line.

Solution:

Parallel lines have the same slope.

Let's find the slope of the given line $Ax + By + C = 0$. We can rewrite this equation in the slope-intercept form, $y = mx + c$, where $m$ is the slope.

From $Ax + By + C = 0$, we isolate the $y$ term:

$By = -Ax - C$

If $B \neq 0$, we can divide by $B$:

$y = -\frac{A}{B}x - \frac{C}{B}$

The slope of the given line is $m = -\frac{A}{B}$.

A line parallel to this line must have the same slope, i.e., $m_{\text{parallel}} = -\frac{A}{B}$.

The equation of a line with slope $-\frac{A}{B}$ can be written in the form $y = -\frac{A}{B}x + k'$, where $k'$ is the y-intercept (a constant, generally different from $-C/B$).

Multiply the equation by $B$ (assuming $B \neq 0$):

$By = -Ax + Bk'$

Rearrange the terms to the general form:

$Ax + By - Bk' = 0$

Let $-Bk'$ be a new constant, say $k$. Then the equation becomes:

$Ax + By + k = 0$

This form represents a line parallel to $Ax + By + C = 0$ for any value of $k$, as long as the lines are distinct (i.e., $k \neq C$). If $k=C$, it is the same line.

Now consider the case when $B=0$. The original equation becomes $Ax + C = 0$ (assuming $A \neq 0$). This is a vertical line with equation $x = -C/A$. A line parallel to a vertical line is also a vertical line with equation $x = k''$ for some constant $k''$. This can be written as $x - k'' = 0$, or $Ax - Ak'' = 0$. Let $k = -Ak''$. The equation is $Ax + k = 0$. If we substitute $B=0$ into the form $Ax + By + k = 0$, we get $Ax + 0y + k = 0$, which simplifies to $Ax + k = 0$. This also covers the case of vertical lines.

Therefore, the general equation of a line parallel to $Ax + By + C = 0$ is of the form $Ax + By + k = 0$, where $k$ is a constant.

Comparing the derived form with the given options:

(A) $Ax + By + k = 0$ - This matches our derived form.

(B) $Bx - Ay + k = 0$ - The slope of this line is $-B/(-A) = B/A$, which is the negative reciprocal of the slope of the original line (unless $A=0$ or $B=0$), indicating perpendicularity, not parallelism (unless $A=0$ or $B=0$, where careful check is needed, but it generally represents a perpendicular line form). For example, if $A=1, B=1$, original slope is $-1$, option B slope is $1/1=1$. If $A=1, B=2$, original slope is $-1/2$, option B slope is $2/1=2$.

(C) $Ax - By + k = 0$ - The slope of this line is $-A/(-B) = A/B$, which is the negative of the original slope (unless $A=0$ or $B=0$), indicating a different direction.

(D) $Bx + Ay + k = 0$ - The slope of this line is $-B/A$. This is generally the slope of a line perpendicular to $Ax+By+C=0$ when $A, B \neq 0$, as $(-\frac{A}{B}) \times (-\frac{B}{A}) = 1$. The correct slope for perpendicularity is $B/A$. Let's recheck option B. Slope of $Bx-Ay+k=0$ is $-B/(-A) = B/A$. This is the correct form for a perpendicular line if $A, B \neq 0$.

The form $Ax + By + k = 0$ ensures that the coefficients of $x$ and $y$ (A and B) are the same as the original line, which implies the same direction vector and thus the same slope (or both are vertical/horizontal together). The only difference is the constant term $C$, replaced by $k$, which determines the specific position of the parallel line.

Final Answer:

The equation of a line parallel to $Ax + By + C = 0$ is of the form $Ax + By + k = 0$.

The correct option is (A) $Ax + By + k = 0$.

Given:

The equation of a line is $Ax + By + C = 0$, where A, B, and C are constants.

To Find:

The general form of the equation of a line perpendicular to the given line.

Solution:

Two lines are perpendicular if the product of their slopes is -1 (for non-vertical and non-horizontal lines). If one line is horizontal, the perpendicular line is vertical, and vice versa.

Let's find the slope of the given line $Ax + By + C = 0$. We rearrange it into the slope-intercept form $y = mx + c$:

$By = -Ax - C$

If $B \neq 0$, the slope is $m_1 = -\frac{A}{B}$.

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

$-\frac{A}{B} \times m_2 = -1$

$m_2 = \frac{-1}{-\frac{A}{B}} = \frac{B}{A}$ (provided $A \neq 0$).

The equation of a line with slope $\frac{B}{A}$ can be written in the form $y = \frac{B}{A}x + k'$ for some constant $k'$.

Multiply by $A$ (assuming $A \neq 0$):

$Ay = Bx + Ak'$

Rearrange the terms:

$Bx - Ay + Ak' = 0$

Let $k = Ak'$. The equation is $Bx - Ay + k = 0$.

Now let's consider the special cases:

• If $A=0$ (and $B \neq 0$), the original line is $By + C = 0$, which is $y = -C/B$ (a horizontal line). A line perpendicular to this is a vertical line of the form $x = k''$, or $x - k'' = 0$. If we substitute $A=0$ into $Bx - Ay + k = 0$, we get $Bx - 0y + k = 0$, which is $Bx + k = 0$, or $x = -k/B$. Since $B \neq 0$, this is a vertical line. The form works.
• If $B=0$ (and $A \neq 0$), the original line is $Ax + C = 0$, which is $x = -C/A$ (a vertical line). A line perpendicular to this is a horizontal line of the form $y = k'''$, or $y - k''' = 0$. If we substitute $B=0$ into $Bx - Ay + k = 0$, we get $0x - Ay + k = 0$, which is $-Ay + k = 0$, or $y = k/A$. Since $A \neq 0$, this is a horizontal line. The form works.

The form $Bx - Ay + k = 0$ covers all cases and represents a line perpendicular to $Ax + By + C = 0$ for any constant $k$ (which determines the specific line among the family of perpendicular lines).

Comparing the derived form with the given options:

(A) $Ax + By + k = 0$: Same coefficients for $x$ and $y$ as the original line; represents parallel lines.

(B) $Bx - Ay + k = 0$: Swapped coefficients of $x$ and $y$, with one of them negated; represents perpendicular lines.

(C) $Ax - By + k = 0$: Same coefficient for $x$, negated coefficient for $y$; represents a line with slope $A/B$. Not perpendicular unless $A/B = -1/(-A/B)$, which means $A/B = B/A$, so $A^2 = B^2$, $A=\pm B$.

(D) $Bx + Ay + k = 0$: Swapped coefficients, both signs kept the same relative to each other; represents a line with slope $-B/A$. Not perpendicular unless $-B/A = -1/(-A/B)$, which means $-B/A = B/A$, so $B=0$ or $A$ is undefined, or $B=-B$, $B=0$.

The form $Bx - Ay + k = 0$ correctly represents a line perpendicular to $Ax + By + C = 0$.

Final Answer:

The equation of a line perpendicular to $Ax + By + C = 0$ is of the form $Bx - Ay + k = 0$.

The correct option is (B) $Bx - Ay + k = 0$.

Given:

The required line passes through the point $(x_1, y_1) = (1, 1)$.

The required line is perpendicular to the line $L_1: x + y = 1$.

To Find:

The equation of the required line.

Solution:

First, find the slope of the given line $L_1: x + y = 1$.

We can rewrite the equation in the slope-intercept form $y = mx + c$:

$y = -x + 1$

The slope of the line $L_1$ is $m_1 = -1$.

Let the slope of the required line be $m_2$. Since the required line is perpendicular to $L_1$, the product of their slopes must be -1 (assuming neither line is vertical or horizontal, which is the case here as $m_1 = -1$).

$m_1 \times m_2 = -1$

$-1 \times m_2 = -1$

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

$m_2 = 1$

The slope of the required line is 1.

Now, we have a point $(x_1, y_1) = (1, 1)$ on the required line and its slope $m_2 = 1$. We can use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute the values:

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

Simplify the equation:

$y - 1 = x - 1$

Move all terms to one side to get the equation in the general form $Ax + By + C = 0$ or similar forms presented in the options:

$y - x - 1 + 1 = 0$

$y - x = 0$

This equation can also be written as $x - y = 0$ by multiplying the entire equation by -1.

Comparing the derived equation with the given options:

(A) $x - y = 0$

(B) $x + y = 2$

(C) $x - y = 2$

(D) $x + y = 0$

The derived equation $x - y = 0$ matches option (A).

Final Answer:

The equation of the line passing through $(1, 1)$ and perpendicular to $x + y = 1$ is $x - y = 0$.

The correct option is (A) $x - y = 0$.

Given:

Two parallel lines with equations $L_1: Ax + By + C_1 = 0$ and $L_2: Ax + By + C_2 = 0$.

Note that for the lines to be parallel, the coefficients of $x$ and $y$ must be the same or proportional. Here, they are already given as the same, $A$ and $B$. We assume $A$ and $B$ are not both zero.

To Find:

The distance between the two parallel lines $L_1$ and $L_2$.

Solution:

To find the distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, we can use a standard formula derived from finding the distance of a point on one line from the other line.

Consider line $L_1: Ax + By + C_1 = 0$. We can find any point on this line. For example, if $A \neq 0$, we can find the x-intercept by setting $y=0$, which gives $Ax + C_1 = 0$, so $x = -C_1/A$. The point would be $(-C_1/A, 0)$. If $B \neq 0$, we can find the y-intercept by setting $x=0$, which gives $By + C_1 = 0$, so $y = -C_1/B$. The point would be $(0, -C_1/B)$.

Let's take a general point $(x_0, y_0)$ on the line $L_1$. Then $Ax_0 + By_0 + C_1 = 0$.

The distance of this point $(x_0, y_0)$ from the line $L_2: Ax + By + C_2 = 0$ is given by the formula for the distance of a point from a line:

$d = \frac{|Ax_0 + By_0 + C_2|}{\sqrt{A^2 + B^2}}$

Since $(x_0, y_0)$ lies on $L_1$, we know that $Ax_0 + By_0 = -C_1$. Substitute this into the numerator:

$d = \frac{|(-C_1) + C_2|}{\sqrt{A^2 + B^2}}$

$d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}$

Since $|C_2 - C_1| = |C_1 - C_2|$, the formula can also be written as:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

This formula gives the distance between the two parallel lines.

Comparing the derived formula with the given options:

(A) $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$ - This matches the standard formula.

(B) $\frac{|C_1 + C_2|}{\sqrt{A^2 + B^2}}$ - Incorrect sign in the numerator.

(C) $\frac{|A+B|}{\sqrt{C_1^2 + C_2^2}}$ - Incorrect formula; A and B are coefficients, $C_1$ and $C_2$ are constants.

(D) $\frac{|C_1 - C_2|}{A+B}$ - Incorrect denominator; the denominator should be $\sqrt{A^2 + B^2}$.

The correct option is (A).

Final Answer:

The distance between the two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.

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

Given:

The equation of the line is $y = x$.

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

To Find:

The angle between the line $y = x$ and the positive x-axis (angle of inclination).

Solution:

The angle between a line and the positive x-axis is called the angle of inclination. The slope of a line is equal to the tangent of its angle of inclination $\theta$.

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

Comparing $y = x$ with $y = mx + c$, we see that the slope is $m = 1$ and the y-intercept is $c = 0$.

The slope $m$ is related to the angle of inclination $\theta$ by the formula:

$m = \tan \theta$

Substitute the slope $m = 1$ into the formula:

$1 = \tan \theta$

To find the angle $\theta$, we need to find the inverse tangent of 1:

$\theta = \tan^{-1}(1)$

The angle whose tangent is 1 is $45^\circ$.

So, $\theta = 45^\circ$.

The angle between the line $y = x$ and the positive x-axis is $45^\circ$.

Comparing the calculated angle with the given options:

(A) $0^\circ$

(B) $45^\circ$

(C) $90^\circ$

(D) $135^\circ$

The calculated angle is $45^\circ$, which matches option (B).

Final Answer:

The angle between the line $y=x$ and the x-axis is $45^\circ$.

The correct option is (B) $45^\circ$.

Given:

Several statements about the slope of a line.

To Determine:

Which of the given statements about the slope of a line are correct.

Solution:

Let's analyze each statement based on the definition and properties of the slope of a line.

The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$, provided $x_1 \neq x_2$. The slope is also the tangent of the angle $\theta$ the line makes with the positive x-axis, $m = \tan \theta$, for $0^\circ \leq \theta < 180^\circ$, $\theta \neq 90^\circ$.

(A) The slope of a horizontal line is 0.

A horizontal line is parallel to the x-axis. If $\theta$ is the angle of inclination, then $\theta = 0^\circ$. The slope is $m = \tan(0^\circ) = 0$. Alternatively, for any two distinct points $(x_1, c)$ and $(x_2, c)$ on a horizontal line, where $x_1 \neq x_2$, the slope is $m = \frac{c - c}{x_2 - x_1} = \frac{0}{x_2 - x_1} = 0$. Thus, this statement is True.

(B) The slope of a vertical line is undefined.

A vertical line is parallel to the y-axis and perpendicular to the x-axis. If $\theta$ is the angle of inclination, then $\theta = 90^\circ$. The tangent of $90^\circ$ is undefined, so the slope $m = \tan(90^\circ)$ is undefined. Alternatively, for any two distinct points $(c, y_1)$ and $(c, y_2)$ on a vertical line, where $y_1 \neq y_2$, the slope is $m = \frac{y_2 - y_1}{c - c} = \frac{y_2 - y_1}{0}$. Division by zero is undefined. Thus, this statement is True.

(C) If two lines are perpendicular, the product of their slopes is -1 (unless one is vertical).

Let the slopes of the two lines be $m_1$ and $m_2$. If the lines are perpendicular and neither is vertical or horizontal, the relationship between their slopes is $m_1 m_2 = -1$. If one line is horizontal (slope 0), the perpendicular line is vertical (undefined slope), and the product is not defined. The statement includes the important caveat "unless one is vertical", which addresses this special case. Thus, this statement is True.

(D) If two lines are parallel, their slopes are equal.

If two non-vertical lines are parallel, they have the same angle of inclination, so their slopes are equal ($m_1 = m_2$). If both lines are vertical, their slopes are both undefined, but they are parallel. The statement "their slopes are equal" holds for all parallel lines except for the case of two vertical lines, where the slopes are undefined but they are parallel. However, in the context of multiple-choice questions on slopes, the standard property stated is that equal slopes imply parallel lines (or they are the same line), and parallel lines have equal slopes (except for vertical lines). The statement is generally accepted as true in this context, implying non-vertical lines or understanding "equal" as referring to the defined value. If we strictly interpret "equal slopes" as having the same numerical value, it applies only to non-vertical parallel lines. If we consider that both having undefined slopes makes them "equal" in some sense, it might include vertical lines. However, the typical property taught is $m_1=m_2$ for parallel lines (non-vertical). Given the options, it's likely referring to the standard case. Let's assume the standard interpretation where this applies to non-vertical lines or implies the case of both having undefined slopes. Thus, this statement is generally considered True in the context of properties of slopes, especially when contrasted with other options.

Considering all statements, (A), (B), and (C) are standard, precise definitions and properties. (D) is also a standard property, typically stated for non-vertical lines, or implicitly covering the vertical case where both slopes are undefined. In most standard curricula, all four statements are presented as correct properties of lines and slopes.

Final Answer:

Based on standard definitions and properties in coordinate geometry, all four statements are correct.

The correct options are (A), (B), (C), and (D).

Given:

The equation of the line is $(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$.

The line is parallel to the x-axis.

To Find:

The value of $k$.

Solution:

A line is parallel to the x-axis if and only if its slope is 0, provided the line is not the x-axis itself.

The general form of a linear equation is $Ax + By + C = 0$. The slope of this line is given by $m = -\frac{A}{B}$, provided $B \neq 0$.

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

$A = k-3$

$B = -(4-k^2) = k^2 - 4$

$C = k^2 - 7k + 6$

For the line to be parallel to the x-axis, its slope must be 0.

Slope $m = -\frac{A}{B} = 0$.

This requires the numerator $A$ to be zero, while the denominator $B$ must be non-zero.

So, we set $A = 0$:

$k-3 = 0$

Solving for $k$, we get:

$k = 3$

Now we must check if $B \neq 0$ when $k = 3$.

$B = k^2 - 4$

Substitute $k = 3$ into the expression for $B$:

$B = (3)^2 - 4$

$B = 9 - 4$

$B = 5$

Since $B = 5 \neq 0$ when $k=3$, the slope of the line is $m = -\frac{0}{5} = 0$. This confirms that the line is horizontal.

We should also check if the line is the x-axis itself, which has the equation $y=0$. The x-axis is parallel to the x-axis. The equation $Ax+By+C=0$ represents the x-axis if $A=0$, $C=0$, and $B \neq 0$.

We found $A=0$ when $k=3$, and $B=5 \neq 0$. Let's check $C$ for $k=3$:

$C = k^2 - 7k + 6$

Substitute $k=3$:

$C = (3)^2 - 7(3) + 6$

$C = 9 - 21 + 6$

$C = -12 + 6$

$C = -6$

Since $C = -6 \neq 0$, the equation is $0x + 5y - 6 = 0$, or $5y = 6$, or $y = 6/5$. This is a horizontal line parallel to the x-axis, but not the x-axis itself.

The condition for being parallel to the x-axis is $A=0$ and $B \neq 0$. We found $k=3$ satisfies this condition.

Comparing the calculated value of $k$ with the given options:

(A) 3

(B) $\pm 2$ (If $k=\pm 2$, $B=0$, the line is vertical unless $A$ is also 0, which is not the case here. Vertical lines are perpendicular to the x-axis.)

(C) 4

(D) $-3$

The calculated value $k=3$ matches option (A).

Final Answer:

The value of $k$ for which the line is parallel to the x-axis is 3.

The correct option is (A) 3.

Given:

Line 1: $L_1: x+y-1=0$

Line 2: $L_2: 2x+3y-5=0$

Line 3: $L_3: x-y+1=0$

The required line passes through the intersection of $L_1$ and $L_2$ and is perpendicular to $L_3$.

To Find:

The equation of the required line.

Solution:

First, we find the point of intersection of lines $L_1$ and $L_2$. We can solve the system of equations:

$x+y-1=0 \quad$ ...(1)

$2x+3y-5=0 \quad$ ...(2)

From equation (1), we can express $x$ in terms of $y$ (or vice versa):

$x = 1-y \quad$ ...(3)

Substitute equation (3) into equation (2):

$2(1-y) + 3y - 5 = 0$

$2 - 2y + 3y - 5 = 0$

$y - 3 = 0$

$y = 3$

Now substitute the value of $y$ back into equation (3) to find $x$:

$x = 1 - 3$

$x = -2$

The point of intersection of $L_1$ and $L_2$ is $(-2, 3)$. Let this point be $P(-2, 3)$. The required line passes through this point.

Next, we find the slope of the line $L_3: x-y+1=0$. We rearrange this equation into the slope-intercept form $y = mx + c$:

$-y = -x - 1$

$y = x + 1$

The slope of line $L_3$ is $m_3 = 1$.

The required line is perpendicular to $L_3$. If $m_{req}$ is the slope of the required line, then the product of their slopes is -1 (since neither line is vertical or horizontal):

$m_3 \times m_{req} = -1$

$1 \times m_{req} = -1$

$m_{req} = -1$

The required line has a slope of -1 and passes through the point $(-2, 3)$. We can now use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$.

Substitute the point $(x_1, y_1) = (-2, 3)$ and the slope $m = -1$:

$y - 3 = -1(x - (-2))$

$y - 3 = -1(x + 2)$

$y - 3 = -x - 2$

Rearrange the terms to the general form $Ax + By + C = 0$:

$x + y - 3 + 2 = 0$

$x + y - 1 = 0$

This is the equation of the line passing through the intersection of $L_1$ and $L_2$ and perpendicular to $L_3$.

Comparing the derived equation $x+y-1=0$ with the given options:

(A) $x+y-3=0$

(B) $x-y+3=0$

(C) $x+y+3=0$

(D) $x-y-3=0$

The derived equation $x+y-1=0$ does not match any of the provided options.

Discussion on Options / Likely Typo:

The calculated intersection point $(-2, 3)$ and the required slope $-1$ are correct based on the problem statement. The resulting equation is $x+y-1=0$.

Let's examine the options with slope $-1$. These are (A) $x+y-3=0$ and (C) $x+y+3=0$.

If the equation was intended to be option (A) $x+y-3=0$, this line has the correct slope $(-1)$. It would pass through the intersection point if the intersection point satisfied $x+y-3=0$. For the calculated point $(-2, 3)$, we have $-2+3-3 = -2 \neq 0$.

Consider the possibility that there was a typo in the first line's equation, and it was meant to be $x+y-3=0$ instead of $x+y-1=0$. If we find the intersection of $x+y-3=0$ and $2x+3y-5=0$:

$x = 3-y$

$2(3-y) + 3y - 5 = 0$

$6 - 2y + 3y - 5 = 0$

$y + 1 = 0$

$y = -1$

$x = 3 - (-1) = 4$

The intersection point would be $(4, -1)$. A line through $(4, -1)$ with slope $-1$ is $y - (-1) = -1(x - 4) \implies y + 1 = -x + 4 \implies x + y - 3 = 0$. This matches option (A).

It is highly probable that the equation of the first line was mistyped and should have been $x+y-3=0$ to lead to option (A) as the correct answer.

Final Answer:

Based on the given equations, the calculated equation of the line is $x+y-1=0$, which is not among the options. However, given the multiple-choice format and the options provided, it appears there is a typo in the question. If the first line was $x+y-3=0$ instead of $x+y-1=0$, the correct answer would be $x+y-3=0$.

Assuming the likely intended question leading to one of the options:

The correct option (based on probable intended question) is (A) $x+y-3=0$.

Solution:

We are given two linear equations, and we need to find the point $(x, y)$ where these lines intersect. The point of intersection is the solution that satisfies both equations simultaneously.

The given equations are:

We can solve this system of equations using the elimination method. Notice that the coefficients of $y$ in the two equations are $-1$ and $+1$, which are opposite in sign.

Add equation (1) and equation (2):

Combine like terms:

Add 3 to both sides:

Divide by 3:

Now substitute the value of $x = 1$ into either equation (1) or (2) to solve for $y$. Let's use equation (2) as it is simpler:

Substitute $x=1$:

Combine constant terms:

Add 3 to both sides:

The solution to the system of equations is $x = 1$ and $y = 3$. Therefore, the point of intersection of the two lines is $(1, 3)$.

We can verify this by substituting $(1, 3)$ into both original equations:

For equation (1): $2x - y + 1 = 0$

For equation (2): $x + y - 4 = 0$

Since the point $(1, 3)$ satisfies both equations, it is the correct point of intersection.

Comparing this result with the given options, we find that option (A) is $(1, 3)$.

The point of intersection of the lines $2x - y + 1 = 0$ and $x + y - 4 = 0$ is $(1, 3)$.

Solution:

The given equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$. This is the intercept form of a linear equation, where $a$ is the x-intercept and $b$ is the y-intercept.

To find the points where the line intersects the coordinate axes, we set the other coordinate to zero.

Intersection with the x-axis: Set $y = 0$ in the equation $\frac{x}{a} + \frac{y}{b} = 1$.

$\frac{x}{a} + \frac{0}{b} = 1$

$\frac{x}{a} = 1$

$x = a$

The point of intersection with the x-axis is $(a, 0)$.

Intersection with the y-axis: Set $x = 0$ in the equation $\frac{x}{a} + \frac{y}{b} = 1$.

$\frac{0}{a} + \frac{y}{b} = 1$

$\frac{y}{b} = 1$

$y = b$

The point of intersection with the y-axis is $(0, b)$.

The triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ and the coordinate axes has its vertices at the origin $(0, 0)$, the x-intercept $(a, 0)$, and the y-intercept $(0, b)$.

This triangle is a right-angled triangle with the right angle at the origin $(0, 0)$.

The length of the base of the triangle along the x-axis is the distance from $(0, 0)$ to $(a, 0)$, which is $|a|$. We use the absolute value because length is always non-negative.

The height of the triangle along the y-axis is the distance from $(0, 0)$ to $(0, b)$, which is $|b|$. We use the absolute value because height is always non-negative.

The area of a triangle is given by the formula: Area = $\frac{1}{2} \times \text{base} \times \text{height}$.

Substituting the base and height of the triangle formed by the line and the axes:

Area $= \frac{1}{2} \times |a| \times |b|$

Area $= \frac{1}{2} |ab|$

Comparing this result with the given options, we find that the area matches option (B).

The area of the triangle formed by the line $\frac{x}{a} + \frac{y}{b} = 1$ with the coordinate axes is $\frac{1}{2}|ab|$.

Solution:

Let A and B be two fixed points and P be a variable point such that $PA^2 + PB^2 = k$, where $k$ is a constant.

Let M be the midpoint of the line segment AB.

According to Apollonius' Theorem, in $\triangle PAB$ with median PM to side AB:

We are given $PA^2 + PB^2 = k$.

Since M is the midpoint of AB, $AM = \frac{1}{2}AB$. Therefore, $AM^2 = (\frac{1}{2}AB)^2 = \frac{1}{4}AB^2$.

Substitute these into the Apollonius' Theorem equation:

Rearranging to solve for $PM^2$:

Since A and B are fixed points, AB is a constant length, so $AB^2$ is a constant. $k$ is also a given constant.

Therefore, $\frac{2k - AB^2}{4}$ is a constant value. Let this constant be $R^2$, where $R \ge 0$ for a real locus.

This shows that the distance of the point P from the fixed point M (the midpoint of AB) is a constant R.

The locus of a point that is at a constant distance from a fixed point is a circle.

The center of the circle is M, and the radius is R.

The locus of a point $P$ such that $PA^2 + PB^2 = k$ is a circle (or a point or empty set depending on $k$).

The correct option is (B) A circle.

Solution:

The equation of the given line is $x - \sqrt{3}y + 7 = 0$.

To find the angle made by the line with the positive x-axis, we need to find its slope. The slope of a line is related to the angle of inclination $\theta$ by the formula $m = \tan \theta$.

We can rewrite the equation of the line in the slope-intercept form, which is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

Starting with $x - \sqrt{3}y + 7 = 0$, isolate the term with $y$:

Divide both sides by $-\sqrt{3}$:

Separate the terms:

Comparing this equation with the slope-intercept form $y = mx + c$, we identify the slope $m$ as $\frac{1}{\sqrt{3}}$.

The slope $m$ is equal to the tangent of the angle $\theta$ the line makes with the positive x-axis:

We know that the angle whose tangent is $\frac{1}{\sqrt{3}}$ is $30^\circ$.

The angle made by the line $x - \sqrt{3}y + 7 = 0$ with the positive x-axis is $30^\circ$.

Comparing this result with the given options, we find that option (A) is $30^\circ$.

The correct option is (A) $30^\circ$.

Solution:

The given equation of the line is $x \cos \alpha + y \sin \alpha = p$.

Method 1: Using the Normal Form Property

The equation $x \cos \alpha + y \sin \alpha = p$ is already in the normal form of a linear equation. In this form, the constant term $p$ represents the perpendicular distance of the line from the origin $(0, 0)$, and $\alpha$ is the angle that the normal (perpendicular from the origin to the line) makes with the positive x-axis.

The distance from the origin to the line $x \cos \alpha + y \sin \alpha = p$ is given by $|p|$. We use the absolute value because distance is always non-negative.

Method 2: Using the General Distance Formula

The general formula for the distance of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is:

We can rewrite the given line equation $x \cos \alpha + y \sin \alpha = p$ in the form $Ax + By + C = 0$ by moving the constant term to the left side:

Here, we have $A = \cos \alpha$, $B = \sin \alpha$, and $C = -p$. The point is the origin, so $(x_0, y_0) = (0, 0)$.

Substitute these values into the distance formula:

Using the fundamental trigonometric identity $\cos^2 \alpha + \sin^2 \alpha = 1$, we substitute this into the denominator:

Since the absolute value of a negative number is the corresponding positive number, $|-p| = |p|$.

Both methods confirm that the distance of the origin from the line $x \cos \alpha + y \sin \alpha = p$ is $|p|$.

Let's examine the given options:

(A) $|p|$ - This matches our result.

(B) $|p| / \sqrt{\cos^2 \alpha + \sin^2 \alpha} = |p|$ - This expression is equal to $|p|$ because $\sqrt{\cos^2 \alpha + \sin^2 \alpha} = \sqrt{1} = 1$. This option presents the result derived from the distance formula and simplifies it, also arriving at $|p|$. While mathematically correct and equal to the distance, option (A) is the most direct and simplified expression for the distance.

(C) $\sqrt{\cos^2 \alpha + \sin^2 \alpha} = 1$ - This is not the distance from the origin (unless $|p|=1$).

(D) 0 - This would mean the line passes through the origin, which is only true if $p=0$.

Based on the standard representation of the distance from the origin in the normal form of a line and the simplified result from the distance formula, the distance is $|p|$. Option (A) provides this direct value.

The correct option is (A) $|p|$.

Solution:

The general equation of a straight line is given by $Ax + By + C = 0$.

We are given that $B = 0$. Substitute this value into the general equation:

For this equation to represent a straight line, at least one of the coefficients A or B must be non-zero. Since we are given $B=0$, it implies that $A \neq 0$.

If $A \neq 0$, we can solve the equation $Ax + C = 0$ for $x$:

Let $k = -\frac{C}{A}$. Since A and C are constants (and $A \neq 0$), $k$ is a constant.

The equation of the line simplifies to $x = k$.

An equation of the form $x = k$ represents a vertical line in the Cartesian coordinate system. All points on this line have the same x-coordinate $k$, regardless of their y-coordinate.

A vertical line is parallel to the y-axis and perpendicular to the x-axis.

Let's examine the given options:

(A) Parallel to the x-axis: A line parallel to the x-axis is horizontal and has the equation $y = \text{constant}$ (which occurs when $A=0, B \neq 0$). This is incorrect.

(B) Parallel to the y-axis: A line parallel to the y-axis is vertical and has the equation $x = \text{constant}$ (which occurs when $B=0, A \neq 0$). This is correct.

(C) Passing through the origin: A line passes through the origin $(0, 0)$ if $A(0) + B(0) + C = 0$, which simplifies to $C = 0$. When $B=0$, the equation is $Ax + C = 0$. This line passes through the origin only if $C=0$ (in which case it is the y-axis, $x=0$). If $C \neq 0$, it does not pass through the origin. So, this is not generally true when $B=0$. Incorrect.

(D) Perpendicular to the x-axis: A line perpendicular to the x-axis is a vertical line. This is the same as being parallel to the y-axis. This is also a correct geometric description.

Both options (B) and (D) describe a vertical line, which is the correct type of line when $B=0$ (and $A \neq 0$). In standard conventions when interpreting $Ax+By+C=0$, a zero coefficient for a variable indicates parallelism to the axis of that variable. The coefficient of $y$ is zero ($B=0$), so the line is parallel to the y-axis.

The line is parallel to the y-axis and also perpendicular to the x-axis. Option (B) directly states the parallelism to the y-axis based on the zero coefficient of y.

The correct option is (B) Parallel to the y-axis.

Solution:

We are given that the line passes through the point $(1, -2)$ and has an undefined slope.

Recall that the slope of a straight line is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$ for two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ on the line.

A line has an undefined slope when the denominator of the slope formula is zero, i.e., $x_2 - x_1 = 0$. This means $x_1 = x_2$.

If $x_1 = x_2$ for any two distinct points on the line, it means that all points on the line have the same x-coordinate.

A line where all points have the same x-coordinate is a vertical line.

The equation of a vertical line is of the form $x = c$, where $c$ is the constant x-coordinate of every point on the line.

We are given that the line passes through the point $(1, -2)$. This means that the x-coordinate of this point, which is 1, must be the constant x-coordinate for all points on the line.

Therefore, the equation of the line is $x = 1$.

Let's check the options:

(A) $x = 1$: This is the equation of a vertical line passing through points with x-coordinate 1. The point $(1, -2)$ has an x-coordinate of 1, so it lies on this line. This matches our result.

(B) $y = -2$: This is the equation of a horizontal line passing through points with y-coordinate -2. It has a slope of 0, not undefined.

(C) $x = -2$: This is the equation of a vertical line passing through points with x-coordinate -2. The point $(1, -2)$ does not lie on this line.

(D) $y = 1$: This is the equation of a horizontal line passing through points with y-coordinate 1. It has a slope of 0, not undefined.

The equation of the line passing through $(1, -2)$ and having an undefined slope is $x = 1$.

The correct option is (A) $x = 1$.

Solution:

Assertion (A): The point $(0, 0)$ lies on the line $2x + 3y = 0$.

To check if a point lies on a line, we substitute the coordinates of the point into the equation of the line. If the equation holds true, the point lies on the line.

Substitute $x = 0$ and $y = 0$ into the equation $2x + 3y = 0$:

Since the equation $0 = 0$ is a true statement, the point $(0, 0)$ lies on the line $2x + 3y = 0$.

Thus, Assertion (A) is True.

Reason (R): Substituting $(0, 0)$ into the equation gives $2(0) + 3(0) = 0$, which is a true statement.

This statement describes the process of checking if the point $(0, 0)$ lies on the line and correctly performs the substitution, arriving at $2(0) + 3(0) = 0$, which simplifies to $0 = 0$. The statement correctly concludes that this result is a true statement.

Thus, Reason (R) is True.

Now we evaluate if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the point $(0, 0)$ lies on the line. Reason (R) explains that substituting the coordinates of $(0, 0)$ into the line's equation results in a true statement ($0=0$). This is the standard method used to verify if a point lies on a line, and the fact that it results in a true statement is precisely why the point lies on the line.

Therefore, Reason (R) is the correct explanation for Assertion (A).

Based on the analysis:

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

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

Solution:

We are given two linear equations:

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

The slope of a line in the general form $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$ (provided $B \neq 0$).

For line (1), $2x - py + 1 = 0$, we have $A_1 = 2$ and $B_1 = -p$.

The slope of line (1) is $m_1 = -\frac{A_1}{B_1} = -\frac{2}{-p}$.

For line (2), $3x + 2y - 5 = 0$, we have $A_2 = 3$ and $B_2 = 2$.

The slope of line (2) is $m_2 = -\frac{A_2}{B_2} = -\frac{3}{2}$.

For the lines to be perpendicular, the product of their slopes must be $-1$:

Substitute the slopes we found:

Multiply the fractions:

Simplify the fraction on the left side:

Multiply both sides by $-1$:

Multiply both sides by $p$:

The value of $p$ that makes the lines perpendicular is 3.

We assumed $p \neq 0$. If $p=0$, line (1) becomes $2x + 1 = 0$, or $x = -\frac{1}{2}$, which is a vertical line (undefined slope). Line (2) has slope $m_2 = -\frac{3}{2}$. A vertical line is perpendicular to a line with a defined, non-zero slope. However, $p=0$ is not among the options. Our derived value $p=3$ is in the options and satisfies the condition $m_1 \cdot m_2 = -1$.

The correct option is (A) 3.

Solution:

The given equation of the line is $4x - 5y = 20$.

The intercept form of a straight line is given by $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.

To convert the given equation into the intercept form, we need to make the right-hand side of the equation equal to 1.

Divide both sides of the equation $4x - 5y = 20$ by 20:

Separate the terms on the left side:

Simplify the fractions:

To match the standard intercept form $\frac{x}{a} + \frac{y}{b} = 1$, we rewrite the equation as:

Comparing this with $\frac{x}{a} + \frac{y}{b} = 1$, we find that the x-intercept is $a = 5$ and the y-intercept is $b = -4$.

The intercept form of the line $4x - 5y = 20$ is $\frac{x}{5} + \frac{y}{-4} = 1$.

Comparing this result with the given options, we find that option (A) matches our derived form.

The correct option is (A) $\frac{x}{5} + \frac{y}{-4} = 1$.

Solution:

The incenter of a triangle is the point where the angle bisectors of the triangle intersect. It is also the center of the incircle (the circle inscribed within the triangle that is tangent to all three sides).

The coordinates of the incenter of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, and with side lengths $a$, $b$, and $c$ opposite to the vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ respectively, are given by the formula:

Let's check the given options:

(A) $(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c})$: This formula matches the standard formula for the coordinates of the incenter, where $a, b, c$ are the lengths of the sides opposite to the vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ respectively.

(B) $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$: This is the formula for the coordinates of the centroid of a triangle, not the incenter.

(C) $(\frac{ax_2+bx_3+cx_1}{a+b+c}, \frac{ay_2+by_3+cy_1}{a+b+c})$: This formula incorrectly pairs the side lengths with the vertex coordinates.

(D) $(\frac{ax_1+by_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c})$: This formula also incorrectly pairs side lengths and vertex coordinates in the numerator, specifically in the y-coordinate expression.

Based on the standard formula, option (A) correctly represents the coordinates of the incenter.

The correct option is (A) $(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c})$.

Solution:

Shifting the origin is a type of coordinate transformation where the axes are translated to a new origin point while remaining parallel to the original axes. If the original coordinates of a point are $(x, y)$ and the new origin is at $(h, k)$ in the original coordinate system, then the new coordinates $(X, Y)$ of the point are related by the formulas:

$X = x - h$

$Y = y - k$

Alternatively, the old coordinates can be expressed in terms of the new coordinates and the shift as:

$x = X + h$

$y = Y + k$

Let's analyze each statement:

(A) Shifting the origin changes the shape of the geometric figure.

When we shift the origin, we are essentially moving the entire coordinate system without stretching, shrinking, or rotating it. This type of transformation is an isometry, which preserves distances and angles. Consequently, the shape and size of any geometric figure remain unchanged. The equation representing the figure will change in terms of the new coordinates, but the figure itself is congruent to the original one.

For example, the equation of a circle $x^2 + y^2 = r^2$ centered at the origin $(0,0)$ in the old system becomes $(X+h)^2 + (Y+k)^2 = r^2$ in the new system, which is a circle centered at $(-h, -k)$ in the new system, but its shape and radius remain the same.

Therefore, this statement is False.

(B) Shifting the origin changes the distance between two points.

Let two points be $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ in the original system. The distance between them is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

In the new system, the coordinates are $P_1(X_1, Y_1)$ and $P_2(X_2, Y_2)$, where $X_1 = x_1 - h$, $Y_1 = y_1 - k$, $X_2 = x_2 - h$, $Y_2 = y_2 - k$.

The distance between them in the new system is $D = \sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2}$.

Substitute the new coordinates:

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

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

So, $D = d$. The distance between two points remains the same after shifting the origin.

Therefore, this statement is False.

(C) Shifting the origin changes the coordinates of the points.

As shown by the transformation formulas, $X = x - h$ and $Y = y - k$. If the origin is actually shifted (i.e., $h \neq 0$ or $k \neq 0$), then for a point $(x, y)$, its new coordinates $(X, Y)$ will be different unless $x=h$ and $y=k$ (the new origin itself). For any point other than the new origin, its coordinate values will change.

For example, if we shift the origin from $(0,0)$ to $(2,3)$, the point originally at $(5,4)$ will have new coordinates $X = 5 - 2 = 3$ and $Y = 4 - 3 = 1$, so its new coordinates are $(3,1)$. The original coordinates $(5,4)$ are different from the new coordinates $(3,1)$.

Therefore, this statement is True.

(D) Shifting the origin changes the slope of a line.

The slope of a line passing through two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

In the new system, the slope is $M = \frac{Y_2 - Y_1}{X_2 - X_1}$.

Using the transformation formulas:

$M = \frac{(y_2 - k) - (y_1 - k)}{(x_2 - h) - (x_1 - h)}$

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

So, $M = m$. The slope of a line remains the same after shifting the origin.

Therefore, this statement is False.

Based on the analysis of each statement, only statement (C) is true.

The correct option is (C) Shifting the origin changes the coordinates of the points.

Solution:

Let the two points be $A = (x_1, y_1)$ and $B = (x_2, y_2)$.

Let the point $P = (h, k)$ divide the line segment AB externally in the ratio $m:n$.

The formula for the coordinates of a point $(x, y)$ that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ externally in the ratio $m:n$ is given by:

We are asked for the value of $h$, which is the x-coordinate of the point $(h, k)$.

Using the formula for the x-coordinate:

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

(A) $\frac{mx_2 - nx_1}{m-n}$: This matches our derived formula for $h$.

(B) $\frac{mx_2 + nx_1}{m+n}$: This is the formula for internal division.

(C) $\frac{nx_1 - mx_2}{m-n}$: This formula has the numerator terms in a different order with signs flipped compared to the standard external division formula for $x$.

(D) $\frac{nx_1 + mx_2}{m+n}$: This is related to the internal division formula but with the ratio terms associated with different points.

The formula for the x-coordinate $h$ when the point divides the segment externally in the ratio $m:n$ is $\frac{mx_2 - nx_1}{m-n}$.

The correct option is (A) $\frac{mx_2 - nx_1}{m-n}$.

Solution:

We are looking for the general equation of a straight line that passes through the origin $(0, 0)$.

The general equation of a straight line is $Ax + By + C = 0$, where A, B, and C are constants, and A and B are not both zero.

If a point lies on a line, its coordinates must satisfy the equation of the line.

For the line to pass through the origin $(0, 0)$, the coordinates $x = 0$ and $y = 0$ must satisfy the equation $Ax + By + C = 0$.

Substitute $x = 0$ and $y = 0$ into the equation:

This shows that if a line passes through the origin, the constant term C in its general equation must be zero.

Therefore, the equation of a line passing through the origin is of the form $Ax + By + 0 = 0$, which simplifies to $Ax + By = 0$.

Let's check the given options:

(A) $Ax + By + C = 0$ where $C \neq 0$: As shown, if the line passes through the origin, $C$ must be 0. So, this form represents lines that do not pass through the origin. Incorrect.

(B) $Ax + By = 0$: This form has the constant term $C=0$. As shown, substituting $(0, 0)$ gives $A(0) + B(0) = 0$, which is $0=0$, a true statement. This form represents all lines passing through the origin. Correct.

(C) $y = mx + c$ where $c \neq 0$: This is the slope-intercept form, where $c$ is the y-intercept. If $c \neq 0$, the line crosses the y-axis at $(0, c)$, which is not the origin. For a line to pass through the origin, the y-intercept must be 0, so $c=0$. Incorrect.

(D) $\frac{x}{a} + \frac{y}{b} = 1$ where $a \neq 0, b \neq 0$: This is the intercept form. Substituting $(0, 0)$ gives $\frac{0}{a} + \frac{0}{b} = 1$, which simplifies to $0 = 1$, a false statement. This form represents lines that have non-zero intercepts on both axes, hence they do not pass through the origin. Incorrect.

The equation of a line passing through the origin is of the form $Ax + By = 0$.

The correct option is (B) $Ax + By = 0$.

Solution:

Let the two given points be $P_1 = (x_1, y_1) = (a \cos \theta, a \sin \theta)$ and $P_2 = (x_2, y_2) = (a \cos \phi, a \sin \phi)$.

The slope $m$ of the line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

Substitute the coordinates of $P_1$ and $P_2$ into the slope formula:

Factor out $a$ from the numerator and the denominator:

Assuming $a \neq 0$ and the points are distinct (so the denominator is not zero), we can cancel $a$:

Now, we use the sum-to-product trigonometric identities:

Apply these identities with $A = \phi$ and $B = \theta$:

Substitute these expressions back into the slope formula:

Assuming the points are distinct, $\phi \neq \theta + 2k\pi$ for any integer $k$. This implies $\frac{\phi-\theta}{2} \neq k\pi$, so $\sin \left(\frac{\phi-\theta}{2}\right) \neq 0$. We can cancel the term $2 \sin \left(\frac{\phi-\theta}{2}\right)$ from the numerator and denominator:

Using the definition of the cotangent function, $\cot x = \frac{\cos x}{\sin x}$:

Since addition is commutative, $\phi+\theta = \theta+\phi$.

Comparing this result with the given options, we find that option (D) matches our derived slope.

The correct option is (D) $-\cot(\frac{\theta + \phi}{2})$.

Solution:

We are given the coordinates of a point on the line $(x_1, y_1) = (2, 3)$.

We are also given the inclination of the line $\theta = 45^\circ$.

The slope $m$ of a line with inclination $\theta$ is given by $m = \tan \theta$.

Substitute the given inclination:

We know that $\tan(45^\circ) = 1$.

The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ can be found using the point-slope form:

Substitute the point $(2, 3)$ and the slope $m = 1$ into the formula:

Simplify the equation:

To convert this to the general form $Ax + By + C = 0$, move all terms to one side:

Now, let's check the given options:

(A) $y - 3 = 1(x - 2) \implies y - 3 = x - 2 \implies x - y + 1 = 0$: This matches the equation we derived.

(B) $y - 2 = 1(x - 3)$: This equation represents a line passing through $(3, 2)$ with slope 1. It does not pass through $(2, 3)$.

(C) $x + y - 5 = 0$: The slope of this line is $m = -\frac{A}{B} = -\frac{1}{1} = -1$. The inclination is $\arctan(-1) = 135^\circ$ (or $-45^\circ$), not $45^\circ$.

(D) $x - y - 1 = 0$: The slope of this line is $m = -\frac{A}{B} = -\frac{1}{-1} = 1$. However, let's check if it passes through $(2, 3)$: $2 - 3 - 1 = -2 \neq 0$. So, $(2, 3)$ does not lie on this line.

The equation of the line passing through $(2, 3)$ and having an inclination of $45^\circ$ is $x - y + 1 = 0$, which is derived from the equation in option (A).

The correct option is (A) $y - 3 = 1(x - 2) \implies y - 3 = x - 2 \implies x - y + 1 = 0$.

Solution:

The formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

For this formula to be valid and yield a well-defined value for $\tan \theta$, the following conditions must be met:

1. The slopes $m_1$ and $m_2$ must be defined. The slope of a line is defined as $m = \tan \alpha$, where $\alpha$ is the angle of inclination. The tangent function is undefined for angles of $90^\circ$ or $270^\circ$. Lines with these inclinations are vertical lines. Therefore, for the slopes $m_1$ and $m_2$ to be defined finite values, neither line can be vertical.

2. The denominator of the fraction, $1 + m_1 m_2$, must not be equal to zero, as division by zero is undefined. So, we must have $1 + m_1 m_2 \neq 0$, which means $m_1 m_2 \neq -1$.

The condition $m_1 m_2 = -1$ (when $m_1$ and $m_2$ are defined) is the condition for the two lines to be perpendicular. If the lines are perpendicular, the angle between them is $\theta = 90^\circ$. The value of $\tan(90^\circ)$ is undefined. Thus, the formula is not directly applicable in the case of perpendicular lines.

Therefore, the formula $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$ is valid when both slopes $m_1$ and $m_2$ are defined (neither line is vertical) and the denominator $1 + m_1 m_2$ is not zero (the lines are not perpendicular). The absolute value is used to give the tangent of the acute angle between the lines.

Let's analyze the options:

(A) Both lines are parallel. If $m_1=m_2$, the formula gives $\tan \theta = 0$, which implies $\theta = 0^\circ$. The formula works, but this option doesn't state the necessary general conditions for the formula's validity (e.g., that slopes must be defined).

(B) Both lines are perpendicular. If $m_1 m_2 = -1$, the denominator is 0, making the expression undefined. The formula is not valid in this specific form to calculate $\tan 90^\circ$.

(C) Neither line is vertical and $1 + m_1 m_2 \neq 0$. This condition ensures that $m_1$ and $m_2$ are defined, and the denominator $1 + m_1 m_2$ is non-zero. These are precisely the requirements for the expression $\frac{m_1 - m_2}{1 + m_1 m_2}$ to be a defined real number, allowing the calculation of $\tan \theta$.

(D) At least one line is vertical. If a line is vertical, its slope is undefined, so the formula cannot be applied directly using numerical values for $m_1$ and $m_2$.

The conditions required for the formula $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$ to be valid are that the slopes $m_1$ and $m_2$ are defined and the denominator $1 + m_1 m_2$ is not zero.

The correct option is (C) Neither line is vertical and $1 + m_1 m_2 \neq 0$.

Solution:

The distance of a point from the x-axis is the length of the perpendicular segment from the point to the x-axis.

Let the point be $P(x_1, y_1)$. The x-axis is the line where the y-coordinate is always 0. The equation of the x-axis is $y = 0$.

Consider the point $P(x_1, y_1)$. The point on the x-axis that is closest to $P$ is the point with the same x-coordinate but a y-coordinate of 0. Let this point on the x-axis be $Q(x_1, 0)$.

The distance between $P(x_1, y_1)$ and $Q(x_1, 0)$ is given by the distance formula:

The square root of $y_1^2$ is $|y_1|$, which is the absolute value of $y_1$. We use the absolute value because distance is always a non-negative quantity.

Alternatively, we can use the formula for the distance of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$, which is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

The x-axis is the line $y = 0$, which can be written as $0x + 1y + 0 = 0$. Here, $A=0$, $B=1$, $C=0$. The point is $(x_1, y_1)$.

The distance is:

The distance of the point $(x_1, y_1)$ from the x-axis is the absolute value of its y-coordinate.

Comparing this with the options:

(A) $x_1$: This is the x-coordinate.

(B) $y_1$: This is the y-coordinate, which can be negative.

(C) $|x_1|$: This is the absolute value of the x-coordinate (distance from y-axis).

(D) $|y_1|$: This is the absolute value of the y-coordinate (distance from x-axis).

The distance of a point $(x_1, y_1)$ from the x-axis is $|y_1|$.

Solution:

We are given two linear equations in slope-intercept form:

Here, $m_1$ is the slope of the first line, and $c_1$ is its y-intercept. Similarly, $m_2$ is the slope of the second line, and $c_2$ is its y-intercept.

Two lines are parallel if and only if they have the same slope.

So, the condition for the two lines to be parallel is $m_1 = m_2$.

Let's consider the role of the y-intercepts, $c_1$ and $c_2$:

• If $m_1 = m_2$ and $c_1 = c_2$, the two equations are identical. This means the lines are coincident (they are the same line). Coincident lines are considered a special case of parallel lines (they are parallel and share all points).
• If $m_1 = m_2$ and $c_1 \neq c_2$, the lines have the same slope but different y-intercepts. This means they never intersect and are strictly parallel (non-coincident).

The question asks for the condition for the lines to be parallel. The fundamental condition is having the same slope ($m_1 = m_2$). However, the options provide conditions involving both slopes and intercepts.

Let's examine the options:

(A) $m_1 = m_2$ and $c_1 = c_2$: This is the condition for the lines to be coincident.

(B) $m_1 = m_2$ and $c_1 \neq c_2$: This is the condition for the lines to be parallel and distinct.

(C) $m_1 m_2 = -1$: This is the condition for the lines to be perpendicular (assuming neither slope is zero or undefined).

(D) $m_1 = -m_2$: This is not a general condition for parallel or perpendicular lines.

In the context of multiple-choice questions distinguishing between coincident and non-coincident cases, the term "parallel" often implicitly refers to the case of distinct parallel lines, where the slopes are equal but the intercepts are different. Option (B) precisely describes this situation.

The lines $y=m_1x+c_1$ and $y=m_2x+c_2$ are parallel if their slopes are equal. If they also have different y-intercepts, they are distinct parallel lines. Option (B) gives this specific condition for distinct parallel lines.

The correct option is (B) $m_1 = m_2$ and $c_1 \neq c_2$.

Solution:

Let the equations of the two lines be $L_1 = 0$ and $L_2 = 0$. These can be written in the general form as:

We are looking for the equation of the family of lines that all pass through the point of intersection of $L_1 = 0$ and $L_2 = 0$.

Let $(x_0, y_0)$ be the point of intersection of the lines $L_1=0$ and $L_2=0$. Since $(x_0, y_0)$ lies on both lines, it satisfies both equations:

Consider the equation formed by a linear combination of $L_1$ and $L_2$ with a parameter $\lambda$:

Substituting the expressions for $L_1$ and $L_2$:

Rearranging the terms:

This is the equation of a straight line for any real value of $\lambda$, provided that $(A_1 + \lambda A_2)$ and $(B_1 + \lambda B_2)$ are not both zero.

Now, let's check if the point of intersection $(x_0, y_0)$ lies on this new line. Substitute $(x_0, y_0)$ into the equation $L_1 + \lambda L_2 = 0$:

Since the point $(x_0, y_0)$ satisfies the equation $L_1 + \lambda L_2 = 0$ for any value of $\lambda$, every line represented by this equation passes through the intersection point of $L_1 = 0$ and $L_2 = 0$.

Thus, $L_1 + \lambda L_2 = 0$ represents the family of lines passing through the intersection of $L_1 = 0$ and $L_2 = 0$. The parameter $\lambda$ takes on different real values to represent different lines in the family. Note that this family includes the line $L_1=0$ (when $\lambda=0$), but it does not include the line $L_2=0$ (unless we consider the limit as $\lambda \to \infty$ or use the form $\mu L_1 + \lambda L_2 = 0$). However, option (C) is the standard representation of the family.

Comparing this with the given options:

(A) $L_1 + L_2 = 0$: This is one specific line from the family ($\lambda = 1$).

(B) $L_1 \cdot L_2 = 0$: This equation is satisfied if either $L_1=0$ or $L_2=0$. This represents the two original lines, not a family of lines passing through their intersection.

(C) $L_1 + \lambda L_2 = 0$, where $\lambda$ is a parameter: This is the correct form for the family of lines passing through the intersection.

(D) $L_1 = \lambda L_2$: This can be written as $L_1 - \lambda L_2 = 0$, which is equivalent to $L_1 + (-\lambda) L_2 = 0$. Letting $\mu = -\lambda$, this is $L_1 + \mu L_2 = 0$, which is the same form as (C).

Option (C) is the standard representation of the family of lines passing through the intersection of $L_1 = 0$ and $L_2 = 0$.

The correct option is (C) $L_1 + \lambda L_2 = 0$, where $\lambda$ is a parameter.

Solution:

Let the two points be $A = (2, 3)$ and $B = (4, 5)$.

A perpendicular bisector of a line segment is a line that is perpendicular to the segment and passes through its midpoint.

Step 1: Find the midpoint of the line segment AB.

The coordinates of the midpoint M of a segment joining $(x_1, y_1)$ and $(x_2, y_2)$ are given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

Midpoint $M = \left(\frac{2+4}{2}, \frac{3+5}{2}\right)$

$M = \left(\frac{6}{2}, \frac{8}{2}\right)$

$M = (3, 4)$

Step 2: Find the slope of the line segment AB.

The slope $m_{AB}$ of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y_2 - y_1}{x_2 - x_1}$.

$m_{AB} = \frac{5 - 3}{4 - 2}$

$m_{AB} = \frac{2}{2}$

$m_{AB} = 1$

Step 3: Find the slope of the perpendicular bisector.

If two lines are perpendicular (and neither is vertical or horizontal), the product of their slopes is $-1$. Let $m_{\perp}$ be the slope of the perpendicular bisector.

$m_{AB} \cdot m_{\perp} = -1$

$1 \cdot m_{\perp} = -1$

$m_{\perp} = -1$

Step 4: Find the equation of the perpendicular bisector.

We have the slope $m_{\perp} = -1$ and a point on the line, the midpoint $(3, 4)$. We can use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

Substitute $(x_1, y_1) = (3, 4)$ and $m = -1$:

Simplify the equation:

Rearrange the terms to get the general form $Ax + By + C = 0$:

Comparing the obtained equation with the given options:

(A) $x+y-7=0$: This matches our equation.

(B) $x-y+1=0$: The slope is $1$, not $-1$.

(C) $x+y-1=0$: The slope is $-1$, but checking the midpoint $(3,4)$: $3+4-1 = 6 \neq 0$, so it doesn't pass through the midpoint.

(D) $x-y-1=0$: The slope is $1$, not $-1$.

The equation of the perpendicular bisector of the line segment joining $(2, 3)$ and $(4, 5)$ is $x + y - 7 = 0$.

The correct option is (A) $x+y-7=0$.

Solution:

The vertices of the triangle are given as $A(0, 0)$, $B(2, 0)$, and $C(0, 3)$.

Let's examine the coordinates of the vertices:

• The line segment AB connects $(0, 0)$ and $(2, 0)$. This segment lies on the x-axis. Its length is the distance between $(0,0)$ and $(2,0)$, which is $\sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2$.
• The line segment AC connects $(0, 0)$ and $(0, 3)$. This segment lies on the y-axis. Its length is the distance between $(0,0)$ and $(0,3)$, which is $\sqrt{(0-0)^2 + (3-0)^2} = \sqrt{9} = 3$.

Since the segment AB lies on the x-axis and the segment AC lies on the y-axis, and the x-axis and y-axis are perpendicular to each other, the angle between the sides AB and AC at vertex A is $90^\circ$.

Thus, the triangle ABC is a right-angled triangle with the right angle at the vertex A $(0, 0)$.

The orthocenter of a triangle is the point where the three altitudes of the triangle intersect.

In a right-angled triangle, the two legs (the sides forming the right angle) are themselves altitudes.

• The altitude from vertex C to side AB is the line segment AC (since AC is perpendicular to AB). The line containing this altitude is the y-axis, with the equation $x=0$.
• The altitude from vertex B to side AC is the line segment AB (since AB is perpendicular to AC). The line containing this altitude is the x-axis, with the equation $y=0$.

The intersection of the altitudes from B and C is the intersection of the lines $x=0$ and $y=0$, which is the point $(0, 0)$.

The third altitude, from the right-angle vertex A to the hypotenuse BC, must also pass through this intersection point.

Therefore, the orthocenter of a right-angled triangle is located at the vertex where the right angle is formed.

In this triangle, the right angle is at the vertex $(0, 0)$.

So, the orthocenter of the triangle with vertices $(0, 0), (2, 0), (0, 3)$ is $(0, 0)$.

Comparing this result with the given options:

(A) $(1, 1.5)$ - This is the midpoint of the hypotenuse (circumcenter in a right triangle) or the centroid $(\frac{0+2+0}{3}, \frac{0+0+3}{3}) = (\frac{2}{3}, 1)$.

(B) $(0, 0)$ - This matches our result.

(C) $(2, 3)$ - These are coordinates of other vertices.

(D) $(0, 0)$ (Since it's a right-angled triangle at the origin) - This matches our result and provides the correct reasoning.

The correct option is (D) $(0, 0)$ (Since it's a right-angled triangle at the origin).

Solution:

The general equation of a straight line is $Ax + By + C = 0$. We analyze each statement:

(A) If $C=0$, the line passes through the origin. Substituting $(0,0)$: $A(0) + B(0) + C = 0 \implies C = 0$. If $C=0$, the equation is satisfied by $(0,0)$. This is True.

(B) If $A=0$, the line is parallel to the x-axis. The equation becomes $By + C = 0$. If $B \neq 0$, this is $y = -C/B$, a horizontal line parallel to the x-axis. If $A=0, B=0$, it's not a line. Assuming it's a line ($B \neq 0$), this is True.

(C) If $B=0$, the line is parallel to the y-axis. The equation becomes $Ax + C = 0$. If $A \neq 0$, this is $x = -C/A$, a vertical line parallel to the y-axis. If $A=0, B=0$, it's not a line. Assuming it's a line ($A \neq 0$), this is True.

(D) The slope of the line is $-A/B$ (if $B \neq 0$). Rearranging $Ax + By + C = 0$ for $y$ (when $B \neq 0$): $By = -Ax - C \implies y = -\frac{A}{B}x - \frac{C}{B}$. The slope is the coefficient of $x$, which is $-\frac{A}{B}$. This is True.

All four statements (A), (B), (C), and (D) are mathematically true statements regarding the line $Ax + By + C = 0$ under the condition that $A$ and $B$ are not both zero.

Since the question asks for a false statement and all options appear to be true, there is likely an error in the question or the provided options.

Based on standard mathematical definitions, none of the statements are false.

Solution:

We are given two lines:

We can find the slopes of these lines by rearranging them into the slope-intercept form $y = mx + c$, where $m$ is the slope.

For Line 1: $x - y = 0 \implies y = x$.

The slope of Line 1 is $m_1 = 1$.

For Line 2: $x + y = 0 \implies y = -x$.

The slope of Line 2 is $m_2 = -1$.

Now, we check the relationship between the slopes $m_1$ and $m_2$. Calculate their product:

When the product of the slopes of two non-vertical lines is $-1$, the lines are perpendicular.

Perpendicular lines intersect at an angle of $90^\circ$.

Alternatively, we can use the formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:

Substitute $m_1 = 1$ and $m_2 = -1$:

The expression $\frac{2}{0}$ is undefined. The tangent of an angle is undefined for angles of $90^\circ$ (and $270^\circ$, etc.).

Thus, $\theta = 90^\circ$.

The angle between the lines $x - y = 0$ and $x + y = 0$ is $90^\circ$.

Comparing this result with the given options, we find that option (C) is $90^\circ$.

The correct option is (C) $90^\circ$.

Solution:

Let the equation of the line be in the intercept form:

where $a$ is the x-intercept and $b$ is the y-intercept.

We are given that the line makes equal intercepts on the axes. This means the x-intercept and the y-intercept are equal.

Substitute $b=a$ into equation (1):

Multiply both sides by $a$ to simplify (assuming $a \neq 0$, otherwise it wouldn't be an intercept form):

We are also given that the line passes through the point $(1, 2)$. This means that the coordinates of this point must satisfy the equation of the line.

Substitute $x=1$ and $y=2$ into equation (2):

Solve for $a$:

Now substitute the value of $a$ back into equation (2) to get the equation of the line:

This equation represents a line passing through $(1, 2)$ and having an x-intercept of 3 (set $y=0$) and a y-intercept of 3 (set $x=0$).

Compare the derived equation with the given options:

(A) $x+y=3$: Matches our equation.

(B) $x-y=-1$: Check point $(1,2)$: $1-2 = -1$. The point lies on this line. Check intercepts: $x$-intercept (set $y=0$) is $x=-1$, $y$-intercept (set $x=0$) is $-y=-1 \implies y=1$. Intercepts are $-1$ and $1$, which are not equal.

(C) $x+y=1$: Check point $(1,2)$: $1+2=3 \neq 1$. Point does not lie on this line.

(D) $x-y=1$: Check point $(1,2)$: $1-2=-1 \neq 1$. Point does not lie on this line.

The equation of the line passing through $(1, 2)$ and making equal intercepts on the axes is $x + y = 3$.

The correct option is (A) $x+y=3$.

Solution:

Let the two points be $P_1(2, -3)$ and $P_2(5, 6)$.

Let the x-axis divide the line segment joining $P_1$ and $P_2$ at the point $P(x, y)$.

Since the point P lies on the x-axis, its y-coordinate must be 0. So, $P = (x, 0)$.

Let the ratio in which the point $P$ divides the line segment $P_1P_2$ be $m:n$.

Using the section formula for the y-coordinate, the coordinates of the point P that divides the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ is given by:

In this case, $(x_1, y_1) = (2, -3)$, $(x_2, y_2) = (5, 6)$, and the y-coordinate of the dividing point is $y = 0$.

Substitute these values into the formula:

Multiply both sides by $(m+n)$ (assuming $m+n \neq 0$):

Rearrange the equation to find the ratio $\frac{m}{n}$:

So, the ratio $m:n$ is $1:2$.

Since the ratio $\frac{m}{n}$ is positive, the division is internal.

The x-axis divides the line segment joining $(2, -3)$ and $(5, 6)$ in the ratio $1:2$ internally.

Compare the result with the given options:

(A) 1 : 2 internally: Matches our result.

(B) 1 : 2 externally: The ratio is correct, but the type of division is incorrect.

(C) 2 : 1 internally: The ratio is incorrect.

(D) 2 : 1 externally: The ratio and the type of division are incorrect.

The correct option is (A) 1 : 2 internally.

Solution:

We are asked to find the locus of a point that is equidistant from two intersecting lines.

Let the two intersecting lines be $L_1$ and $L_2$. These lines intersect at a point, and they form four angles at this point of intersection.

Consider a point P. The distance from P to a line is defined as the length of the perpendicular segment from P to the line.

The locus of points equidistant from two lines is the set of points where the perpendicular distance to the first line is equal to the perpendicular distance to the second line.

Geometrically, the set of points equidistant from two intersecting lines consists of the lines that bisect the angles formed by the two lines. There are two such lines, which are the angle bisectors of the angles formed by $L_1$ and $L_2$.

These two angle bisectors are perpendicular to each other and pass through the point of intersection of the original lines $L_1$ and $L_2$.

Thus, the locus of a point equidistant from two intersecting lines is a pair of straight lines (the angle bisectors).

Let's consider the options:

(A) A straight line: This is incorrect, as there are two angle bisectors.

(B) A pair of straight lines (angle bisectors): This matches our geometric understanding.

(C) A parabola: The locus of a point equidistant from a fixed point (focus) and a fixed line (directrix) is a parabola.

(D) A circle: The locus of a point equidistant from a fixed point (center) is a circle.

The locus of a point that is equidistant from two intersecting lines is a pair of straight lines, specifically the angle bisectors of the angles formed by the given lines.

The correct option is (B) A pair of straight lines (angle bisectors).

Solution:

Let the original coordinate system have its origin at $O(0, 0)$. Let the coordinates of a point P in this system be $(x, y)$.

The origin is shifted to a new point $O'(h, k)$ in the original coordinate system.

Let the new coordinate system have its origin at $O'(h, k)$ and its axes parallel to the original axes. Let the coordinates of the point P in this new system be $(x', y')$.

Consider the position vector of the point P from the original origin O, denoted by $\vec{OP}$. In the original system, this vector is $(x, y)$.

Consider the position vector of the new origin O' from the original origin O, denoted by $\vec{OO'}$. In the original system, this vector is $(h, k)$.

Consider the position vector of the point P from the new origin O', denoted by $\vec{O'P}$. In the new system, this vector is $(x', y')$.

From vector addition, the position vector of P from the original origin is the sum of the position vector of the new origin from the original origin and the position vector of P from the new origin:

In terms of coordinates:

Equating the corresponding components, we get the relationship between the old coordinates $(x, y)$, the new coordinates $(x', y')$, and the shift $(h, k)$:

We can also rearrange these equations to express the new coordinates in terms of the old coordinates and the shift:

Both sets of equations represent the correct relationship between the old and new coordinates when the origin is shifted. Let's compare these with the given options:

(A) $x = x' + h, y = y' + k$: This matches the first form of the relationship we derived.

(B) $x' = x + h, y' = y + k$: This is incorrect.

(C) $x = x' - h, y = y' - k$: This is incorrect.

(D) $x' = x - h, y' = y - k$: This matches the second form of the relationship we derived.

Both options (A) and (D) express the correct relationship. However, it is a common convention when describing coordinate transformations to express the new coordinates in terms of the old coordinates and the transformation parameters. Option (D) follows this convention by expressing $(x', y')$ in terms of $(x, y)$, $(h, k)$. Option (A) expresses $(x, y)$ in terms of $(x', y')$, $(h, k)$.

Given the context "the new coordinates of a point P(x, y) become (x', y')", it is most natural to provide the formula for the new coordinates $(x', y')$.

The correct relationship, typically expressed as the new coordinates in terms of the old, is $x' = x - h$ and $y' = y - k$.

The correct option is (D) $x' = x - h, y' = y - k$.

Solution:

We are given the coordinates of a point on the line $(x_1, y_1) = (-2, 3)$.

We are also given that the line makes an angle of $60^\circ$ with the positive x-axis. This is the inclination of the line, $\theta = 60^\circ$.

The slope $m$ of a line with inclination $\theta$ is given by the formula:

Substitute the given inclination:

The value of $\tan(60^\circ)$ is $\sqrt{3}$.

Now we use the point-slope form of the equation of a straight line. The equation of a line passing through the point $(x_1, y_1)$ with slope $m$ is:

Substitute the given point $(-2, 3)$ and the calculated slope $m = \sqrt{3}$ into the formula:

Simplify the expression in the parenthesis:

This is the equation of the line.

Compare the derived equation with the given options:

(A) $y - 3 = \sqrt{3}(x + 2)$: This matches the equation we found.

(B) $y + 3 = \sqrt{3}(x - 2)$: The point is incorrectly used.

(C) $y - 3 = -\sqrt{3}(x + 2)$: The slope sign is incorrect.

(D) $y + 3 = -\sqrt{3}(x - 2)$: The point and slope sign are incorrect.

The equation of the line passing through $(-2, 3)$ and making an angle of $60^\circ$ with the positive x-axis is $y - 3 = \sqrt{3}(x + 2)$.

The correct option is (A) $y - 3 = \sqrt{3}(x + 2)$.

Solution:

The given family of lines is represented by the equation $y = mx + c$.

In this equation, $m$ is the slope of the line, and $c$ is the y-intercept.

The problem states that $m$ is a constant, while $c$ is a parameter (meaning $c$ can take different values for different lines in the family).

Since the slope $m$ is the same for all lines in this family, and $c$ can vary, the lines will have the same slope but generally different y-intercepts.

Lines that have the same slope are parallel to each other.

• If $c_1 \neq c_2$, the lines $y = mx + c_1$ and $y = mx + c_2$ are distinct parallel lines.
• If $c_1 = c_2$, the lines are coincident (the same line), which is considered a special case of parallel lines.

Thus, the family of lines $y = mx + c$ where $m$ is a constant represents a set of lines that are parallel to each other.

Let's look at the options:

(A) Parallel to each other: This is consistent with our finding that the slope $m$ is constant.

(B) Passing through the origin: A line passes through the origin $(0,0)$ if $0 = m(0) + c$, which means $c=0$. The family includes the line $y=mx$ (when $c=0$), which passes through the origin, but it also includes lines with $c \neq 0$, which do not pass through the origin.

(C) Perpendicular to each other: Lines are perpendicular if the product of their slopes is $-1$. Since the slope is a constant $m$, the product of slopes of any two lines in the family is $m \cdot m = m^2$. Unless $m^2 = -1$ (which is not possible for a real slope) or $m=0$ and one line is vertical (which isn't covered by $y=mx+c$), the lines are not perpendicular.

(D) All having the same y-intercept $c$: The problem implies that $c$ is a parameter that varies, while $m$ is fixed. If $c$ were constant and $m$ were the parameter, then all lines would have the same y-intercept.

The family of lines $y = mx + c$ with constant $m$ and varying $c$ represents a set of parallel lines.

The correct option is (A) Parallel to each other.

Solution:

We are asked to find the equation of a line that is parallel to the line $2x - 3y + 5 = 0$ and passes through the point $(1, 2)$.

Step 1: Find the slope of the given line.

The equation of the given line is $2x - 3y + 5 = 0$. This is in the general form $Ax + By + C = 0$. The slope of a line in this form (where $B \neq 0$) is given by $m = -\frac{A}{B}$.

For the line $2x - 3y + 5 = 0$, we have $A = 2$ and $B = -3$.

Step 2: Determine the slope of the required line.

Two lines are parallel if and only if they have the same slope.

Since the required line is parallel to the given line, its slope $m_{\text{required}}$ is equal to the slope of the given line.

Step 3: Use the point-slope form to find the equation of the line.

We have the slope of the required line $m = \frac{2}{3}$ and a point it passes through $(x_1, y_1) = (1, 2)$. The point-slope form of the equation of a line is:

Substitute the values:

Step 4: Convert the equation to the general form and compare with options.

Multiply both sides by 3 to clear the fraction:

Distribute:

Move all terms to one side (e.g., to the right side):

This is the equation of the required line.

Comparing this equation with the given options, we see that option (A) matches our derived equation.

Let's verify that the line $2x - 3y + 4 = 0$ passes through $(1, 2)$: $2(1) - 3(2) + 4 = 2 - 6 + 4 = 0$. The point lies on the line.

The correct option is (A) $2x - 3y + 4 = 0$.

Solution:

The equation of a line can be expressed in the normal form, which is $x \cos \alpha + y \sin \alpha = p$.

In this form:

• $p$ is the length of the perpendicular from the origin to the line.
• $\alpha$ is the angle which the perpendicular from the origin to the line makes with the positive direction of the x-axis.

We are given:

• The perpendicular distance from the origin is 5. So, $p = 5$.
• The angle made by the perpendicular with the positive x-axis is $30^\circ$. So, $\alpha = 30^\circ$.

Substitute these values of $p$ and $\alpha$ into the normal form equation:

This is the equation of the line.

Let's compare this equation with the given options:

(A) $x \cos 30^\circ + y \sin 30^\circ = 5$: This directly matches the equation we derived from the normal form.

(B) $x \sin 30^\circ + y \cos 30^\circ = 5$: This swaps the cosine and sine terms, implying the angle $\alpha$ would be $90^\circ - 30^\circ = 60^\circ$ (if the form was $x \cos \alpha + y \sin \alpha = p$).

(C) $x \cos 30^\circ + y \sin 30^\circ = -5$: This has the correct angle $\alpha$ but the sign of the distance $p$ is negative. While $p$ can technically be a directed distance, in standard normal form for distance, it's usually taken as positive, and the angle $\alpha$ handles the orientation.

(D) $x \cos 60^\circ + y \sin 60^\circ = 5$: This uses $\alpha = 60^\circ$ instead of the given $30^\circ$.

The equation of the line is precisely given by substituting the values of $p$ and $\alpha$ into the normal form.

The correct option is (A) $x \cos 30^\circ + y \sin 30^\circ = 5$.

Solution:

We are given the parametric equations of a point $(x, y)$ in terms of a parameter $t$:

To find the locus of the point, we need to eliminate the parameter $t$ from these equations and obtain a relationship between $x$ and $y$.

From equation (1), we can express $t$ in terms of $x$:

Now, substitute the expression for $t$ from equation (3) into equation (2):

Simplify the equation:

Rearrange the equation into the general form $Ax + By + C = 0$:

This equation is a linear equation in $x$ and $y$ of the form $Ax + By + C = 0$, where $A=2$, $B=-1$, and $C=-5$. Since A and B are not both zero, this equation represents a straight line in the Cartesian plane.

Therefore, the locus of the point $(x, y)$ for any real number $t$ is a straight line.

Comparing our result with the given options:

(A) A point: Incorrect.

(B) A straight line: Correct.

(C) A parabola: Incorrect (a parabola involves one variable being a quadratic function of the other).

(D) A circle: Incorrect (a circle involves $x^2 + y^2$ or $(x-h)^2 + (y-k)^2$).

The correct option is (B) A straight line.

Given:

Point A with coordinates $(3, 5)$.

Point B with coordinates $(-1, 2)$.

To Find:

The distance between points A and B.

Solution:

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

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

Here, we have:

$(x_1, y_1) = (3, 5)$

$(x_2, y_2) = (-1, 2)$

Substitute these values into the distance formula:

$d = \sqrt{(-1 - 3)^2 + (2 - 5)^2}$

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

$d = \sqrt{16 + 9}$

$d = \sqrt{25}$

$d = 5$

Thus, the distance between points A and B is 5 units.

Given:

The coordinates of the two points are $(4, -3)$ and $(9, 7)$.

The line segment joining these points is divided internally in the ratio $3:2$.

To Find:

The coordinates of the point of division.

Solution:

Let the two given points be $A(x_1, y_1) = (4, -3)$ and $B(x_2, y_2) = (9, 7)$.

Let the point dividing the line segment AB internally in the ratio $m:n = 3:2$ be $P(x, y)$.

We use the section formula for internal division, which states that the coordinates $(x, y)$ of a point that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ internally are given by:

$x = \frac{mx_2 + nx_1}{m+n}$

$y = \frac{my_2 + ny_1}{m+n}$

Here, $x_1 = 4$, $y_1 = -3$, $x_2 = 9$, $y_2 = 7$, $m = 3$, and $n = 2$.

Substitute the values into the formulas:

$x = \frac{(3)(9) + (2)(4)}{3+2}$

$x = \frac{27 + 8}{5}$

$x = \frac{35}{5}$

$x = 7$

Now, for the y-coordinate:

$y = \frac{(3)(7) + (2)(-3)}{3+2}$

$y = \frac{21 - 6}{5}$

$y = \frac{15}{5}$

$y = 3$

Therefore, the coordinates of the point that divides the line segment joining $(4, -3)$ and $(9, 7)$ internally in the ratio 3:2 are $(7, 3)$.

Given:

The vertices of the triangle are A(2, 3), B(-1, 0), and C(2, -4).

To Find:

The area of the triangle ABC.

Solution:

Let the coordinates of the vertices be $(x_1, y_1) = (2, 3)$, $(x_2, y_2) = (-1, 0)$, and $(x_3, y_3) = (2, -4)$.

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

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the given coordinates into the formula:

Area $= \frac{1}{2} |2(0 - (-4)) + (-1)((-4) - 3) + 2(3 - 0)|$

Area $= \frac{1}{2} |2(0 + 4) + (-1)(-7) + 2(3)|$

Area $= \frac{1}{2} |2(4) + 7 + 6|$

Area $= \frac{1}{2} |8 + 7 + 6|$

Area $= \frac{1}{2} |21|$

Area $= \frac{21}{2}$

Area $= 10.5$

Therefore, the area of the triangle with the given vertices is 10.5 square units.

Given:

A fixed point A with coordinates $(3, 4)$.

The distance from a point on the locus to A is always 5 units.

To Find:

The equation of the locus of the point.

Solution:

Let P$(x, y)$ be any point on the locus.

The distance between the point P$(x, y)$ and the fixed point A$(3, 4)$ is given to be 5.

Using the distance formula, the distance PA is:

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

Substituting the coordinates of P and A, we get:

$PA = \sqrt{(x - 3)^2 + (y - 4)^2}$

According to the problem statement, $PA = 5$.

So, we have:

$\sqrt{(x - 3)^2 + (y - 4)^2} = 5$

To eliminate the square root, square both sides of the equation:

$(x - 3)^2 + (y - 4)^2 = 5^2$

$(x - 3)^2 + (y - 4)^2 = 25$

Now, expand the squared terms:

$(x^2 - 2(x)(3) + 3^2) + (y^2 - 2(y)(4) + 4^2) = 25$

$(x^2 - 6x + 9) + (y^2 - 8y + 16) = 25$

Remove the parentheses and rearrange the terms:

$x^2 + y^2 - 6x - 8y + 9 + 16 = 25$

$x^2 + y^2 - 6x - 8y + 25 = 25$

Subtract 25 from both sides of the equation:

$x^2 + y^2 - 6x - 8y + 25 - 25 = 0$

$x^2 + y^2 - 6x - 8y = 0$

This is the equation of the locus of the point whose distance from (3, 4) is always 5. This equation represents a circle centered at (3, 4) with a radius of 5.

Given:

Original coordinates of the point: $(x, y) = (-5, 1)$.

New origin coordinates after shifting: $(h, k) = (2, 3)$.

To Find:

The new coordinates of the point $(-5, 1)$ relative to the new origin $(2, 3)$.

Solution:

Let the original coordinates of a point be $(x, y)$ and the coordinates of the new origin be $(h, k)$.

If the origin is shifted to $(h, k)$, the new coordinates $(x', y')$ of the point are given by the transformation formulas:

$x' = x - h$

$y' = y - k$

In this problem, we have:

$x = -5$, $y = 1$

$h = 2$, $k = 3$

Substitute these values into the transformation formulas to find the new coordinates $(x', y')$:

$x' = -5 - 2$

$x' = -7$

$y' = 1 - 3$

$y' = -2$

Therefore, the new coordinates of the point $(-5, 1)$ when the origin is shifted to $(2, 3)$ are $(-7, -2)$.

Given:

Two points on the line are $A(3, -2)$ and $B(-1, 4)$.

To Find:

The slope of the line passing through points A and B.

Solution:

Let the coordinates of the first point be $(x_1, y_1) = (3, -2)$ and the coordinates of the second point be $(x_2, y_2) = (-1, 4)$.

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

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

Substitute the given coordinates into the slope formula:

$m = \frac{4 - (-2)}{-1 - 3}$

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

$m = \frac{6}{-4}$

Simplify the fraction:

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

Therefore, the slope of the line passing through the points (3, -2) and (-1, 4) is $-\frac{3}{2}$.

Given:

A point on the line is $(x_1, y_1) = (2, -3)$.

The slope of the line is $m = -\frac{1}{2}$.

To Find:

The equation of the line.

Solution:

We can use the point-slope form of the equation of a line, which is given by:

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

Substitute the given point $(x_1, y_1) = (2, -3)$ and the slope $m = -\frac{1}{2}$ into this formula:

$y - (-3) = -\frac{1}{2}(x - 2)$

Simplify the left side:

$y + 3 = -\frac{1}{2}(x - 2)$

To eliminate the fraction, multiply both sides of the equation by 2:

$2(y + 3) = 2 \left(-\frac{1}{2}(x - 2)\right)$

$2y + 6 = -(x - 2)$

$2y + 6 = -x + 2$

Now, rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

Add $x$ to both sides:

$x + 2y + 6 = 2$

Subtract 2 from both sides:

$x + 2y + 6 - 2 = 0$

$x + 2y + 4 = 0$

This is the equation of the line passing through the point (2, -3) with a slope of $-\frac{1}{2}$.

Given:

Two points on the line are $A(1, 2)$ and $B(-3, 4)$.

To Find:

The equation of the line passing through points A and B.

Solution:

Let the coordinates of the two points be $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (-3, 4)$.

We can use the two-point form of the equation of a line, which is given by:

$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Substitute the given coordinates into this formula:

$y - 2 = \frac{4 - 2}{-3 - 1}(x - 1)$

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

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

To eliminate the fraction, multiply both sides of the equation by 2:

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

$2y - 4 = -x + 1$

Now, rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

Add $x$ to both sides:

$x + 2y - 4 = 1$

Subtract 1 from both sides:

$x + 2y - 4 - 1 = 0$

$x + 2y - 5 = 0$

This is the equation of the line passing through the points (1, 2) and (-3, 4).

Alternate Solution (using slope-intercept form):

First, find the slope of the line passing through $(1, 2)$ and $(-3, 4)$.

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2}$

Now use the point-slope form with point $(1, 2)$ and slope $m = -\frac{1}{2}$:

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

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

Multiply by 2:

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

$2y - 4 = -x + 1$

Rearrange to $Ax + By + C = 0$ form:

$x + 2y - 4 - 1 = 0$

$x + 2y - 5 = 0$

The equation of the line is $x + 2y - 5 = 0$.

Given:

The x-intercept of the line is $a = 3$.

The y-intercept of the line is $b = -2$.

To Find:

The equation of the line.

Solution:

We can use the intercept form of the equation of a line, which is given by:

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

where $a$ is the x-intercept and $b$ is the y-intercept.

Substitute the given values of $a = 3$ and $b = -2$ into the formula:

$\frac{x}{3} + \frac{y}{-2} = 1$

This can be written as:

$\frac{x}{3} - \frac{y}{2} = 1$

To eliminate the denominators, find the least common multiple (LCM) of 3 and 2, which is 6.

Multiply both sides of the equation by 6:

$6 \left(\frac{x}{3}\right) - 6 \left(\frac{y}{2}\right) = 6(1)$

$2x - 3y = 6$

To write the equation in the general form $Ax + By + C = 0$, move the constant term to the left side:

$2x - 3y - 6 = 0$

This is the equation of the line with x-intercept 3 and y-intercept -2.

Alternate Solution (using two points):

An x-intercept of 3 means the line passes through the point $(3, 0)$.

A y-intercept of -2 means the line passes through the point $(0, -2)$.

Let the two points be $A(x_1, y_1) = (3, 0)$ and $B(x_2, y_2) = (0, -2)$.

We can use the two-point form of the equation of a line:

$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

Substitute the coordinates of points A and B:

$\frac{y - 0}{x - 3} = \frac{-2 - 0}{0 - 3}$

$\frac{y}{x - 3} = \frac{-2}{-3}$

$\frac{y}{x - 3} = \frac{2}{3}$

Cross-multiply:

$3y = 2(x - 3)$

$3y = 2x - 6$

Rearrange into the general form $Ax + By + C = 0$:

$0 = 2x - 3y - 6$

or

$2x - 3y - 6 = 0$

This confirms the result obtained using the intercept form.

Given:

The coordinates of the point are $(x_1, y_1) = (3, -5)$.

The equation of the line is $3x - 4y - 26 = 0$.

To Find:

The perpendicular distance of the point $(3, -5)$ from the line $3x - 4y - 26 = 0$.

Solution:

The equation of the given line is in the form $Ax + By + C = 0$, where $A = 3$, $B = -4$, and $C = -26$.

The coordinates of the given point are $(x_1, y_1) = (3, -5)$.

The perpendicular distance $d$ of a point $(x_1, y_1)$ from a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Substitute the values of $A$, $B$, $C$, $x_1$, and $y_1$ into the formula:

$d = \frac{|(3)(3) + (-4)(-5) + (-26)|}{\sqrt{3^2 + (-4)^2}}$

$d = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}}$

$d = \frac{|29 - 26|}{\sqrt{25}}$

$d = \frac{|3|}{5}$

$d = \frac{3}{5}$

Therefore, the distance of the point (3, -5) from the line $3x - 4y - 26 = 0$ is $\frac{3}{5}$ units.

Given:

Equation of the first line: $y = \sqrt{3}x + 7$.

Equation of the second line: $y = \frac{1}{\sqrt{3}}x - 2$.

To Find:

The angle between the two lines.

Solution:

The equations of the lines are given in the form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

From the first equation, $y = \sqrt{3}x + 7$, the slope of the first line is $m_1 = \sqrt{3}$.

From the second equation, $y = \frac{1}{\sqrt{3}}x - 2$, the slope of the second line is $m_2 = \frac{1}{\sqrt{3}}$.

The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

Substitute the values of $m_1$ and $m_2$ into the formula:

$\tan \theta = \left|\frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)}\right|$

Simplify the numerator and the denominator:

Numerator: $\sqrt{3} - \frac{1}{\sqrt{3}} = \frac{(\sqrt{3})(\sqrt{3}) - 1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

Denominator: $1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right) = 1 + 1 = 2$

Now substitute these simplified values back into the tangent formula:

$\tan \theta = \left|\frac{\frac{2}{\sqrt{3}}}{2}\right|$

$\tan \theta = \left|\frac{2}{\sqrt{3}} \times \frac{1}{2}\right|$

$\tan \theta = \left|\frac{\cancel{2}}{\sqrt{3} \times \cancel{2}}\right|$

$\tan \theta = \left|\frac{1}{\sqrt{3}}\right|$

$\tan \theta = \frac{1}{\sqrt{3}}$

We know that $\tan 30^\circ = \frac{1}{\sqrt{3}}$ or $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

Therefore, the angle $\theta$ is $30^\circ$ or $\frac{\pi}{6}$ radians.

The angle between the lines is $30^\circ$ (or $\frac{\pi}{6}$).

Given:

The equation of the given line is $2x + 3y = 5$.

The line we need to find passes through the point $(-1, 2)$.

The line we need to find is parallel to the given line.

To Find:

The equation of the line parallel to $2x + 3y = 5$ and passing through $(-1, 2)$.

Solution:

First, find the slope of the given line $2x + 3y = 5$.

The equation of a line in the form $Ax + By + C = 0$ has a slope given by $m = -\frac{A}{B}$.

For the line $2x + 3y - 5 = 0$, we have $A = 2$, $B = 3$, and $C = -5$.

The slope of the given line is $m_1 = -\frac{2}{3}$.

Parallel lines have the same slope.

So, the slope of the line we need to find is $m = m_1 = -\frac{2}{3}$.

Now we have the slope $m = -\frac{2}{3}$ and a point on the line $(x_1, y_1) = (-1, 2)$.

We can use the point-slope form of the equation of a line:

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

Substitute the values of $m$, $x_1$, and $y_1$:

$y - 2 = -\frac{2}{3}(x - (-1))$

$y - 2 = -\frac{2}{3}(x + 1)$

To eliminate the denominator, multiply both sides of the equation by 3:

$3(y - 2) = -2(x + 1)$

$3y - 6 = -2x - 2$

Rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

Add $2x$ to both sides:

$2x + 3y - 6 = -2$

Add 2 to both sides:

$2x + 3y - 6 + 2 = 0$

$2x + 3y - 4 = 0$

This is the equation of the line parallel to $2x + 3y = 5$ and passing through the point $(-1, 2)$.

Given:

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

The line we need to find passes through the point $(1, -3)$.

The line we need to find is perpendicular to the given line.

To Find:

The equation of the line perpendicular to $x - 2y = 4$ and passing through $(1, -3)$.

Solution:

First, find the slope of the given line $x - 2y = 4$.

Rewrite the equation in the form $y = mx + c$ to easily find the slope:

$-2y = -x + 4$

$y = \frac{-x + 4}{-2}$

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

The slope of the given line is $m_1 = \frac{1}{2}$.

If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the required line be $m_2$. Then,

$m_1 \times m_2 = -1$

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

Multiply both sides by 2:

$m_2 = -2$

Now we have the slope of the required line, $m_2 = -2$, and a point on the line $(x_1, y_1) = (1, -3)$.

We can use the point-slope form of the equation of a line:

$y - y_1 = m_2(x - x_1)$

Substitute the values of $m_2$, $x_1$, and $y_1$:

$y - (-3) = -2(x - 1)$

Simplify the equation:

$y + 3 = -2x + 2$

Rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

Add $2x$ to both sides:

$2x + y + 3 = 2$

Subtract 2 from both sides:

$2x + y + 3 - 2 = 0$

$2x + y + 1 = 0$

This is the equation of the line perpendicular to $x - 2y = 4$ and passing through the point $(1, -3)$.

Given:

The coordinates of the two endpoints of the line segment are $A(-3, 10)$ and $B(6, -8)$.

The point of division is $P(-1, 6)$.

To Find:

The ratio in which the point P divides the line segment AB.

Solution:

Let the point $P(-1, 6)$ divide the line segment joining $A(-3, 10)$ and $B(6, -8)$ in the ratio $m_1 : m_2$.

Using the section formula for internal division, the coordinates of the point P are given by:

$P(x, y) = \left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right)$

Here, $(x_1, y_1) = (-3, 10)$, $(x_2, y_2) = (6, -8)$, and $(x, y) = (-1, 6)$.

We can use either the x-coordinate or the y-coordinate to find the ratio $m_1 : m_2$.

Using the x-coordinate:

$-1 = \frac{m_1 (6) + m_2 (-3)}{m_1 + m_2}$

$-1(m_1 + m_2) = 6m_1 - 3m_2$

$-m_1 - m_2 = 6m_1 - 3m_2$

Rearrange the terms to group $m_1$ and $m_2$:

$3m_2 - m_2 = 6m_1 + m_1$

$2m_2 = 7m_1$

To find the ratio $\frac{m_1}{m_2}$, divide both sides by $7m_2$ (assuming $m_2 \neq 0$):

$\frac{2}{7} = \frac{m_1}{m_2}$

So, the ratio $\frac{m_1}{m_2} = \frac{2}{7}$. This means $m_1 : m_2 = 2 : 7$.

Let's verify this using the y-coordinate:

$6 = \frac{m_1 (-8) + m_2 (10)}{m_1 + m_2}$

$6(m_1 + m_2) = -8m_1 + 10m_2$

$6m_1 + 6m_2 = -8m_1 + 10m_2$

Rearrange the terms:

$6m_1 + 8m_1 = 10m_2 - 6m_2$

$14m_1 = 4m_2$

To find the ratio $\frac{m_1}{m_2}$, divide both sides by $14m_2$ (assuming $m_2 \neq 0$):

$\frac{14}{4} = \frac{m_2}{m_1}$

$\frac{7}{2} = \frac{m_2}{m_1}$

This means $\frac{m_1}{m_2} = \frac{2}{7}$, which gives the ratio $m_1 : m_2 = 2 : 7$.

Since the ratio is positive (2:7), the point P divides the line segment AB internally.

The ratio in which the point (-1, 6) divides the line segment joining (-3, 10) and (6, -8) is 2:7.

Alternate Method (using ratio $k:1$):

Let the point $P(-1, 6)$ divide the line segment joining $A(-3, 10)$ and $B(6, -8)$ in the ratio $k:1$.

Using the section formula, the coordinates of P are:

$P(x, y) = \left(\frac{k x_2 + 1 x_1}{k + 1}, \frac{k y_2 + 1 y_1}{k + 1}\right)$

Here, $(x_1, y_1) = (-3, 10)$, $(x_2, y_2) = (6, -8)$, and $(x, y) = (-1, 6)$.

Using the x-coordinate:

$-1 = \frac{k(6) + 1(-3)}{k + 1}$

$-1(k + 1) = 6k - 3$

$-k - 1 = 6k - 3$

Add $k$ to both sides and add 3 to both sides:

$3 - 1 = 6k + k$

$2 = 7k$

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

Using the y-coordinate:

$6 = \frac{k(-8) + 1(10)}{k + 1}$

$6(k + 1) = -8k + 10$

$6k + 6 = -8k + 10$

Add $8k$ to both sides and subtract 6 from both sides:

$6k + 8k = 10 - 6$

$14k = 4$

$k = \frac{4}{14}$

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

Both coordinates give $k = \frac{2}{7}$. The ratio is $k:1 = \frac{2}{7} : 1$.

Multiply both parts of the ratio by 7 to get integer form:

$\left(\frac{2}{7} \times 7\right) : (1 \times 7) = 2 : 7$

Since $k = \frac{2}{7}$ is positive, the division is internal.

The ratio in which the point (-1, 6) divides the line segment joining (-3, 10) and (6, -8) is 2:7 internally.

Given:

The vertices of the triangle are A(1, 4), B(5, 2), and C(6, 3).

To Find:

The coordinates of the centroid of the triangle ABC.

Solution:

Let the coordinates of the vertices of the triangle be $(x_1, y_1) = (1, 4)$, $(x_2, y_2) = (5, 2)$, and $(x_3, y_3) = (6, 3)$.

The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is the point $G(x, y)$ given by the formula:

$x = \frac{x_1 + x_2 + x_3}{3}$

$y = \frac{y_1 + y_2 + y_3}{3}$

Substitute the coordinates of the given vertices into the formulas:

For the x-coordinate of the centroid:

$x = \frac{1 + 5 + 6}{3}$

$x = \frac{12}{3}$

$x = 4$

For the y-coordinate of the centroid:

$y = \frac{4 + 2 + 3}{3}$

$y = \frac{9}{3}$

$y = 3$

Therefore, the coordinates of the centroid of the triangle with vertices (1, 4), (5, 2), and (6, 3) are $(4, 3)$.

Given:

The three points are A(1, -1), B(5, 2), and C(9, 5).

To Prove:

The points A, B, and C are collinear.

Proof:

Points are collinear if they lie on the same straight line.

One way to show that three points are collinear is to demonstrate that the slope between the first two points is equal to the slope between the second and third points.

Let the points be $A(x_1, y_1) = (1, -1)$, $B(x_2, y_2) = (5, 2)$, and $C(x_3, y_3) = (9, 5)$.

The slope of the line segment joining two points $(x_a, y_a)$ and $(x_b, y_b)$ is given by $m = \frac{y_b - y_a}{x_b - x_a}$.

Calculate the slope of the line segment AB:

$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}$

$m_{AB} = \frac{2 - (-1)}{5 - 1}$

$m_{AB} = \frac{2 + 1}{4}$

$m_{AB} = \frac{3}{4}$

Calculate the slope of the line segment BC:

$m_{BC} = \frac{y_3 - y_2}{x_3 - x_2}$

$m_{BC} = \frac{5 - 2}{9 - 5}$

$m_{BC} = \frac{3}{4}$

Since the slope of AB ($m_{AB}$) is equal to the slope of BC ($m_{BC}$), and point B is common to both line segments AB and BC, the points A, B, and C lie on the same straight line.

Therefore, the points (1, -1), (5, 2), and (9, 5) are collinear.

Alternate Proof (using Area of Triangle):

Three points are collinear if the area of the triangle formed by them 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:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Using the points $A(1, -1)$, $B(5, 2)$, and $C(9, 5)$:

Area $= \frac{1}{2} |1(2 - 5) + 5(5 - (-1)) + 9(-1 - 2)|$

Area $= \frac{1}{2} |1(-3) + 5(5 + 1) + 9(-3)|$

Area $= \frac{1}{2} |-3 + 5(6) - 27|$

Area $= \frac{1}{2} |-3 + 30 - 27|$

Area $= \frac{1}{2} |27 - 27|$

Area $= \frac{1}{2} |0|$

Area $= 0$

Since the area of the triangle formed by the points A, B, and C is 0, the points are collinear.

Given:

The endpoints of the line segment are $A(1, 0)$ and $B(2, 3)$.

The required line is perpendicular to the line segment AB.

The required line divides the line segment AB in the ratio $1:2$ internally.

To Find:

The equation of the line.

Solution:

Let the point that divides the line segment joining $A(1, 0)$ and $B(2, 3)$ in the ratio $1:2$ internally be $P(x, y)$.

Using the section formula for internal division, with $(x_1, y_1) = (1, 0)$, $(x_2, y_2) = (2, 3)$, $m_1 = 1$, and $m_2 = 2$:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{(1)(2) + (2)(1)}{1 + 2} = \frac{2 + 2}{3} = \frac{4}{3}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{(1)(3) + (2)(0)}{1 + 2} = \frac{3 + 0}{3} = \frac{3}{3} = 1$

So, the point of division is $P(\frac{4}{3}, 1)$. The required line passes through this point.

Next, find the slope of the line segment AB.

The slope of the line segment joining $A(1, 0)$ and $B(2, 3)$ is:

$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3$

The required line is perpendicular to the line segment AB.

If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the required line be $m$. Then,

$m \times m_{AB} = -1$

$m \times 3 = -1$

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

Now we have the slope of the required line, $m = -\frac{1}{3}$, and a point on the line, $P(\frac{4}{3}, 1)$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

$y - 1 = -\frac{1}{3}(x - \frac{4}{3})$

Multiply both sides by 3 to clear the denominator on the right side:

$3(y - 1) = -1(x - \frac{4}{3})$

$3y - 3 = -x + \frac{4}{3}$

Multiply both sides by 3 again to clear the remaining denominator:

$3(3y - 3) = 3(-x + \frac{4}{3})$

$9y - 9 = -3x + 4$

Rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

$3x + 9y - 9 - 4 = 0$

$3x + 9y - 13 = 0$

The equation of the line which is perpendicular to the line segment joining A(1, 0) and B(2, 3) and divides it in the ratio 1:2 is $3x + 9y - 13 = 0$.

Given:

The perpendicular distance of the line from the origin is $p = 5$.

The angle made by the perpendicular from the origin to the line with the positive x-axis is $\alpha = 30^\circ$.

To Find:

The equation of the line.

Solution:

The equation of a line in the normal form is given by:

$x \cos \alpha + y \sin \alpha = p$

where $p$ is the perpendicular distance from the origin to the line, and $\alpha$ is the angle which the normal (perpendicular from the origin) makes with the positive x-axis.

We are given $p = 5$ and $\alpha = 30^\circ$.

We need to find the values of $\cos 30^\circ$ and $\sin 30^\circ$:

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

$\sin 30^\circ = \frac{1}{2}$

Substitute these values into the normal form equation:

$x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 5$

Multiply the entire equation by 2 to remove the denominators:

$2 \left(x \frac{\sqrt{3}}{2} + y \frac{1}{2}\right) = 2(5)$

$\sqrt{3}x + y = 10$

To write the equation in the general form $Ax + By + C = 0$, move the constant term to the left side:

$\sqrt{3}x + y - 10 = 0$

This is the equation of the line whose perpendicular distance from the origin is 5 and the angle made by the perpendicular with the positive x-axis is $30^\circ$.

Given:

The point from which the perpendicular is drawn is $P(x_1, y_1) = (1, 2)$.

The equation of the line is $L: 3x - 4y + 5 = 0$.

To Find:

The coordinates of the foot of the perpendicular from point P to the line L.

Solution:

Let the foot of the perpendicular from the point $P(1, 2)$ to the line $3x - 4y + 5 = 0$ be $Q(x_2, y_2)$.

The line segment PQ is perpendicular to the given line $L$.

First, find the slope of the given line $L: 3x - 4y + 5 = 0$.

We can rewrite the equation in the form $y = mx + c$ to find the slope:

$-4y = -3x - 5$

$y = \frac{-3x - 5}{-4}$

$y = \frac{3}{4}x + \frac{5}{4}$

The slope of the given line L is $m_L = \frac{3}{4}$.

The line segment PQ is perpendicular to L. If two lines are perpendicular, the product of their slopes is -1.

Let the slope of the line segment PQ be $m_{PQ}$. Then,

$m_{PQ} \times m_L = -1$

$m_{PQ} \times \frac{3}{4} = -1$

$m_{PQ} = -\frac{4}{3}$

The foot of the perpendicular $Q(x_2, y_2)$ lies on the line segment PQ and also on the line L.

The point Q is the intersection point of the line PQ (passing through P(1, 2) with slope $m_{PQ} = -\frac{4}{3}$) and the line L ($3x - 4y + 5 = 0$).

Find the equation of the line PQ using the point-slope form $y - y_1 = m(x - x_1)$, with $P(1, 2)$ and $m_{PQ} = -\frac{4}{3}$:

$y - 2 = -\frac{4}{3}(x - 1)$

Multiply by 3:

$3(y - 2) = -4(x - 1)$

$3y - 6 = -4x + 4$

Rearrange into general form:

$4x + 3y - 6 - 4 = 0$

$4x + 3y - 10 = 0$

So, we need to solve the system of linear equations for the intersection point Q$(x_2, y_2)$:

$3x - 4y + 5 = 0$ \$(1)\$

$4x + 3y - 10 = 0$ \$(2)\$

Multiply equation (1) by 3 and equation (2) by 4 to eliminate y:

$3 \times (3x - 4y + 5 = 0) \implies 9x - 12y + 15 = 0$ \$(3)\$

$4 \times (4x + 3y - 10 = 0) \implies 16x + 12y - 40 = 0$ \$(4)\$

Add equation (3) and equation (4):

$(9x - 12y + 15) + (16x + 12y - 40) = 0 + 0$

$9x + 16x - 12y + 12y + 15 - 40 = 0$

$25x - 25 = 0$

$25x = 25$

$x = \frac{25}{25}$

$x = 1$

Now substitute the value of $x = 1$ into either equation (1) or (2) to find y.

Using equation (1):

$3(1) - 4y + 5 = 0$

$3 - 4y + 5 = 0$

$8 - 4y = 0$

$8 = 4y$

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

$y = 2$

The coordinates of the intersection point Q are $(1, 2)$.

It seems the foot of the perpendicular is the point itself. Let's check if the point (1, 2) lies on the line $3x - 4y + 5 = 0$.

Substitute $x=1$ and $y=2$ into the line equation:

$3(1) - 4(2) + 5 = 3 - 8 + 5 = -5 + 5 = 0$

Since $3(1) - 4(2) + 5 = 0$, the point $(1, 2)$ lies on the line $3x - 4y + 5 = 0$.

If the point lies on the line, the foot of the perpendicular from the point to the line is the point itself.

Therefore, the foot of the perpendicular from the point (1, 2) to the line $3x - 4y + 5 = 0$ is (1, 2).

Given:

The equations of the four lines are:

$L_1: 2x - y + 1 = 0$

$L_2: 2x - y - 3 = 0$

$L_3: 3x + y + 2 = 0$

$L_4: 3x + y - 4 = 0$

To Find:

The area of the parallelogram formed by these four lines.

Solution:

Observe the slopes of the given lines:

The slope of $L_1$ ($2x - y + 1 = 0$) is $m_1 = -\frac{2}{-1} = 2$.

The slope of $L_2$ ($2x - y - 3 = 0$) is $m_2 = -\frac{2}{-1} = 2$.

Since $m_1 = m_2$, lines $L_1$ and $L_2$ are parallel.

The slope of $L_3$ ($3x + y + 2 = 0$) is $m_3 = -\frac{3}{1} = -3$.

The slope of $L_4$ ($3x + y - 4 = 0$) is $m_4 = -\frac{3}{1} = -3$.

Since $m_3 = m_4$, lines $L_3$ and $L_4$ are parallel.

Since we have two pairs of parallel lines with different slopes (2 and -3), these four lines form a parallelogram.

The general form of a pair of parallel lines is $a_1 x + b_1 y + c_1 = 0$ and $a_1 x + b_1 y + c_2 = 0$.

The other pair of parallel lines is $a_2 x + b_2 y + d_1 = 0$ and $a_2 x + b_2 y + d_2 = 0$.

The area of the parallelogram formed by these four lines is given by the formula:

Area $= \frac{|(c_1 - c_2)(d_1 - d_2)|}{|a_1 b_2 - a_2 b_1|}$

From the given equations, we can identify the coefficients:

For the first pair of parallel lines ($L_1$ and $L_2$):

$2x - y + 1 = 0 \implies a_1 = 2, b_1 = -1, c_1 = 1$

$2x - y - 3 = 0 \implies a_1 = 2, b_1 = -1, c_2 = -3$

For the second pair of parallel lines ($L_3$ and $L_4$):

$3x + y + 2 = 0 \implies a_2 = 3, b_2 = 1, d_1 = 2$

$3x + y - 4 = 0 \implies a_2 = 3, b_2 = 1, d_2 = -4$

Now, substitute these values into the area formula:

Calculate $(c_1 - c_2)$:

$c_1 - c_2 = 1 - (-3) = 1 + 3 = 4$

Calculate $(d_1 - d_2)$:

$d_1 - d_2 = 2 - (-4) = 2 + 4 = 6$

Calculate $(a_1 b_2 - a_2 b_1)$:

$a_1 b_2 - a_2 b_1 = (2)(1) - (-1)(3) = 2 - (-3) = 2 + 3 = 5$

Now, calculate the area:

Area $= \frac{|(4)(6)|}{|5|}$

Area $= \frac{|24|}{5}$

Area $= \frac{24}{5}$

The area of the parallelogram formed by the given lines is $\frac{24}{5}$ square units.

Given:

The equation of the line is $\frac{x}{3} + \frac{y}{4} = 1$.

The distance from the required points on the x-axis to the line is 4 units.

To Find:

The coordinates of the points on the x-axis that satisfy the given condition.

Solution:

First, let's rewrite the equation of the line in the standard form $Ax + By + C = 0$.

The given equation is $\frac{x}{3} + \frac{y}{4} = 1$.

To eliminate the denominators, we multiply the entire equation by the least common multiple (LCM) of 3 and 4, which is 12.

$12 \left(\frac{x}{3} + \frac{y}{4}\right) = 12(1)$

$12 \times \frac{x}{3} + 12 \times \frac{y}{4} = 12$

$4x + 3y = 12$

Rearranging the terms, we get the equation in the standard form:

$4x + 3y - 12 = 0$

Here, $A = 4$, $B = 3$, and $C = -12$.

Let the point on the x-axis be $P(x_1, y_1)$. Since the point is on the x-axis, its y-coordinate is 0. So, the coordinates of the point are $(x, 0)$.

The perpendicular distance $d$ of a point $(x_1, y_1)$ from a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

We are given that the distance $d = 4$ units, and the point is $(x, 0)$. Substitute the values of $A$, $B$, $C$, $x_1 = x$, and $y_1 = 0$ into the distance formula:

$4 = \frac{|(4)(x) + (3)(0) + (-12)|}{\sqrt{4^2 + 3^2}}$

$4 = \frac{|4x + 0 - 12|}{\sqrt{16 + 9}}$

$4 = \frac{|4x - 12|}{\sqrt{25}}$

$4 = \frac{|4x - 12|}{5}$

Multiply both sides by 5:

$4 \times 5 = |4x - 12|$

$20 = |4x - 12|$

The absolute value equation $|4x - 12| = 20$ implies two possible cases:

Case 1: $4x - 12 = 20$

$4x = 20 + 12$

$4x = 32$

$x = \frac{32}{4}$

$x = 8$

The point on the x-axis is $(8, 0)$.

Case 2: $4x - 12 = -20$

$4x = -20 + 12$

$4x = -8$

$x = \frac{-8}{4}$

$x = -2$

The point on the x-axis is $(-2, 0)$.

Therefore, the points on the x-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units are $(8, 0)$ and $(-2, 0)$.

Given:

Point P with coordinates $(0, -4)$.

Point B with coordinates $(8, 0)$.

The line passes through the origin $O(0, 0)$.

To Find:

The slope of the line passing through the origin and the midpoint of the line segment joining P and B.

Solution:

First, find the coordinates of the midpoint of the line segment joining P$(0, -4)$ and B$(8, 0)$.

The midpoint $M(x, y)$ of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by the midpoint formula:

$x = \frac{x_1 + x_2}{2}$

$y = \frac{y_1 + y_2}{2}$

Here, $(x_1, y_1) = (0, -4)$ and $(x_2, y_2) = (8, 0)$.

Midpoint x-coordinate: $x = \frac{0 + 8}{2} = \frac{8}{2} = 4$

Midpoint y-coordinate: $y = \frac{-4 + 0}{2} = \frac{-4}{2} = -2$

The midpoint of the line segment PB is $M(4, -2)$.

The line we need to find the slope of passes through the origin $O(0, 0)$ and the midpoint $M(4, -2)$.

The slope of a line passing through two points $(x_a, y_a)$ and $(x_b, y_b)$ is given by the formula:

$m = \frac{y_b - y_a}{x_b - x_a}$

Using the origin $(x_a, y_a) = (0, 0)$ and the midpoint $(x_b, y_b) = (4, -2)$:

$m = \frac{-2 - 0}{4 - 0}$

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

Simplify the fraction:

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

Therefore, the slope of the line which passes through the origin and the midpoint of the line segment joining P(0, -4) and B(8, 0) is $-\frac{1}{2}$.

Given:

The vertices of the triangle are A(1, 2), B(2, -1), and C(3, 1).

To Find:

The coordinates of the orthocentre of triangle ABC.

Solution:

The orthocentre of a triangle is the point of intersection of its altitudes.

An altitude is a line segment from a vertex perpendicular to the opposite side.

Let the vertices be $A(x_1, y_1) = (1, 2)$, $B(x_2, y_2) = (2, -1)$, and $C(x_3, y_3) = (3, 1)$.

First, we find the slopes of two sides of the triangle.

Slope of side AB ($m_{AB}$):

$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 2}{2 - 1} = \frac{-3}{1} = -3$

Slope of side BC ($m_{BC}$):

$m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{1 - (-1)}{3 - 2} = \frac{1 + 1}{1} = 2$

Slope of side AC ($m_{AC}$):

$m_{AC} = \frac{y_3 - y_1}{x_3 - x_1} = \frac{1 - 2}{3 - 1} = \frac{-1}{2}$

Next, we find the equations of two altitudes. Let's consider the altitude from vertex A to side BC, and the altitude from vertex B to side AC.

Equation of the altitude from A to BC:

This altitude passes through A(1, 2) and is perpendicular to BC. The slope of BC is $m_{BC} = 2$. The slope of the altitude from A ($m_A$) is the negative reciprocal of $m_{BC}$ (since the product of slopes of perpendicular lines is -1).

$m_A = -\frac{1}{m_{BC}} = -\frac{1}{2}$

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with point A(1, 2) and slope $m_A = -\frac{1}{2}$:

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

Multiply both sides by 2:

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

$2y - 4 = -x + 1$

Rearrange into the general form $Ax + By + C = 0$:

$x + 2y - 4 - 1 = 0$

$x + 2y - 5 = 0$

Equation of the altitude from B to AC:

This altitude passes through B(2, -1) and is perpendicular to AC. The slope of AC is $m_{AC} = -\frac{1}{2}$. The slope of the altitude from B ($m_B$) is the negative reciprocal of $m_{AC}$.

$m_B = -\frac{1}{m_{AC}} = -\frac{1}{-\frac{1}{2}} = 2$

Using the point-slope form, $y - y_1 = m(x - x_1)$, with point B(2, -1) and slope $m_B = 2$:

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

$y + 1 = 2x - 4$

Rearrange into the general form:

$2x - y - 4 - 1 = 0$

$2x - y - 5 = 0$

The orthocentre is the point of intersection of these two altitudes. We need to solve the system of equations (1) and (2):

$x + 2y - 5 = 0$

$2x - y - 5 = 0$

From equation (2), we can express $y$ in terms of $x$:

$-y = -2x + 5 \implies y = 2x - 5$

Substitute this expression for $y$ into equation (1):

$x + 2(2x - 5) - 5 = 0$

$x + 4x - 10 - 5 = 0$

$5x - 15 = 0$

$5x = 15$

$x = \frac{15}{5} = 3$

Now substitute the value of $x = 3$ back into the expression for $y$:

$y = 2(3) - 5 = 6 - 5 = 1$

The point of intersection is $(3, 1)$.

Thus, the coordinates of the orthocentre of the triangle are $(3, 1)$.

Note that the orthocentre $(3, 1)$ is the same as vertex C. This indicates that the triangle ABC is a right-angled triangle with the right angle at vertex C.

Let's check the dot product of vectors $\vec{CA}$ and $\vec{CB}$:

$\vec{CA} = A - C = (1 - 3, 2 - 1) = (-2, 1)$

$\vec{CB} = B - C = (2 - 3, -1 - 1) = (-1, -2)$

$\vec{CA} \cdot \vec{CB} = (-2)(-1) + (1)(-2) = 2 - 2 = 0$

Since the dot product is 0, $\vec{CA}$ is perpendicular to $\vec{CB}$, which means $\angle ACB = 90^\circ$. The orthocentre of a right-angled triangle is at the vertex where the right angle is located.

Given:

Line 1: $L_1: x - y + 1 = 0$

Line 2: $L_2: 2x - 3y + 5 = 0$

The required line passes through the intersection of $L_1$ and $L_2$.

The required line is parallel to the line $L_3: 3x + 2y = 7$.

To Find:

The equation of the required straight line.

Solution:

First, find the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$.

We can solve the system of equations:

$x - y + 1 = 0$ \$(1)\$

$2x - 3y + 5 = 0$ \$(2)\$

From equation (1), we can express $x$ in terms of $y$:

$x = y - 1$

Substitute this expression for $x$ into equation (2):

$2(y - 1) - 3y + 5 = 0$

$2y - 2 - 3y + 5 = 0$

$-y + 3 = 0$

$-y = -3$

$y = 3$

Now substitute the value of $y = 3$ back into the expression for $x$:

$x = 3 - 1 = 2$

The point of intersection of the two lines is $(2, 3)$. The required line passes through this point.

Next, find the slope of the line $3x + 2y = 7$.

Rewrite the equation in the form $y = mx + c$ to find the slope:

$2y = -3x + 7$

$y = -\frac{3}{2}x + \frac{7}{2}$

The slope of the line $3x + 2y = 7$ is $m_3 = -\frac{3}{2}$.

The required line is parallel to $3x + 2y = 7$. Parallel lines have the same slope.

So, the slope of the required line is $m = m_3 = -\frac{3}{2}$.

Now we have the slope of the required line, $m = -\frac{3}{2}$, and a point on the line, $(x_1, y_1) = (2, 3)$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

$y - 3 = -\frac{3}{2}(x - 2)$

Multiply both sides by 2 to clear the denominator:

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

$2y - 6 = -3x + 6$

Rearrange the terms to get the equation in the general form $Ax + By + C = 0$:

Add $3x$ to both sides:

$3x + 2y - 6 = 6$

Subtract 6 from both sides:

$3x + 2y - 6 - 6 = 0$

$3x + 2y - 12 = 0$

This is the equation of the straight line passing through the intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$ and parallel to the line $3x + 2y = 7$.

Alternate Solution (using family of lines):

The equation of a line passing through the intersection of the lines $L_1 = 0$ and $L_2 = 0$ is given by $L_1 + \lambda L_2 = 0$, where $\lambda$ is a constant.

So, the equation of the required line is of the form:

$(x - y + 1) + \lambda (2x - 3y + 5) = 0$

Rearrange this equation to group x and y terms:

$x - y + 1 + 2\lambda x - 3\lambda y + 5\lambda = 0$

$(1 + 2\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda) = 0$

This equation is in the form $Ax + By + C = 0$, where $A = 1 + 2\lambda$ and $B = -1 - 3\lambda$.

The slope of this line is $m = -\frac{A}{B} = -\frac{1 + 2\lambda}{-1 - 3\lambda} = \frac{1 + 2\lambda}{1 + 3\lambda}$.

The required line is parallel to $3x + 2y = 7$. The slope of this line is $m_3 = -\frac{3}{2}$.

Since the lines are parallel, their slopes are equal:

$\frac{1 + 2\lambda}{1 + 3\lambda} = -\frac{3}{2}$

Cross-multiply:

$2(1 + 2\lambda) = -3(1 + 3\lambda)$

$2 + 4\lambda = -3 - 9\lambda$

Group $\lambda$ terms on one side and constants on the other:

$4\lambda + 9\lambda = -3 - 2$

$13\lambda = -5$

$\lambda = -\frac{5}{13}$

Now substitute the value of $\lambda = -\frac{5}{13}$ back into the equation of the family of lines:

$(x - y + 1) - \frac{5}{13}(2x - 3y + 5) = 0$

Multiply the entire equation by 13 to remove the denominator:

$13(x - y + 1) - 5(2x - 3y + 5) = 0$

$13x - 13y + 13 - 10x + 15y - 25 = 0$

Combine like terms:

$(13x - 10x) + (-13y + 15y) + (13 - 25) = 0$

$3x + 2y - 12 = 0$

This is the same equation obtained using the first method.

Given:

The equations of three lines are:

$L_1: 2x - 3y + 5 = 0$

$L_2: 3x + 4y - 7 = 0$

$L_3: 5x + y - 2 = 0$

To Prove:

The three given lines are concurrent.

To Find:

The point of concurrency.

Solution:

Three lines are said to be concurrent if they all intersect at a single point. To show that the lines are concurrent, we can find the point of intersection of any two lines and then check if the third line passes through that point.

Let's find the intersection point of lines $L_1$ and $L_2$. We need to solve the system of equations:

We can use the method of elimination or substitution to solve this system.

Multiply equation (1) by 4 and equation (2) by 3 to make the coefficients of $y$ equal in magnitude but opposite in sign:

Now, add equation (3) and equation (4):

$(8x - 12y + 20) + (9x + 12y - 21) = 0 + 0$

$8x + 9x - 12y + 12y + 20 - 21 = 0$

$17x - 1 = 0$

$17x = 1$

$x = \frac{1}{17}$

Now substitute the value of $x = \frac{1}{17}$ into equation (1) to find $y$:

$2\left(\frac{1}{17}\right) - 3y + 5 = 0$

$\frac{2}{17} - 3y + 5 = 0$

Multiply by 17 to clear the denominator:

$17\left(\frac{2}{17}\right) - 17(3y) + 17(5) = 0$

$2 - 51y + 85 = 0$

$87 - 51y = 0$

$51y = 87$

$y = \frac{87}{51}$

Simplify the fraction for $y$ by dividing both numerator and denominator by their greatest common divisor, which is 3:

$y = \frac{\cancel{87}^{29}}{\cancel{51}_{17}} = \frac{29}{17}$

The point of intersection of lines $L_1$ and $L_2$ is $(\frac{1}{17}, \frac{29}{17})$.

Now, we check if this point lies on the third line $L_3: 5x + y - 2 = 0$. Substitute $x = \frac{1}{17}$ and $y = \frac{29}{17}$ into the equation of $L_3$:

$5\left(\frac{1}{17}\right) + \left(\frac{29}{17}\right) - 2$

$= \frac{5}{17} + \frac{29}{17} - 2$

$= \frac{5 + 29}{17} - 2$

$= \frac{34}{17} - 2$

$= 2 - 2 = 0$

Since substituting the coordinates of the intersection point of $L_1$ and $L_2$ into the equation of $L_3$ satisfies the equation (results in 0), the third line also passes through the same point.

Therefore, the lines $2x - 3y + 5 = 0$, $3x + 4y - 7 = 0$, and $5x + y - 2 = 0$ are concurrent.

The point of concurrency is the intersection point we found, which is $(\frac{1}{17}, \frac{29}{17})$.

Given:

A point through which the required line passes is $(x_1, y_1) = (2, 3)$.

The angle between the required line and the given line is $\theta = 45^\circ$.

The equation of the given line is $3x - 4y + 5 = 0$.

To Find:

The equation(s) of the line(s) satisfying the given conditions.

Solution:

First, find the slope of the given line $3x - 4y + 5 = 0$.

The equation of a line in the form $Ax + By + C = 0$ has a slope $m = -\frac{A}{B}$.

For the line $3x - 4y + 5 = 0$, we have $A = 3$, $B = -4$, and $C = 5$.

The slope of the given line is $m_1 = -\frac{3}{-4} = \frac{3}{4}$.

Let $m_2$ be the slope of the required line.

The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$

We are given $\theta = 45^\circ$, so $\tan 45^\circ = 1$. Also, $m_1 = \frac{3}{4}$. Substitute these values into the formula:

$1 = \left|\frac{m_2 - \frac{3}{4}}{1 + \frac{3}{4} m_2}\right|$

Simplify the expression inside the absolute value:

$\frac{m_2 - \frac{3}{4}}{1 + \frac{3}{4} m_2} = \frac{\frac{4m_2 - 3}{4}}{\frac{4 + 3m_2}{4}} = \frac{4m_2 - 3}{4 + 3m_2}$

So, we have:

$1 = \left|\frac{4m_2 - 3}{4 + 3m_2}\right|$

This gives two possible cases:

Case 1: $\frac{4m_2 - 3}{4 + 3m_2} = 1$

$4m_2 - 3 = 4 + 3m_2$

$4m_2 - 3m_2 = 4 + 3$

$m_2 = 7$

Case 2: $\frac{4m_2 - 3}{4 + 3m_2} = -1$

$4m_2 - 3 = -(4 + 3m_2)$

$4m_2 - 3 = -4 - 3m_2$

$4m_2 + 3m_2 = -4 + 3$

$7m_2 = -1$

$m_2 = -\frac{1}{7}$

There are two possible slopes for the required line: $7$ and $-\frac{1}{7}$. Since the line passes through the point $(2, 3)$, we can use the point-slope form $y - y_1 = m(x - x_1)$ to find the equation for each case.

Equation for the line with slope $m_2 = 7$ and passing through (2, 3):

$y - 3 = 7(x - 2)$

$y - 3 = 7x - 14$

Rearranging to the general form $Ax + By + C = 0$:

$7x - y - 14 + 3 = 0$

$7x - y - 11 = 0$

Equation for the line with slope $m_2 = -\frac{1}{7}$ and passing through (2, 3):

$y - 3 = -\frac{1}{7}(x - 2)$

Multiply both sides by 7:

$7(y - 3) = -1(x - 2)$

$7y - 21 = -x + 2$

Rearranging to the general form $Ax + By + C = 0$:

$x + 7y - 21 - 2 = 0$

$x + 7y - 23 = 0$

Therefore, there are two lines passing through the point (2, 3) and making an angle of $45^\circ$ with the line $3x - 4y + 5 = 0$. Their equations are $7x - y - 11 = 0$ and $x + 7y - 23 = 0$.

Given:

Point A with coordinates $(1, 2)$.

Point B with coordinates $(3, -4)$.

The sum of the squares of the distances from a point P on the locus to A and B is 50.

To Find:

The equation of the locus of point P.

Solution:

Let P$(x, y)$ be any point on the locus.

The distance between P$(x, y)$ and A$(1, 2)$ is $PA$. The square of the distance is $PA^2$.

Using the distance formula, $PA^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:

$PA^2 = (x - 1)^2 + (y - 2)^2$

The distance between P$(x, y)$ and B$(3, -4)$ is $PB$. The square of the distance is $PB^2$.

$PB^2 = (x - 3)^2 + (y - (-4))^2 = (x - 3)^2 + (y + 4)^2$

According to the problem statement, the sum of the squares of the distances from P to A and B is 50:

$PA^2 + PB^2 = 50$

Substitute the expressions for $PA^2$ and $PB^2$:

$(x - 1)^2 + (y - 2)^2 + (x - 3)^2 + (y + 4)^2 = 50$

Expand the squared terms:

$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (x^2 - 6x + 9) + (y^2 + 8y + 16) = 50$

Combine like terms:

$(x^2 + x^2) + (y^2 + y^2) + (-2x - 6x) + (-4y + 8y) + (1 + 4 + 9 + 16) = 50$

$2x^2 + 2y^2 - 8x + 4y + 30 = 50$

Move the constant term to the left side:

$2x^2 + 2y^2 - 8x + 4y + 30 - 50 = 0$

$2x^2 + 2y^2 - 8x + 4y - 20 = 0$

Divide the entire equation by 2 to simplify:

$\frac{2x^2}{2} + \frac{2y^2}{2} - \frac{8x}{2} + \frac{4y}{2} - \frac{20}{2} = \frac{0}{2}$

$x^2 + y^2 - 4x + 2y - 10 = 0$

This is the equation of the locus of the point P. This equation represents a circle.

Given:

Equation of the first line: $L_1: 3x - 4y + 7 = 0$.

Equation of the second line: $L_2: 3x - 4y + 5 = 0$.

To Find:

The distance between the two parallel lines.

Solution:

The two lines are in the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$. For the given lines, we have:

$A = 3$

$B = -4$

$C_1 = 7$

$C_2 = 5$

The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

Substitute the values of $A$, $B$, $C_1$, and $C_2$ into the formula:

$d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}}$

$d = \frac{|2|}{\sqrt{9 + 16}}$

$d = \frac{2}{\sqrt{25}}$

$d = \frac{2}{5}$

Therefore, the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$ is $\frac{2}{5}$ units.

Alternate Solution (using a point on one line):

Pick any point on one of the lines, say $L_1: 3x - 4y + 7 = 0$.

Let $x = 1$. Then $3(1) - 4y + 7 = 0 \implies 3 - 4y + 7 = 0 \implies 10 - 4y = 0 \implies 4y = 10 \implies y = \frac{10}{4} = \frac{5}{2}$.

So, the point $(1, \frac{5}{2})$ lies on $L_1$.

Now find the distance of this point $(x_1, y_1) = (1, \frac{5}{2})$ from the other line $L_2: 3x - 4y + 5 = 0$.

The equation of $L_2$ is $3x - 4y + 5 = 0$, so $A = 3$, $B = -4$, and $C = 5$.

The distance $d$ is given by the formula:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

$d = \frac{|(3)(1) + (-4)(\frac{5}{2}) + 5|}{\sqrt{3^2 + (-4)^2}}$

$d = \frac{|3 - 10 + 5|}{\sqrt{9 + 16}}$

$d = \frac{|-7 + 5|}{\sqrt{25}}$

$d = \frac{|-2|}{5}$

$d = \frac{2}{5}$

This confirms the result obtained using the direct formula for the distance between parallel lines.

Let the equation of a line be $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

Given:

The line passes through the point $(x_1, y_1) = (0, 2)$.

The line makes an angle $\theta = \frac{2\pi}{3}$ with the positive x-axis.

To Find:

1. The equation of the line passing through $(0, 2)$ with angle $\frac{2\pi}{3}$.

2. The equation of the line parallel to the first line and cutting the y-axis at a distance of 2 units below the origin (i.e., passing through $(0, -2)$).

Solution:

The slope $m$ of a line making an angle $\theta$ with the positive x-axis is given by $m = \tan(\theta)$.

The first line passes through the point $(x_1, y_1) = (0, 2)$ and has a slope $m_1 = -\sqrt{3}$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$:

This is the equation of the first line.

Now, we need to find the equation of a line parallel to the first line and cutting the y-axis at a distance of 2 units below the origin.

A line parallel to the first line will have the same slope.

Cutting the y-axis at a distance of 2 units below the origin means the line passes through the point $(0, -2)$.

Using the point-slope form with $(x_2, y_2) = (0, -2)$ and slope $m_2 = -\sqrt{3}$:

This is the equation of the second line.

Answer:

The equation of the line passing through $(0, 2)$ making an angle $\frac{2\pi}{3}$ with the positive x-axis is $\mathbf{\sqrt{3}x + y - 2 = 0}$.

The equation of the line parallel to the first line and cutting the y-axis at a distance of 2 units below the origin is $\mathbf{\sqrt{3}x + y + 2 = 0}$.

Let the three given lines be:

To find the area of the triangle formed by these lines, we first need to find the vertices of the triangle, which are the points of intersection of these lines taken pairwise.

Finding the Vertices:

Intersection of Line 1 and Line 2:

Add equation (1) and equation (2):

Substitute $x=1$ into equation (1):

The first vertex is $A = (1, 2)$.

Intersection of Line 1 and Line 3:

Subtract equation (1) from equation (3):

Substitute $x=3$ into equation (1):

The second vertex is $B = (3, 0)$.

Intersection of Line 2 and Line 3:

From equation (2), $y = x + 1$. Substitute this into equation (3):

Substitute $x = \frac{5}{3}$ into $y = x + 1$:

The third vertex is $C = \left(\frac{5}{3}, \frac{8}{3}\right)$.

The vertices of the triangle are $A(1, 2)$, $B(3, 0)$, and $C\left(\frac{5}{3}, \frac{8}{3}\right)$.

Calculating the Area:

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

Let $(x_1, y_1) = (1, 2)$, $(x_2, y_2) = (3, 0)$, and $(x_3, y_3) = \left(\frac{5}{3}, \frac{8}{3}\right)$.

Answer:

The area of the triangle formed by the given lines is $\mathbf{\frac{4}{3}}$ square units.

Let the given point be $P(x_1, y_1) = (3, 8)$.

Let the equation of the given line be $L: x + 3y = 7$, which can be written as $x + 3y - 7 = 0$.

Comparing this with the standard form $ax + by + c = 0$, we have $a = 1$, $b = 3$, and $c = -7$.

To Find:

The image of the point $P(3, 8)$ with respect to the line $x + 3y = 7$. Let the image point be $P'(x_2, y_2)$.

Solution:

The image of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$ is the point $(x_2, y_2)$ given by the formula:

Substitute the values of $(x_1, y_1) = (3, 8)$ and $a = 1$, $b = 3$, $c = -7$ into the formula:

First, calculate the value of the right-hand side:

Now, equate the first part of the formula to -4:

Next, equate the second part of the formula to -4:

So, the image of the point $(3, 8)$ with respect to the line $x + 3y = 7$ is the point $(-1, -4)$.

Answer:

The image of the point (3, 8) with respect to the line $x + 3y = 7$ is $\mathbf{(-1, -4)}$.

Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.

Given:

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

The sum of its intercepts on the axes is 9.

To Find:

The equation of the line.

Solution:

Since the line passes through the point $(2, 2)$, substituting these coordinates into the intercept form of the equation, we get:

From equation (i), we can express $b$ in terms of $a$:

Substitute this expression for $b$ into equation (ii):

Find a common denominator and combine the terms on the left side:

Multiply both sides by $(9a - a^2)$:

Rearrange the equation into a quadratic form:

Factor the quadratic equation:

This gives two possible values for $a$:

Case 1: $a = 3$

Using equation (i), $b = 9 - a = 9 - 3 = 6$.

The equation of the line is $\frac{x}{3} + \frac{y}{6} = 1$.

Multiply the equation by the LCM of 3 and 6, which is 6, to clear the denominators:

Case 2: $a = 6$

Using equation (i), $b = 9 - a = 9 - 6 = 3$.

The equation of the line is $\frac{x}{6} + \frac{y}{3} = 1$.

Multiply the equation by the LCM of 6 and 3, which is 6, to clear the denominators:

Both lines satisfy the given conditions.

Answer:

There are two possible equations for the line:

$2x + y = 6$

$x + 2y = 6$

Let the required line be L. The perpendicular from the origin O(0, 0) to the line L meets it at point P(-2, 9).

Given:

Point on the line where the perpendicular from the origin meets it: $P(-2, 9)$.

Origin: $O(0, 0)$.

To Find:

The equation of the line L.

Solution:

The line segment OP is the perpendicular from the origin to the line L. Therefore, the line segment OP is perpendicular to the line L.

First, we find the slope of the line segment OP.

The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Using points $O(0, 0)$ and $P(-2, 9)$, the slope of OP is:

Since the line L is perpendicular to the line segment OP, the product of their slopes is -1 (provided neither line is vertical or horizontal).

Let the slope of the required line L be $m_L$.

The required line L passes through the point $P(-2, 9)$ and has a slope $m_L = \frac{2}{9}$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (-2, 9)$ and $m = \frac{2}{9}$:

Multiply both sides by 9 to eliminate the fraction:

Rearrange the equation into the standard form $Ax + By + C = 0$:

Answer:

The equation of the line is $\mathbf{2x - 9y + 85 = 0}$.

Let the slope of the first line be $m_1$ and the slope of the second line be $m_2$.

Let the angle between the two lines be $\theta$.

Given:

To Find:

The slope of the other line ($m_2$).

Solution:

The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

Substitute the given values $\theta = \frac{\pi}{4}$ and $m_1 = \frac{1}{2}$ into the formula:

We know that $\tan\left(\frac{\pi}{4}\right) = 1$. So,

This implies that the expression inside the absolute value can be either 1 or -1.

Case 1: $\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = 1$

Collect terms involving $m_2$ on one side and constant terms on the other:

Multiply both sides by 2:

Case 2: $\frac{m_2 - \frac{1}{2}}{1 + \frac{1}{2} m_2} = -1$

Collect terms involving $m_2$ on one side and constant terms on the other:

Multiply both sides by $\frac{2}{3}$:

Thus, there are two possible values for the slope of the other line.

Answer:

The slope of the other line can be either 3 or $-\frac{1}{3}$.