Chapter 7 Coordinate Geometry (Additional Questions)

Solution:

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To find the distance of a point $(x, y)$ from the x-axis, we consider the absolute value of its y-coordinate.

The given point is $(4, -3)$.

Here, the x-coordinate is $4$ and the y-coordinate is $-3$.

The distance of the point $(4, -3)$ from the x-axis is the absolute value of the y-coordinate.

Distance = $|y|$

Distance = $|-3|$

Distance = $3$ units

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The distance of the point $(4, -3)$ from the x-axis is 3 units.

This corresponds to option (C).

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The final answer is $\boxed{3 \text{ units}}$.

The correct option is (C).

Solution:

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Given:

The coordinates of the endpoints of the line segment are $(x_1, y_1) = (2, 5)$ and $(x_2, y_2) = (7, 10)$.

The line segment is divided internally in the ratio $m_1 : m_2 = 2 : 3$.

To Find:

The coordinates of the point $(x, y)$ which divides the line segment internally in the given ratio.

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Formula:

We use the section formula for internal division.

If a point $(x, y)$ divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$, then the coordinates of the point are given by:

$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$

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Calculation:

Substitute the given values into the section formula:

$x = \frac{(2)(7) + (3)(2)}{2 + 3}$

$x = \frac{14 + 6}{5}$

$x = \frac{20}{5}$

$x = 4$

Now, calculate the y-coordinate:

$y = \frac{(2)(10) + (3)(5)}{2 + 3}$

$y = \frac{20 + 15}{5}$

$y = \frac{35}{5}$

$y = 7$

So, the coordinates of the point are $(4, 7)$.

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Conclusion:

The coordinates of the point which divides the line segment joining $(2, 5)$ and $(7, 10)$ internally in the ratio $2:3$ are $(4, 7)$.

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

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The final answer is $\boxed{(4, 7)}$.

The correct option is (A).

Solution:

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Given:

The vertices of the triangle are A(1, -1), B(-4, 6), and C(-3, -5).

Let $(x_1, y_1) = (1, -1)$

Let $(x_2, y_2) = (-4, 6)$

Let $(x_3, y_3) = (-3, -5)$

To Find:

The area of the triangle with the given vertices.

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Formula:

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

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Calculation:

Substitute the coordinates of the vertices into the formula:

Area $= \frac{1}{2} |1(6 - (-5)) + (-4)((-5) - (-1)) + (-3)((-1) - 6)|$

Area $= \frac{1}{2} |1(6 + 5) - 4(-5 + 1) - 3(-1 - 6)|$

Area $= \frac{1}{2} |1(11) - 4(-4) - 3(-7)|$

Area $= \frac{1}{2} |11 + 16 + 21|$

Area $= \frac{1}{2} |48|$

Area $= \frac{1}{2} \times 48$

Area $= 24$ square units

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Conclusion:

The area of the triangle with vertices A(1, -1), B(-4, 6), and C(-3, -5) is 24 square units.

Comparing this result with the given options, we see that option (A) matches our calculated area.

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The final answer is $\boxed{24 \text{ square units}}$.

The correct option is (A).

Solution:

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Given:

The two points are A(-5, 7) and B(-1, 3).

Let $(x_1, y_1) = (-5, 7)$

Let $(x_2, y_2) = (-1, 3)$

To Find:

The distance between points A and B.

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Formula:

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

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Calculation:

Substitute the coordinates of points A and B into the distance formula:

Distance $= \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$

Distance $= \sqrt{(-1 + 5)^2 + (-4)^2}$

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

Distance $= \sqrt{16 + 16}$

Distance $= \sqrt{32}$

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Conclusion:

The distance between the points A(-5, 7) and B(-1, 3) is $\sqrt{32}$ units.

Comparing this result with the given options, we see that option (A) matches our calculated distance.

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The final answer is $\boxed{\sqrt{32} \text{ units}}$.

The correct option is (A).

Solution:

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Given:

The two points are $(x_1, y_1) = (a, b)$ and $(x_2, y_2) = (-a, -b)$.

To Find:

The midpoint of the line segment joining these two points.

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Formula:

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

$x = \frac{x_1 + x_2}{2}$

$y = \frac{y_1 + y_2}{2}$

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Calculation:

Substitute the given coordinates into the midpoint formula:

$x = \frac{a + (-a)}{2}$

$x = \frac{a - a}{2}$

$x = \frac{0}{2}$

$x = 0$

Now, calculate the y-coordinate:

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

$y = \frac{b - b}{2}$

$y = \frac{0}{2}$

$y = 0$

So, the coordinates of the midpoint are $(0, 0)$.

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Conclusion:

The midpoint of the line segment joining the points $(a, b)$ and $(-a, -b)$ is $(0, 0)$.

Comparing this result with the given options, we see that option (C) matches our calculated midpoint.

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The final answer is $\boxed{(0, 0)}$.

The correct option is (C).

Solution:

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Given:

The coordinates of point A are $(x_1, y_1) = (1, p)$.

The coordinates of point B are $(x_2, y_2) = (4, 0)$.

The distance between A and B is AB = 5 units.

To Find:

The value(s) of p.

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Formula:

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

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Calculation:

We are given that the distance AB is 5 units. Using the distance formula:

$5 = \sqrt{(4 - 1)^2 + (0 - p)^2}$

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

$5 = \sqrt{9 + p^2}$

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

$5^2 = (\sqrt{9 + p^2})^2$

$25 = 9 + p^2$

Now, isolate the $p^2$ term:

$p^2 = 25 - 9$

$p^2 = 16$

Take the square root of both sides to find the value(s) of p:

$p = \pm \sqrt{16}$

$p = \pm 4$

So, the possible values of p are 4 and -4.

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Conclusion:

The value(s) of p for which the distance between A(1, p) and B(4, 0) is 5 units are $\pm 4$.

Comparing this result with the given options, we see that option (C) matches our values.

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The final answer is $\boxed{\pm 4}$.

The correct option is (C).

Solution:

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Given:

The three points are (1, 5), (2, 3), and (-2, -11).

The points are stated to be collinear.

To Find:

The area of the triangle formed by these points.

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Concept:

Three or more points are said to be collinear if they lie on the same straight line.

When three points are collinear, they do not form a triangle in the geometric sense of having a non-zero area. The "triangle" formed by collinear points is a degenerate triangle, which is essentially a line segment.

The area of a degenerate triangle is always zero.

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Conclusion:

Since the problem explicitly states that the given points are collinear, the area of the triangle formed by these points must be 0.

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

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

If the points are collinear, this expression will always evaluate to 0.

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The final answer is $\boxed{0}$.

The correct option is (C).

Solution:

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Given:

Point A = $(2, -5)$

Point B = $(-2, 9)$

A point on the x-axis is equidistant from A and B.

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To Find:

The coordinates of the point on the x-axis.

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Formula:

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

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Calculation:

Let the point on the x-axis be P$(x, 0)$. Since the point is on the x-axis, its y-coordinate is 0.

The distance PA must be equal to the distance PB.

$PA = PB$

Squaring both sides, we get $PA^2 = PB^2$.

Using the distance formula squared:

$PA^2 = (x - 2)^2 + (0 - (-5))^2$

$PA^2 = (x - 2)^2 + (5)^2$

$PA^2 = x^2 - 4x + 4 + 25$

$PA^2 = x^2 - 4x + 29$

$PB^2 = (x - (-2))^2 + (0 - 9)^2$

$PB^2 = (x + 2)^2 + (-9)^2$

$PB^2 = x^2 + 4x + 4 + 81$

$PB^2 = x^2 + 4x + 85$

Setting $PA^2$ equal to $PB^2$:

$x^2 - 4x + 29 = x^2 + 4x + 85$

Subtract $x^2$ from both sides:

$-4x + 29 = 4x + 85$

Add $4x$ to both sides:

$29 = 8x + 85$

Subtract 85 from both sides:

$29 - 85 = 8x$

$-56 = 8x$

Divide by 8:

$x = \frac{-56}{8}$

$x = -7$

The x-coordinate of the point is -7.

Since the point is on the x-axis, its y-coordinate is 0.

Thus, the coordinates of the point are $(-7, 0)$.

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Conclusion:

The point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9)$ is $(-7, 0)$.

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

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The final answer is $\boxed{(-7, 0)}$.

The correct option is (A).

Solution:

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Given:

The vertices of a 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.

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Formula:

The centroid of a triangle is the point of intersection of its medians. The coordinates of the centroid G$(x, y)$ of a triangle with vertices A$(x_1, y_1)$, B$(x_2, y_2)$, and C$(x_3, y_3)$ are given by the formula:

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

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

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

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Conclusion:

Comparing this formula with the given options, we find that option (C) matches the correct coordinates for the centroid.

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The final answer is $\boxed{(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})}$.

The correct option is (C).

Solution:

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Given:

The vertices of the triangle $\triangle ABO$ are A($a, 0$), B(0, $b$), and O(0, 0).

Let $(x_1, y_1) = (a, 0)$

Let $(x_2, y_2) = (0, b)$

Let $(x_3, y_3) = (0, 0)$

The area of $\triangle ABO$ is given as 20 square units.

To Find:

The relation between $a$ and $b$.

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Formula:

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

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Calculation:

Substitute the coordinates of the vertices A($a, 0$), B(0, $b$), and O(0, 0) into the area formula, and set the area equal to 20:

$20 = \frac{1}{2} |a(b - 0) + 0(0 - 0) + 0(0 - b)|$

$20 = \frac{1}{2} |a(b) + 0 + 0|$

$20 = \frac{1}{2} |ab|$

Multiply both sides of the equation by 2:

$20 \times 2 = 2 \times \frac{1}{2} |ab|$

$40 = |ab|$

Thus, the relation between $a$ and $b$ is $|ab| = 40$.

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Conclusion:

The area of the triangle with vertices A($a, 0$), B(0, $b$), and O(0, 0) being 20 square units implies that $|ab| = 40$.

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

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The final answer is $\boxed{|ab| = 40}$.

The correct option is (B).

Solution:

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A point lies on the y-axis if and only if its x-coordinate is 0.

Let's examine the x-coordinate of each given point:

(A) $(3, 0)$: The x-coordinate is $3$. This point lies on the x-axis.

(B) $(0, -4)$: The x-coordinate is $0$. This point lies on the y-axis.

(C) $(2, 5)$: The x-coordinate is $2$. This point lies in the first quadrant.

(D) $(-1, 0)$: The x-coordinate is $-1$. This point lies on the x-axis.

Therefore, the point with an x-coordinate of 0 is $(0, -4)$.

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Conclusion:

The point $(0, -4)$ lies on the y-axis because its x-coordinate is 0.

Comparing this with the given options, we find that option (B) is the correct answer.

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The final answer is $\boxed{(0, -4)}$.

The correct option is (B).

Solution:

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Given:

The endpoints of the line segment are A$(5, -6)$ and B$(-1, -4)$.

Let $(x_1, y_1) = (5, -6)$ and $(x_2, y_2) = (-1, -4)$.

The line segment is divided by the y-axis.

To Find:

The ratio in which the y-axis divides the line segment AB, and whether the division is internal or external.

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Formula:

We use the section formula. If a point P$(x, y)$ divides the line segment joining A$(x_1, y_1)$ and B$(x_2, y_2)$ in the ratio $m_1 : m_2$, then its coordinates are given by:

$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$

If a point lies on the y-axis, its x-coordinate is 0. Let the ratio be $k : 1$.

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Calculation:

Let the point of intersection of the line segment and the y-axis be P$(0, y)$.

Using the section formula for the x-coordinate with ratio $k:1$:

$0 = \frac{k(x_2) + 1(x_1)}{k + 1}$

$0 = \frac{k(-1) + 1(5)}{k + 1}$

$0 = \frac{-k + 5}{k + 1}$

Since the denominator $k+1$ cannot be zero, the numerator must be zero:

$-k + 5 = 0$

$5 = k$

The ratio $k:1$ is $5:1$.

Since the value of $k = 5$ is positive, the point divides the line segment internally.

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Conclusion:

The y-axis divides the line segment joining $(5, -6)$ and $(-1, -4)$ in the ratio $5:1$ internally.

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

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The final answer is $\boxed{5:1 \text{ internally}}$.

The correct option is (B).

Solution:

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Given:

The point is $(p, q)$.

The origin is the point $(0, 0)$.

Let $(x_1, y_1) = (p, q)$ and $(x_2, y_2) = (0, 0)$.

To Find:

The distance between the point $(p, q)$ and the origin $(0, 0)$.

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Formula:

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

For the distance from the origin $(0, 0)$ to a point $(x, y)$, the formula simplifies to:

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

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Calculation:

Using the simplified distance formula for the point $(p, q)$ from the origin $(0, 0)$:

Distance $= \sqrt{(p - 0)^2 + (q - 0)^2}$

Distance $= \sqrt{p^2 + q^2}$

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Conclusion:

The distance of the point $(p, q)$ from the origin is $\sqrt{p^2 + q^2}$.

Comparing this result with the given options, we find that option (C) matches our calculated distance.

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The final answer is $\boxed{\sqrt{p^2+q^2}}$.

The correct option is (C).

Solution:

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Given:

Coordinates of one end of the diameter: $(x_1, y_1) = (2, 3)$.

Coordinates of the center of the circle: $(x_m, y_m) = (-2, 5)$.

The center of the circle is the midpoint of its diameter.

To Find:

The coordinates of the other end of the diameter, let's call it $(x_2, y_2)$.

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Formula:

If $(x_m, y_m)$ is the midpoint of the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$, then the midpoint formula is:

$x_m = \frac{x_1 + x_2}{2}$

$y_m = \frac{y_1 + y_2}{2}$

We can rearrange these formulas to find the coordinates of the other endpoint $(x_2, y_2)$:

$x_2 = 2x_m - x_1$

$y_2 = 2y_m - y_1$

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Calculation:

Substitute the given values into the rearranged midpoint formulas:

$x_2 = 2(-2) - 2$

$x_2 = -4 - 2$

$x_2 = -6$

Now, calculate the y-coordinate:

$y_2 = 2(5) - 3$

$y_2 = 10 - 3$

$y_2 = 7$

So, the coordinates of the other end of the diameter are $(-6, 7)$.

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Conclusion:

The coordinates of the other end of the diameter are $(-6, 7)$.

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

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The final answer is $\boxed{(-6, 7)}$.

The correct option is (A).

Solution:

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Let's analyze the given Assertion (A) and Reason (R).

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

This statement is the standard distance formula in coordinate geometry. It is a fundamental result used to calculate the distance between any two points in a Cartesian plane.

Thus, Assertion (A) is True.

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Reason (R): This formula is derived using the Pythagorean theorem in a right triangle formed by the points and their projections on the axes.

Consider two points P$(x_1, y_1)$ and Q$(x_2, y_2)$. We can construct a right-angled triangle by drawing a line through P parallel to the x-axis and a line through Q parallel to the y-axis. These lines intersect at a point R. The coordinates of R will be $(x_2, y_1)$ or $(x_1, y_2)$ depending on how the triangle is formed. Let's use $(x_2, y_1)$.

The vertices of the right triangle are P$(x_1, y_1)$, R$(x_2, y_1)$, and Q$(x_2, y_2)$.

The length of the leg PR (horizontal distance) is the absolute difference of the x-coordinates: $PR = |x_2 - x_1|$.

The length of the leg QR (vertical distance) is the absolute difference of the y-coordinates: $QR = |y_2 - y_1|$.

The distance between P and Q is the hypotenuse PQ.

By the Pythagorean theorem, in the right-angled triangle PRQ:

$PQ^2 = PR^2 + QR^2$

$PQ^2 = (|x_2 - x_1|)^2 + (|y_2 - y_1|)^2$

Since the square of a number is the same as the square of its absolute value, we have:

$PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Taking the square root of both sides, we get the distance formula:

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

This derivation confirms that the distance formula is indeed derived using the Pythagorean theorem as described in the Reason.

Thus, Reason (R) is True.

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

The Reason explains *how* the distance formula (stated in the Assertion) is derived. The derivation process using the Pythagorean theorem is the fundamental basis for the distance formula in a Cartesian coordinate system.

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

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

Comparing this conclusion with the given options, we find that option (A) is correct.

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The final answer is $\boxed{\text{Both A and R are true and R is the correct explanation of A.}}$.

The correct option is (A).

Solution:

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

Assertion (A): If three points are collinear, the area of the triangle formed by them is 0.

When three points lie on the same straight line, they do not form a proper triangle with a non-zero area. Such a configuration is called a degenerate triangle, and its area is always zero. This statement is a fundamental property in coordinate geometry relating collinearity and the area of a triangle.

Thus, Assertion (A) is True.

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Reason (R): The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.

This statement provides the standard formula for calculating the area of a triangle when the coordinates of its vertices are known. This formula is widely used in coordinate geometry.

Thus, Reason (R) is True.

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

The Reason provides the formula for the area of a triangle. The Assertion states that the area is 0 if the points are collinear.

In coordinate geometry, one of the methods to check if three points are collinear is to calculate the area of the triangle formed by these points using the formula given in Reason (R).

If the points are collinear, substituting their coordinates into the formula $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ will result in a value of 0.

Therefore, the fact that the area of the triangle is 0 for collinear points (Assertion A) is directly verified and explained by the behaviour of the area formula (Reason R) when the points are collinear. The formula is the mathematical tool that confirms this property.

Hence, Reason (R) is the correct explanation of Assertion (A).

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

Comparing this conclusion with the given options, we find that option (A) is the correct answer.

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The final answer is $\boxed{\text{Both A and R are true and R is the correct explanation of A.}}$.

The correct option is (A).

Solution:

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Let's identify each formula and match it with the corresponding concept.

(i) $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ is the formula used to calculate the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$. This is the Distance Formula.

So, (i) matches with (c).

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(ii) $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ is the formula used to find the coordinates of the midpoint of the line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$. This is the Midpoint Formula.

So, (ii) matches with (d).

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(iii) $(\frac{m_1 x_2 + m_2 x_1}{m_1+m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1+m_2})$ is the formula used to find the coordinates of a point that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$. This is the Section Formula (Internal).

So, (iii) matches with (a).

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(iv) $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ is the formula used to calculate the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$. This is the Area of a Triangle formula using coordinates.

So, (iv) matches with (b).

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Based on the matches:

(i) - (c)

(ii) - (d)

(iii) - (a)

(iv) - (b)

Comparing this with the given options, option (A) matches our findings.

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The final answer is $\boxed{\text{(i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)}}$.

The correct option is (A).

Solution:

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Given:

The coordinates of the school (S) are $(x_1, y_1) = (2, 5)$.

The coordinates of the library (L) are $(x_2, y_2) = (7, 10)$.

To Find:

The straight-line distance between the school (S) and the library (L).

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Formula:

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

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Calculation:

Substitute the coordinates of points S and L into the distance formula:

Distance SL $= \sqrt{(7 - 2)^2 + (10 - 5)^2}$

Distance SL $= \sqrt{(5)^2 + (5)^2}$

Distance SL $= \sqrt{25 + 25}$

Distance SL $= \sqrt{50}$

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Conclusion:

The straight-line distance between the school and the library is $\sqrt{50}$ units.

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

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The final answer is $\boxed{\sqrt{50} \text{ units}}$.

The correct option is (A).

Solution:

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Given:

The coordinates of the school (S) are $(x_1, y_1) = (2, 5)$.

The coordinates of the hospital (H) are $(x_2, y_2) = (-3, 2)$.

The community center (C) is located exactly halfway between S and H, meaning it is the midpoint of the line segment SH.

To Find:

The coordinates of the community center (C), which is the midpoint of SH.

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Formula:

The midpoint $(x_m, y_m)$ of a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the midpoint formula:

$x_m = \frac{x_1 + x_2}{2}$

$y_m = \frac{y_1 + y_2}{2}$

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Calculation:

Substitute the coordinates of S and H into the midpoint formula to find the coordinates of the community center C$(x_C, y_C)$:

$x_C = \frac{2 + (-3)}{2}$

$x_C = \frac{2 - 3}{2}$

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

$x_C = -0.5$

Now, calculate the y-coordinate:

$y_C = \frac{5 + 2}{2}$

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

$y_C = 3.5$

So, the coordinates of the community center are $(-0.5, 3.5)$.

---

Conclusion:

The coordinates of the community center, which is halfway between the school and the hospital, are $(-0.5, 3.5)$.

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

---

The final answer is $\boxed{(-0.5, 3.5)}$.

The correct option is (A).

Solution:

---

Given:

The three points A(1, 2), B(3, 4), and C(x, 6) are collinear.

Let $(x_1, y_1) = (1, 2)$

Let $(x_2, y_2) = (3, 4)$

Let $(x_3, y_3) = (x, 6)$

To Find:

The value of x.

---

Method 1: Using the Area of a Triangle

If three points are collinear, the area of the triangle formed by them is 0.

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

Since the points are collinear, the Area = 0.

$0 = \frac{1}{2} |1(4 - 6) + 3(6 - 2) + x(2 - 4)|$

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

$0 = \frac{1}{2} |-2 + 12 - 2x|$

$0 = \frac{1}{2} |10 - 2x|$

For the area to be 0, the expression inside the absolute value must be 0.

$10 - 2x = 0$

$10 = 2x$

$x = \frac{10}{2}$

$x = 5$

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Method 2: Using Slope

If three points are collinear, the slope between any two pairs of points is equal.

Slope of AB ($m_{AB}$) $= \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 2}{3 - 1} = \frac{2}{2} = 1$

Slope of BC ($m_{BC}$) $= \frac{y_3 - y_2}{x_3 - x_2} = \frac{6 - 4}{x - 3} = \frac{2}{x - 3}$

Since the points are collinear, $m_{AB} = m_{BC}$.

$1 = \frac{2}{x - 3}$

Multiply both sides by $(x - 3)$ (assuming $x \neq 3$, otherwise the points would not be distinct):

$1 \times (x - 3) = 2$

$x - 3 = 2$

$x = 2 + 3$

$x = 5$

---

Conclusion:

Using either the area of a triangle or the slope method, we find that the value of x is 5.

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

---

The final answer is $\boxed{5}$.

The correct option is (B).

Solution:

---

Given:

The vertices of the triangle are A(2, 3), B(-1, 0), and C(2, -4).

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

Let $(x_2, y_2) = (-1, 0)$

Let $(x_3, y_3) = (2, -4)$

To Find:

The coordinates of the centroid of the triangle.

---

Formula:

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

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

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

---

Calculation:

Substitute the coordinates of the vertices into the centroid formula:

For the x-coordinate:

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

$x = \frac{2 - 1 + 2}{3}$

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

$x = 1$

For the y-coordinate:

$y = \frac{3 + 0 + (-4)}{3}$

$y = \frac{3 + 0 - 4}{3}$

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

So, the coordinates of the centroid are $(1, -\frac{1}{3})$.

---

Conclusion:

The centroid of the triangle with vertices A(2, 3), B(-1, 0), and C(2, -4) is $(1, -\frac{1}{3})$.

Comparing this result with the given options, we find that option (C) matches our calculated coordinates.

---

The final answer is $\boxed{(1, -\frac{1}{3})}$.

The correct option is (C).

Solution:

---

Given:

The endpoints of the line segment are A$(2, -2)$ and B$(-7, 4)$.

Let $(x_1, y_1) = (2, -2)$ and $(x_2, y_2) = (-7, 4)$.

The line segment AB is trisected by two points. Let these points be P and Q, such that P is closer to A and Q is closer to B.

This means AP = PQ = QB.

To Find:

The coordinates of the points P and Q.

---

Concept of Trisection:

If P and Q trisect the line segment AB, then P divides AB in the ratio 1 : 2 (AP : PB = 1 : (PQ + QB) = 1 : (AP + AP) = 1 : 2).

Q divides AB in the ratio 2 : 1 (AQ : QB = (AP + PQ) : QB = (QB + QB) : QB = 2 : 1).

---

Formula:

We use the section formula for internal division. If a point $(x, y)$ divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$, then its coordinates are given by:

$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$

---

Calculation for point P (ratio 1 : 2):

Here, $(x_1, y_1) = (2, -2)$, $(x_2, y_2) = (-7, 4)$, $m_1 = 1$, $m_2 = 2$.

x-coordinate of P: $x_P = \frac{(1)(-7) + (2)(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1$

y-coordinate of P: $y_P = \frac{(1)(4) + (2)(-2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = 0$

The coordinates of point P are $(-1, 0)$.

---

Calculation for point Q (ratio 2 : 1):

Here, $(x_1, y_1) = (2, -2)$, $(x_2, y_2) = (-7, 4)$, $m_1 = 2$, $m_2 = 1$.

x-coordinate of Q: $x_Q = \frac{(2)(-7) + (1)(2)}{2 + 1} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4$

y-coordinate of Q: $y_Q = \frac{(2)(4) + (1)(-2)}{2 + 1} = \frac{8 - 2}{3} = \frac{6}{3} = 2$

The coordinates of point Q are $(-4, 2)$.

---

Conclusion:

The coordinates of the points which trisect the line segment joining A(2, -2) and B(-7, 4) are $(-1, 0)$ and $(-4, 2)$.

Comparing these coordinates with the given options, we find that option (A) matches our results.

---

The final answer is $\boxed{(-1, 0) \text{ and } (-4, 2)}$.

The correct option is (A).

Solution:

---

Given:

The endpoints of the line segment are A$(2, -2)$ and B$(3, 7)$.

Let $(x_1, y_1) = (2, -2)$ and $(x_2, y_2) = (3, 7)$.

The equation of the line is $2x + y - 4 = 0$.

To Find:

The ratio in which the line $2x + y - 4 = 0$ divides the line segment joining A and B, and whether the division is internal or external.

---

Formula:

Let the line $2x + y - 4 = 0$ divide the line segment AB at point P$(x, y)$ in the ratio $k:1$.

Using the section formula, the coordinates of P are:

$x = \frac{k x_2 + 1 x_1}{k + 1} = \frac{k(3) + 1(2)}{k + 1} = \frac{3k + 2}{k + 1}$

$y = \frac{k y_2 + 1 y_1}{k + 1} = \frac{k(7) + 1(-2)}{k + 1} = \frac{7k - 2}{k + 1}$

Since the point P lies on the line $2x + y - 4 = 0$, its coordinates must satisfy the equation of the line.

---

Calculation:

Substitute the expressions for $x$ and $y$ into the equation $2x + y - 4 = 0$:

$2\left(\frac{3k + 2}{k + 1}\right) + \left(\frac{7k - 2}{k + 1}\right) - 4 = 0$

Multiply the entire equation by $(k + 1)$ to eliminate the denominators (assuming $k \neq -1$):

$2(3k + 2) + (7k - 2) - 4(k + 1) = 0$

$6k + 4 + 7k - 2 - 4k - 4 = 0$

Combine like terms:

$(6k + 7k - 4k) + (4 - 2 - 4) = 0$

$9k - 2 = 0$

Solve for $k$:

$9k = 2$

$k = \frac{2}{9}$

The ratio is $k:1$, which is $\frac{2}{9}:1$, or $2:9$.

Since the value of $k = \frac{2}{9}$ is positive, the point of division lies between the points A and B. This means the division is internal.

---

Conclusion:

The line $2x + y - 4 = 0$ divides the line segment joining A(2, -2) and B(3, 7) in the ratio $2:9$ internally.

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

---

The final answer is $\boxed{2:9 \text{ internally}}$.

The correct option is (B).

Solution:

---

A point lies on the x-axis if its y-coordinate is equal to zero. The general form of a point on the x-axis is $(x, 0)$, where $x$ can be any real number.

Let's check the y-coordinate of each given point:

---

Analysis of Options:

(A) $(5, 0)$: The y-coordinate is $0$. This point lies on the x-axis.

(B) $(0, 5)$: The y-coordinate is $5$. This point does not lie on the x-axis (it lies on the y-axis).

(C) $(-3, 0)$: The y-coordinate is $0$. This point lies on the x-axis.

(D) $(0, -3)$: The y-coordinate is $-3$. This point does not lie on the x-axis (it lies on the y-axis).

(E) $(0, 0)$: The y-coordinate is $0$. This point (the origin) lies on the x-axis (and also on the y-axis).

---

Conclusion:

The points that lie on the x-axis are those whose y-coordinate is 0. From the given options, these points are $(5, 0)$, $(-3, 0)$, and $(0, 0)$.

Therefore, options (A), (C), and (E) correspond to points that lie on the x-axis.

---

The final answer consists of the options corresponding to points on the x-axis.

The correct options are (A), (C), and (E).

Solution:

---

Given:

The vertices of the triangle are $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (1, 0)$, and $(x_3, y_3) = (0, k)$.

The area of the triangle is given as 5 square units.

To Find:

The value(s) of $k$.

---

Formula:

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

---

Calculation:

Substitute the coordinates of the vertices and the given area into the formula:

$5 = \frac{1}{2} |0(0 - k) + 1(k - 0) + 0(0 - 0)|$

$5 = \frac{1}{2} |0 + k + 0|$

$5 = \frac{1}{2} |k|$

Multiply both sides of the equation by 2:

$5 \times 2 = |k|$

$10 = |k|$

The absolute value of $k$ is 10. This implies that $k$ can be either 10 or -10.

$k = \pm 10$

---

Conclusion:

The value(s) of $k$ for which the area of the triangle with vertices $(0, 0), (1, 0), (0, k)$ is 5 square units are $\pm 10$.

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

---

The final answer is $\boxed{\pm 10}$.

The correct option is (D).

Solution:

---

Given:

The point is $(x, y) = (-6, 8)$.

The origin is the point $(0, 0)$.

To Find:

The distance of the point $(-6, 8)$ from the origin $(0, 0)$.

---

Formula:

The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by the distance formula:

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

---

Calculation:

Substitute the coordinates of the point $(-6, 8)$ into the distance formula from the origin:

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

Distance $= \sqrt{36 + 64}$

Distance $= \sqrt{100}$

Distance $= 10$

The distance of the point $(-6, 8)$ from the origin is 10 units.

---

Conclusion:

The distance of the point $(-6, 8)$ from the origin is 10 units.

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

---

The final answer is $\boxed{10 \text{ units}}$.

The correct option is (A).

Solution:

---

Given:

The vertices of the triangle are A(1, 1), B(2, 4), and C(3, 5).

Let $(x_1, y_1) = (1, 1)$

Let $(x_2, y_2) = (2, 4)$

Let $(x_3, y_3) = (3, 5)$

To Find:

The area of the triangle ABC.

---

Formula:

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

---

Calculation:

Substitute the coordinates of the vertices into the area formula:

Area $= \frac{1}{2} |1(4 - 5) + 2(5 - 1) + 3(1 - 4)|$

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

Area $= \frac{1}{2} |-1 + 8 - 9|$

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

Since the absolute value of -2 is 2:

Area $= \frac{1}{2} \times 2$

Area $= 1$ square unit

---

Conclusion:

The area of the triangle with vertices A(1, 1), B(2, 4), and C(3, 5) is 1 square unit.

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

---

The final answer is $\boxed{1 \text{ square unit}}$.

The correct option is (A).

Solution:

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The Cartesian plane is divided into four quadrants based on the signs of the x and y coordinates.

Quadrant I: x > 0, y > 0 (positive x, positive y)

Quadrant II: x < 0, y > 0 (negative x, positive y)

Quadrant III: x < 0, y < 0 (negative x, negative y)

Quadrant IV: x > 0, y < 0 (positive x, negative y)

Points on the x-axis have a y-coordinate of 0. Points on the y-axis have an x-coordinate of 0. The origin (0, 0) is the intersection of the axes and is not considered to be in any quadrant.

---

Analysis of Options:

(A) $(3, -1)$: The x-coordinate is $3$ (positive) and the y-coordinate is $-1$ (negative). This point lies in Quadrant IV.

(B) $(-2, 5)$: The x-coordinate is $-2$ (negative) and the y-coordinate is $5$ (positive). This point lies in Quadrant II.

(C) $(-4, -3)$: The x-coordinate is $-4$ (negative) and the y-coordinate is $-3$ (negative). This point lies in Quadrant III.

(D) $(6, 2)$: The x-coordinate is $6$ (positive) and the y-coordinate is $2$ (positive). This point lies in Quadrant I.

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Conclusion:

The point with a negative x-coordinate and a positive y-coordinate lies in the second quadrant.

From the options, the point $(-2, 5)$ has an x-coordinate of $-2$ (negative) and a y-coordinate of $5$ (positive).

Therefore, the point $(-2, 5)$ lies in the second quadrant.

Comparing this with the given options, we find that option (B) is the correct answer.

---

The final answer is $\boxed{(-2, 5)}$.

The correct option is (B).

Solution:

---

Given:

The coordinates of Aman's location (A) are $(x_1, y_1) = (4, 6)$.

The coordinates of Bimal's location (B) are $(x_2, y_2) = (1, 5)$.

The coordinates of Charan's location (C) are $(7, 2)$, but these are not needed for this specific question.

To Find:

The square of the straight-line distance between Aman (A) and Bimal (B).

---

Formula:

The square of the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

Distance$^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

(This is derived directly from the distance formula, Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, by squaring both sides).

---

Calculation:

Substitute the coordinates of points A and B into the formula for the square of the distance:

AB$^2 = (1 - 4)^2 + (5 - 6)^2$

AB$^2 = (-3)^2 + (-1)^2$

AB$^2 = 9 + 1$

AB$^2 = 10$

The square of the distance between Aman and Bimal is 10 square units.

---

Conclusion:

The square of the distance between Aman and Bimal is 10.

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

---

The final answer is $\boxed{10}$.

The correct option is (A).

Solution:

---

Given:

The coordinates of Bimal (B) are $(x_1, y_1) = (1, 5)$.

The coordinates of Charan (C) are $(x_2, y_2) = (7, 2)$.

Deepak is located at the midpoint of the line segment joining B and C.

To Find:

The coordinates of Deepak.

---

Formula:

The midpoint $(x_m, y_m)$ of a line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the midpoint formula:

$x_m = \frac{x_1 + x_2}{2}$

$y_m = \frac{y_1 + y_2}{2}$

---

Calculation:

Substitute the coordinates of B and C into the midpoint formula to find Deepak's coordinates $(x_D, y_D)$:

For the x-coordinate:

$x_D = \frac{1 + 7}{2}$

$x_D = \frac{8}{2}$

$x_D = 4$

For the y-coordinate:

$y_D = \frac{5 + 2}{2}$

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

$y_D = 3.5$

So, Deepak's coordinates are $(4, 3.5)$.

---

Conclusion:

Deepak is located at the midpoint of BC, and his coordinates are $(4, 3.5)$.

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

---

The final answer is $\boxed{(4, 3.5)}$.

The correct option is (A).

Solution:

---

Given:

The vertices of the rhombus are given in order as A(3, 0), B(4, 5), C(-1, 4), and D(-2, -1).

Let $(x_1, y_1) = (3, 0)$

Let $(x_2, y_2) = (4, 5)$

Let $(x_3, y_3) = (-1, 4)$

Let $(x_4, y_4) = (-2, -1)$

To Find:

The area of the rhombus ABCD.

---

Method 1: Using the lengths of the diagonals

The area of a rhombus is half the product of the lengths of its diagonals. The diagonals of the rhombus ABCD are AC and BD.

The distance formula between two points $(x_a, y_a)$ and $(x_b, y_b)$ is $\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2}$.

Length of diagonal AC (distance between A(3, 0) and C(-1, 4)):

AC $= \sqrt{(-1 - 3)^2 + (4 - 0)^2}$

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

AC $= \sqrt{16 + 16}$

AC $= \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$ units

Length of diagonal BD (distance between B(4, 5) and D(-2, -1)):

BD $= \sqrt{(-2 - 4)^2 + (-1 - 5)^2}$

BD $= \sqrt{(-6)^2 + (-6)^2}$

BD $= \sqrt{36 + 36}$

BD $= \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$ units

Area of rhombus $= \frac{1}{2} \times \text{length of AC} \times \text{length of BD}$

Area $= \frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2})$

Area $= \frac{1}{2} \times 4 \times 6 \times (\sqrt{2} \times \sqrt{2})$

Area $= \frac{1}{2} \times 24 \times 2$

Area $= 24$ square units

---

Method 2: Using the area formula for a general quadrilateral

The area of a quadrilateral with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)$ taken in order is given by:

Area $= \frac{1}{2} |(x_1y_2 - y_1x_2) + (x_2y_3 - y_2x_3) + (x_3y_4 - y_3x_4) + (x_4y_1 - y_4x_1)|$

Substitute the coordinates of the vertices:

$x_1y_2 - y_1x_2 = (3)(5) - (0)(4) = 15 - 0 = 15$

$x_2y_3 - y_2x_3 = (4)(4) - (5)(-1) = 16 - (-5) = 16 + 5 = 21$

$x_3y_4 - y_3x_4 = (-1)(-1) - (4)(-2) = 1 - (-8) = 1 + 8 = 9$

$x_4y_1 - y_4x_1 = (-2)(0) - (-1)(3) = 0 - (-3) = 0 + 3 = 3$

Sum of the terms $= 15 + 21 + 9 + 3 = 48$

Area $= \frac{1}{2} |48|$

Area $= \frac{1}{2} \times 48$

Area $= 24$ square units

---

Conclusion:

Using both methods, we find that the area of the rhombus is 24 square units.

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

---

The final answer is $\boxed{24 \text{ square units}}$.

The correct option is (A).

Solution:

---

Given:

Point P$(x, y)$ is equidistant from point A$(5, 1)$ and point B$(-1, 5)$.

This means the distance from P to A is equal to the distance from P to B.

To Find:

The relation between $x$ and $y$ that satisfies the given condition.

---

Concept:

If point P$(x, y)$ is equidistant from points A$(x_1, y_1)$ and B$(x_2, y_2)$, then the distance PA is equal to the distance PB.

$PA = PB$

Squaring both sides to eliminate square roots:

$PA^2 = PB^2$

---

Formula:

The square of the distance between two points $(x_a, y_a)$ and $(x_b, y_b)$ is given by $(x_b - x_a)^2 + (y_b - y_a)^2$.

---

Calculation:

Let $(x_1, y_1) = (5, 1)$ and $(x_2, y_2) = (-1, 5)$.

The square of the distance PA is:

$PA^2 = (x - 5)^2 + (y - 1)^2$

The square of the distance PB is:

$PB^2 = (x - (-1))^2 + (y - 5)^2 = (x + 1)^2 + (y - 5)^2$

Setting $PA^2 = PB^2$:

$(x - 5)^2 + (y - 1)^2 = (x + 1)^2 + (y - 5)^2$

Expand the squared terms:

$(x^2 - 10x + 25) + (y^2 - 2y + 1) = (x^2 + 2x + 1) + (y^2 - 10y + 25)$

Remove the parentheses and simplify by cancelling $x^2$, $y^2$, $1$, and $25$ from both sides:

$x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + 2x + 1 + y^2 - 10y + 25$

$-10x - 2y = 2x - 10y$

Move all terms involving $x$ and $y$ to one side:

$-10x - 2x = -10y + 2y$

$-12x = -8y$

Divide both sides by $-4$:

$\frac{-12x}{-4} = \frac{-8y}{-4}$

$3x = 2y$

This is the required relation between $x$ and $y$. This equation represents the perpendicular bisector of the line segment AB.

---

Conclusion:

The relation between $x$ and $y$ such that the point P$(x, y)$ is equidistant from A(5, 1) and B(-1, 5) is $3x = 2y$.

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

---

The final answer is $\boxed{3x = 2y}$.

The correct option is (B).

Solution:

---

Concept:

The coordinates of a point that divides the line segment joining A$(x_1, y_1)$ and B$(x_2, y_2)$ externally in the ratio $m:n$ are given by the external section formula.

The formula for the coordinates $(x, y)$ of such a point is:

$x = \frac{m x_2 - n x_1}{m - n}$

$y = \frac{m y_2 - n y_1}{m - n}$

provided that $m \neq n$.

---

Comparison with Options:

Let's compare the standard external section formula with the given options:

(A) $(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n})$ is the formula for internal division.

(B) $(\frac{m x_2 - n x_1}{m-n}, \frac{m y_2 - n y_1}{m-n})$ matches the standard formula for external division.

(C) $(\frac{n x_1 - m x_2}{m-n}, \frac{n y_1 - m y_2}{m-n})$. This can be written as $(\frac{-(m x_2 - n x_1)}{m-n}, \frac{-(m y_2 - n y_1)}{m-n})$. This would be equivalent to option (B) if the denominator was $n-m$ instead of $m-n$, or if the numerator was also negative. As it stands, it is not the standard form.

(D) $(\frac{m x_1 + n x_2}{m+n}, \frac{m y_1 + n y_2}{m+n})$ is not a standard section formula.

---

Conclusion:

The correct formula for the coordinates of a point dividing the line segment joining A$(x_1, y_1)$ and B$(x_2, y_2)$ externally in the ratio $m:n$ is $(\frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n})$.

This matches option (B).

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The final answer is $\boxed{(\frac{m x_2 - n x_1}{m-n}, \frac{m y_2 - n y_1}{m-n})}$.

The correct option is (B).

Solution:

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Given:

The point is $(x, y) = (-5, -12)$.

The origin is the point $(0, 0)$.

To Find:

The distance of the point $(-5, -12)$ from the origin $(0, 0)$.

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Formula:

The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by the distance formula:

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

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Calculation:

Substitute the coordinates of the point $(-5, -12)$ into the distance formula from the origin:

Distance $= \sqrt{(-5)^2 + (-12)^2}$

Distance $= \sqrt{25 + 144}$

Distance $= \sqrt{169}$

Distance $= 13$

The distance of the point $(-5, -12)$ from the origin is 13 units.

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Conclusion:

The distance of the point $(-5, -12)$ from the origin is 13 units.

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

---

The final answer is $\boxed{13 \text{ units}}$.

The correct option is (A).

Solution:

---

Given:

The three points are A(a, 0), B(0, b), and C(1, 1). They are collinear.

Let $(x_1, y_1) = (a, 0)$

Let $(x_2, y_2) = (0, b)$

Let $(x_3, y_3) = (1, 1)$

To Find:

The relation between $a$ and $b$ for which these points are collinear.

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Concept:

Three points are collinear if and only if the area of the triangle formed by them is zero.

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Formula:

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

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Calculation:

Since the points are collinear, the area of the triangle formed by A(a, 0), B(0, b), and C(1, 1) is 0.

Substitute the coordinates into the area formula and set the area to 0:

$0 = \frac{1}{2} |a(b - 1) + 0(1 - 0) + 1(0 - b)|$

Multiply both sides by 2:

$0 = |a(b - 1) + 0 + 1(-b)|$

$0 = |ab - a - b|$

For the absolute value to be zero, the expression inside must be zero:

$ab - a - b = 0$

Rearrange the terms to find the relation. We can add $a$ and $b$ to both sides:

$ab = a + b$

Assuming $a \neq 0$ and $b \neq 0$, we can divide both sides by $ab$. If $a=0$ or $b=0$, the points would not be collinear unless $(1,1)$ was on the x-axis or y-axis (or origin), which is not the case.

$\frac{ab}{ab} = \frac{a + b}{ab}$

$1 = \frac{a}{ab} + \frac{b}{ab}$

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

Or, written differently:

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

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Alternative Method: Using Slope

If the points are collinear, the slope between any two pairs of points is equal.

Slope of AB ($m_{AB}$) $= \frac{y_2 - y_1}{x_2 - x_1} = \frac{b - 0}{0 - a} = \frac{b}{-a}$ (assuming $a \neq 0$)

Slope of AC ($m_{AC}$) $= \frac{y_3 - y_1}{x_3 - x_1} = \frac{1 - 0}{1 - a} = \frac{1}{1 - a}$ (assuming $a \neq 1$)

Equating the slopes ($m_{AB} = m_{AC}$):

$\frac{b}{-a} = \frac{1}{1 - a}$

Cross-multiply:

$b(1 - a) = -a(1)$

$b - ab = -a$

Add $ab$ and $a$ to both sides:

$a + b = ab$

Dividing by $ab$ (assuming $a \neq 0, b \neq 0$):

$\frac{a}{ab} + \frac{b}{ab} = \frac{ab}{ab}$

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

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

---

Conclusion:

If the points (a, 0), (0, b), and (1, 1) are collinear, the relation between $a$ and $b$ is $\frac{1}{a} + \frac{1}{b} = 1$.

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

---

The final answer is $\boxed{\frac{1}{a} + \frac{1}{b} = 1}$.

The correct option is (B).

Solution:

---

Let's examine each statement:

---

(A) The x-coordinate is also called the abscissa.

This statement is True. In a Cartesian coordinate system, the x-coordinate of a point is indeed referred to as its abscissa.

---

(B) The y-coordinate is also called the ordinate.

This statement is also True. The y-coordinate of a point is referred to as its ordinate.

---

(C) The origin is the intersection of the x-axis and y-axis.

This statement is True. The origin (0, 0) is the point where the x-axis and y-axis intersect.

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(D) A point $(x, y)$ lies on the x-axis if $y \neq 0$.

This statement is False.

A point lies on the x-axis if and only if its y-coordinate is 0. The condition for a point $(x, y)$ to lie on the x-axis is $y = 0$, not $y \neq 0$. If $y \neq 0$, the point does not lie on the x-axis (unless $x=0$, in which case it lies on the y-axis). For example, the point $(3, 0)$ lies on the x-axis ($y=0$), while the point $(3, 2)$ does not ($y \neq 0$).

---

Conclusion:

The false statement is (D).

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The final answer is $\boxed{\text{A point (x, y) lies on the x-axis if } y \neq 0}$.

The correct option is (D).

Solution:

---

Given:

The vertices of parallelogram ABCD are A(1, 2), B(4, y), C(x, 6), and D(3, 5).

Let $(x_A, y_A) = (1, 2)$

Let $(x_B, y_B) = (4, y)$

Let $(x_C, y_C) = (x, 6)$

Let $(x_D, y_D) = (3, 5)$

To Find:

The values of $x$ and $y$.

---

Concept:

In a parallelogram, the diagonals bisect each other. This means that the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

---

Formula:

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

Midpoint $= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

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Calculation:

Find the midpoint of diagonal AC using vertices A(1, 2) and C(x, 6):

Midpoint of AC $= \left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) = \left(\frac{1 + x}{2}, \frac{8}{2}\right) = \left(\frac{1 + x}{2}, 4\right)$

Find the midpoint of diagonal BD using vertices B(4, y) and D(3, 5):

Midpoint of BD $= \left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right) = \left(\frac{7}{2}, \frac{y + 5}{2}\right)$

Since the midpoints coincide, we equate the corresponding coordinates:

Equating the x-coordinates:

$\frac{1 + x}{2} = \frac{7}{2}$

Multiply both sides by 2:

$1 + x = 7$

$x = 7 - 1$

$x = 6$

Equating the y-coordinates:

$4 = \frac{y + 5}{2}$

Multiply both sides by 2:

$4 \times 2 = y + 5$

$8 = y + 5$

$y = 8 - 5$

$y = 3$

Thus, the values of $x$ and $y$ are 6 and 3, respectively.

---

Conclusion:

The values of x and y are 6 and 3. This corresponds to option (A).

---

The final answer is $\boxed{x=6, y=3}$.

The correct option is (A).

Solution:

---

Given:

The endpoints of the line segment are P$(x_1, y_1) = (a, b)$ and Q$(x_2, y_2) = (c, d)$.

The line segment PQ is trisected by two points. Let these points be R and S, such that R is closer to P and S is closer to Q.

This means PR = RS = SQ.

To Find:

The coordinates of the points R and S.

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Concept of Trisection:

If R and S trisect the line segment PQ, then:

Point R divides the line segment PQ internally in the ratio PR : RQ = PR : (RS + SQ) = 1 : (1 + 1) = 1 : 2.

Point S divides the line segment PQ internally in the ratio PS : SQ = (PR + RS) : SQ = (1 + 1) : 1 = 2 : 1.

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Formula:

We use the section formula for internal division. If a point $(x, y)$ divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$, then its coordinates are given by:

$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$

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Calculation for the first point R (Ratio 1:2):

Here, $(x_1, y_1) = (a, b)$, $(x_2, y_2) = (c, d)$, $m_1 = 1$, $m_2 = 2$.

x-coordinate of R: $x_R = \frac{(1)(c) + (2)(a)}{1 + 2} = \frac{c + 2a}{3} = \frac{2a + c}{3}$

y-coordinate of R: $y_R = \frac{(1)(d) + (2)(b)}{1 + 2} = \frac{d + 2b}{3} = \frac{2b + d}{3}$

The coordinates of the first point of trisection (closer to P) are $(\frac{2a + c}{3}, \frac{2b + d}{3})$.

---

Calculation for the second point S (Ratio 2:1):

Here, $(x_1, y_1) = (a, b)$, $(x_2, y_2) = (c, d)$, $m_1 = 2$, $m_2 = 1$.

x-coordinate of S: $x_S = \frac{(2)(c) + (1)(a)}{2 + 1} = \frac{2c + a}{3} = \frac{a + 2c}{3}$

y-coordinate of S: $y_S = \frac{(2)(d) + (1)(b)}{2 + 1} = \frac{2d + b}{3} = \frac{b + 2d}{3}$

The coordinates of the second point of trisection (closer to Q) are $(\frac{a + 2c}{3}, \frac{b + 2d}{3})$.

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Conclusion:

The coordinates of the points of trisection of the line segment joining P(a, b) and Q(c, d) are $(\frac{2a + c}{3}, \frac{2b + d}{3})$ and $(\frac{a + 2c}{3}, \frac{b + 2d}{3})$.

Comparing these coordinates with the given options, we find that option (B) lists both of these coordinate pairs.

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The final answer is $\boxed{(\frac{a+2c}{3}, \frac{b+2d}{3}) \text{ and } (\frac{2a+c}{3}, \frac{2b+d}{3})}$.

The correct option is (B).

Solution:

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Given:

Two vertices of an equilateral triangle are A$(0, 0)$ and B$(a, 0)$. These vertices form the base of the triangle on the x-axis.

The side length of the equilateral triangle is the distance between A and B.

Side length = $\sqrt{(a - 0)^2 + (0 - 0)^2} = \sqrt{a^2} = |a|$.

Assuming $a > 0$, the side length is $a$. If $a < 0$, the side length is $-a$. Let's work with the general case of side length $|a|$.

To Find:

The coordinates of the third vertex, let's call it C$(x, y)$.

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Concept:

In an equilateral triangle, all sides are equal in length.

For a triangle with base on the x-axis, the third vertex lies on the perpendicular bisector of the base. The perpendicular bisector of the segment AB will be a vertical line passing through the midpoint of AB.

The midpoint of the base AB is M$(\frac{0+a}{2}, \frac{0+0}{2}) = (\frac{a}{2}, 0)$.

So, the x-coordinate of the third vertex C must be $\frac{a}{2}$. Thus, C$(\frac{a}{2}, y)$.

The distance from C to A must be equal to the side length, $|a|$.

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Calculation:

Using the distance formula between C$(\frac{a}{2}, y)$ and A$(0, 0)$:

$CA = |a|$

$CA^2 = (|a|)^2 = a^2$

$(\frac{a}{2} - 0)^2 + (y - 0)^2 = a^2$

$(\frac{a}{2})^2 + y^2 = a^2$

$\frac{a^2}{4} + y^2 = a^2$

$y^2 = a^2 - \frac{a^2}{4}$

$y^2 = \frac{4a^2 - a^2}{4}$

$y^2 = \frac{3a^2}{4}$

Taking the square root of both sides:

$y = \pm \sqrt{\frac{3a^2}{4}}$

$y = \pm \frac{\sqrt{3}\sqrt{a^2}}{\sqrt{4}}$

$y = \pm \frac{\sqrt{3}|a|}{2}$

Since the options involve $a$ directly rather than $|a|$, it implies that $a$ is likely treated as a positive quantity representing the side length, or that the formulas provided implicitly assume $a>0$. If we assume the side length is $a$ (i.e., $a>0$), then $|a|=a$, and $y = \pm \frac{\sqrt{3}a}{2} = \pm \frac{a\sqrt{3}}{2}$.

The possible coordinates for the third vertex are $(\frac{a}{2}, \frac{a\sqrt{3}}{2})$ and $(\frac{a}{2}, -\frac{a\sqrt{3}}{2})$. One point is above the x-axis and the other is below.

---

Conclusion:

The possible coordinates for the third vertex are $(\frac{a}{2}, \frac{a\sqrt{3}}{2})$ and $(\frac{a}{2}, -\frac{a\sqrt{3}}{2})$.

Comparing this with the given options, we see that option (A) provides one possibility and option (B) provides the other possibility. Option (D) states that both (A) and (B) are possible.

---

The final answer includes both possible locations for the third vertex.

The correct option is (D).

Solution:

---

Given:

Location A coordinates: $(x_1, y_1) = (2, 5)$.

Location B coordinates: $(x_2, y_2) = (10, 11)$.

The valve V divides the line segment AB internally in the ratio $m_1 : m_2 = 3 : 5$.

To Find:

The coordinates of the valve V$(x, y)$.

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Formula:

We use the section formula for internal division. If a point $(x, y)$ divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$, then its coordinates are given by:

$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$

---

Calculation:

Substitute the given values into the section formula to find the coordinates of V$(x_V, y_V)$:

For the x-coordinate:

$x_V = \frac{(3)(10) + (5)(2)}{3 + 5}$

$x_V = \frac{30 + 10}{8}$

$x_V = \frac{40}{8}$

$x_V = 5$

For the y-coordinate:

$y_V = \frac{(3)(11) + (5)(5)}{3 + 5}$

$y_V = \frac{33 + 25}{8}$

$y_V = \frac{58}{8}$

$y_V = \frac{29}{4}$

$y_V = 7.25$

So, the coordinates of the valve V are $(5, 7.25)$.

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Conclusion:

The calculated coordinates of the valve are $(5, 7.25)$. Comparing this result with the given options, we notice that none of the options exactly match these coordinates.

However, option (A) has the correct x-coordinate (5), while its y-coordinate (8) is close to our calculated y-coordinate (7.25). Given that one coordinate matches exactly, option (A) is the most likely intended answer among the choices provided, possibly due to a slight discrepancy in the question's options.

Based on standard calculation, the coordinates are $(5, 7.25)$. Assuming option (A) is the intended answer, we choose it.

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The calculated answer is $(5, 7.25)$.

Comparing with options, the closest match with an exact coordinate is (A) $(5, 8)$.

The correct option is (A).

The Cartesian plane is a fundamental concept in coordinate geometry. It is a two-dimensional plane formed by two perpendicular number lines, called the axes, that intersect at a point called the origin.

---

The x-axis is the horizontal number line in the Cartesian plane. Points to the right of the origin have positive x-coordinates, and points to the left have negative x-coordinates. The origin is represented by the x-coordinate $0$.

---

The y-axis is the vertical number line in the Cartesian plane. Points above the origin have positive y-coordinates, and points below have negative y-coordinates. The origin is represented by the y-coordinate $0$.

---

The x-axis and the y-axis intersect at a single point called the origin. This point corresponds to the ordered pair $(0, 0)$, where the first coordinate is the x-coordinate and the second coordinate is the y-coordinate.

The distance formula is used to calculate the distance between any two points in a two-dimensional Cartesian plane. Given two points, say $P_1$ with coordinates $(x_1, y_1)$ and $P_2$ with coordinates $(x_2, y_2)$, the distance between them, denoted as $d$, is defined by the following formula:

---

The distance $d$ between $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

Given points:

Point A: $(x_1, y_1) = (2, 3)$

Point B: $(x_2, y_2) = (4, 1)$

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Formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

---

Calculation:

Substitute the coordinates of points A and B into the distance formula:

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

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

$d = \sqrt{4 + 4}$

$d = \sqrt{8}$

Simplify the square root:

$d = \sqrt{4 \times 2}$

$d = \sqrt{4} \times \sqrt{2}$

$d = 2\sqrt{2}$

---

Answer:

The distance between points A$(2, 3)$ and B$(4, 1)$ is $2\sqrt{2}$ units.

Given points:

Point P: $(x_1, y_1) = (-5, 12)$

Origin: $(x_2, y_2) = (0, 0)$

---

Formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

---

Calculation:

Substitute the coordinates of point P and the origin into the distance formula:

$d = \sqrt{(0 - (-5))^2 + (0 - 12)^2}$

$d = \sqrt{(0 + 5)^2 + (-12)^2}$

$d = \sqrt{(5)^2 + (-12)^2}$

$d = \sqrt{25 + 144}$

$d = \sqrt{169}$

$d = 13$

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Answer:

The distance of the point P$(-5, 12)$ from the origin $(0, 0)$ is $13$ units.

The section formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. For internal division, let the two given points be A$(x_1, y_1)$ and B$(x_2, y_2)$. If a point P$(x, y)$ divides the line segment AB internally in the ratio $m_1 : m_2$, then the coordinates of point P are given by the formula:

---

The coordinates $(x, y)$ of the point P are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Given points:

Point 1: $(x_1, y_1) = (4, -3)$

Point 2: $(x_2, y_2) = (8, 5)$

Ratio of internal division: $m_1 : m_2 = 3 : 1$, so $m_1 = 3$ and $m_2 = 1$.

---

Formula:

The coordinates $(x, y)$ of a point that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are given by the section formula:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

---

Calculation:

Substitute the given values into the section formula:

For the x-coordinate:

$x = \frac{(3)(8) + (1)(4)}{3 + 1}$

$x = \frac{24 + 4}{4}$

$x = \frac{28}{4}$

$x = 7$

For the y-coordinate:

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

$y = \frac{15 - 3}{4}$

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

$y = 3$

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Answer:

The coordinates of the point that divides the line segment joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3:1$ internally are $(7, 3)$.

Given points:

Point 1: $(x_1, y_1) = (-1, 7)$

Point 2: $(x_2, y_2) = (4, -3)$

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Formula:

The coordinates of the midpoint $(x, y)$ of a line segment joining the points $(x_1, y_1)$ and $(x_2, y_2)$ are given by the midpoint formula (which is a special case of the section formula for a $1:1$ ratio):

$x = \frac{x_1 + x_2}{2}$

$y = \frac{y_1 + y_2}{2}$

---

Calculation:

Substitute the given coordinates into the midpoint formula:

For the x-coordinate:

$x = \frac{-1 + 4}{2}$

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

For the y-coordinate:

$y = \frac{7 + (-3)}{2}$

$y = \frac{7 - 3}{2}$

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

$y = 2$

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Answer:

The coordinates of the midpoint of the line segment joining the points $(-1, 7)$ and $(4, -3)$ are $(\frac{3}{2}, 2)$.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by a formula derived from the concept of determinants or by extending the idea of the area of a trapezoid. The formula is:

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

---

Alternatively, it can also be written as:

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

---

The absolute value symbol $|\dots|$ is used because the area of a triangle must always be a non-negative value. If the points are collinear (lie on the same straight line), the area calculated by this formula will be $0$.

Given vertices:

$(x_1, y_1) = (1, -1)$

$(x_2, y_2) = (-4, 6)$

$(x_3, y_3) = (-3, -5)$

---

Formula:

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

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

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Calculation:

Substitute the given coordinates into the formula:

Area $= \frac{1}{2} |1(6 - (-5)) + (-4)((-5) - (-1)) + (-3)((-1) - 6)|$

Area $= \frac{1}{2} |1(6 + 5) + (-4)(-5 + 1) + (-3)(-1 - 6)|$

Area $= \frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|$

Area $= \frac{1}{2} |11 + 16 + 21|$

Area $= \frac{1}{2} |48|$

Area $= \frac{1}{2} \times 48$

Area $= 24$

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Answer:

The area of the triangle with vertices $(1, -1)$, $(-4, 6)$, and $(-3, -5)$ is $24$ square units.

To check if the points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are collinear, we can calculate the area of the triangle formed by these points. If the area of the triangle is $0$, then the points are collinear; otherwise, they are not.

---

Given points:

$(x_1, y_1) = (1, 5)$

$(x_2, y_2) = (2, 3)$

$(x_3, y_3) = (-2, -11)$

---

Formula:

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

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

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Calculation:

Substitute the given coordinates into the formula:

Area $= \frac{1}{2} |1(3 - (-11)) + 2((-11) - 5) + (-2)(5 - 3)|$

Area $= \frac{1}{2} |1(3 + 11) + 2(-11 - 5) + (-2)(2)|$

Area $= \frac{1}{2} |1(14) + 2(-16) + (-2)(2)|$

Area $= \frac{1}{2} |14 - 32 - 4|$

Area $= \frac{1}{2} |14 - 36|$

Area $= \frac{1}{2} |-22|$

Area $= \frac{1}{2} \times 22$

Area $= 11$

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Conclusion:

Since the area of the triangle formed by the given points is $11$ square units, which is not equal to $0$, the points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are not collinear.

For the points A$(2, 3)$, B$(4, k)$, and C$(6, -3)$ to be collinear, the area of the triangle formed by these points must be $0$.

---

Given points:

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

$(x_2, y_2) = (4, k)$

$(x_3, y_3) = (6, -3)$

---

Condition for collinearity:

Area of $\triangle ABC = 0$

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Formula:

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

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

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Calculation:

Substitute the given coordinates into the formula and set the area to $0$:

$\frac{1}{2} |2(k - (-3)) + 4(-3 - 3) + 6(3 - k)| = 0$

Multiply by $2$ and remove the absolute value since the expression inside must be $0$:

$2(k + 3) + 4(-6) + 6(3 - k) = 0$

Expand the terms:

$2k + 6 - 24 + 18 - 6k = 0$

Combine like terms:

$(2k - 6k) + (6 - 24 + 18) = 0$

$-4k + 0 = 0$

$-4k = 0$

Divide by $-4$:

$k = 0$

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Answer:

The value of $k$ for which the points A$(2, 3)$, B$(4, k)$, and C$(6, -3)$ are collinear is $k=0$.

Given points:

Point A: $(2, -5)$

Point B: $(-2, 9)$

---

To Find:

The coordinates of a point on the x-axis that is equidistant from points A and B.

---

Let:

Let the point on the x-axis be P$(x, 0)$, since any point on the x-axis has its y-coordinate as $0$.

---

Condition:

The point P is equidistant from A and B. This means the distance PA is equal to the distance PB.

$PA = PB$

---

Formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

---

Calculation:

Using the distance formula, we can write the equation $PA = PB$ as:

$\sqrt{(x - 2)^2 + (0 - (-5))^2} = \sqrt{(x - (-2))^2 + (0 - 9)^2}$

$\sqrt{(x - 2)^2 + (0 + 5)^2} = \sqrt{(x + 2)^2 + (-9)^2}$

$\sqrt{(x - 2)^2 + 5^2} = \sqrt{(x + 2)^2 + (-9)^2}$

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

$(x - 2)^2 + 5^2 = (x + 2)^2 + (-9)^2$

Expand the squared terms using the formula $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$:

$(x^2 - 4x + 4) + 25 = (x^2 + 4x + 4) + 81$

Simplify both sides:

$x^2 - 4x + 29 = x^2 + 4x + 85$

Subtract $x^2$ from both sides:

$-4x + 29 = 4x + 85$

Move all terms involving $x$ to one side and constants to the other side:

$29 - 85 = 4x + 4x$

$-56 = 8x$

Divide by $8$ to solve for $x$:

$x = \frac{-56}{8}$

$x = -7$

---

Coordinates of the point:

The point on the x-axis is P$(x, 0)$. Substituting the value of $x$, we get the coordinates as $(-7, 0)$.

---

Answer:

The point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9)$ is $(-7, 0)$.

Given points:

Point A: $(6, 5)$

Point B: $(-4, 3)$

---

To Find:

The coordinates of a point on the y-axis that is equidistant from points A and B.

---

Let:

Let the point on the y-axis be P$(0, y)$, since any point on the y-axis has its x-coordinate as $0$.

---

Condition:

The point P is equidistant from A and B. This means the distance PA is equal to the distance PB.

$PA = PB$

---

Formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

---

Calculation:

Using the distance formula, we can write the equation $PA = PB$ as:

$\sqrt{(0 - 6)^2 + (y - 5)^2} = \sqrt{(0 - (-4))^2 + (y - 3)^2}$

$\sqrt{(-6)^2 + (y - 5)^2} = \sqrt{(0 + 4)^2 + (y - 3)^2}$

$\sqrt{(-6)^2 + (y - 5)^2} = \sqrt{4^2 + (y - 3)^2}$

$\sqrt{36 + (y - 5)^2} = \sqrt{16 + (y - 3)^2}$

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

$36 + (y - 5)^2 = 16 + (y - 3)^2$

Expand the squared terms using the formula $(a-b)^2 = a^2 - 2ab + b^2$:

$36 + (y^2 - 10y + 25) = 16 + (y^2 - 6y + 9)$

Simplify both sides:

$y^2 - 10y + 61 = y^2 - 6y + 25$

Subtract $y^2$ from both sides:

$-10y + 61 = -6y + 25$

Move all terms involving $y$ to one side and constants to the other side:

$61 - 25 = -6y + 10y$

$36 = 4y$

Divide by $4$ to solve for $y$:

$y = \frac{36}{4}$

$y = 9$

---

Coordinates of the point:

The point on the y-axis is P$(0, y)$. Substituting the value of $y$, we get the coordinates as $(0, 9)$.

---

Answer:

The point on the y-axis which is equidistant from $(6, 5)$ and $(-4, 3)$ is $(0, 9)$.

Given points:

Point A: $(x_1, y_1) = (5, -6)$

Point B: $(x_2, y_2) = (-1, -4)$

---

To Find:

The ratio in which the y-axis divides the line segment AB and the coordinates of the point of intersection.

---

Let:

Let the y-axis divide the line segment joining A$(5, -6)$ and B$(-1, -4)$ at the point P$(x, y)$ in the ratio $k : 1$ internally.

Since the point P lies on the y-axis, its x-coordinate is $0$. So, P$(0, y)$.

---

Formula:

Using the section formula, the coordinates of the point P$(x, y)$ that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1 : m_2$ internally are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

---

Calculation of Ratio:

Here, $m_1 = k$, $m_2 = 1$, $(x_1, y_1) = (5, -6)$, and $(x_2, y_2) = (-1, -4)$. The point of intersection is P$(0, y)$, so $x = 0$.

Using the x-coordinate part of the section formula:

$0 = \frac{k(-1) + 1(5)}{k + 1}$

$0 = \frac{-k + 5}{k + 1}$

For this equation to be true, the numerator must be $0$ (assuming $k+1 \neq 0$, which is true for a finite ratio):

$-k + 5 = 0$

$-k = -5$

$k = 5$

The ratio is $k : 1 = 5 : 1$. Since $k$ is positive, the y-axis divides the line segment internally in the ratio $5:1$.

---

Calculation of Point of Intersection:

Now, we use the y-coordinate part of the section formula with the ratio $m_1 : m_2 = 5 : 1$ ($m_1 = 5$, $m_2 = 1$):

$y = \frac{5(-4) + 1(-6)}{5 + 1}$

$y = \frac{-20 - 6}{6}$

$y = \frac{-26}{6}$

Simplify the fraction:

$y = -\frac{13}{3}$

The coordinates of the point of intersection P are $(0, y)$.

P$(0, -\frac{13}{3})$

---

Answer:

The y-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$ in the ratio $5:1$ internally.

The point of intersection is $(0, -\frac{13}{3})$.

Given points:

Let the line segment be AB, where A$(x_1, y_1) = (4, -1)$ and B$(x_2, y_2) = (-2, -3)$.

---

Concept of Trisection:

Trisection of a line segment means dividing it into three equal parts. Let P and Q be the points of trisection of the line segment AB such that AP = PQ = QB.

This means that point P divides the line segment AB internally in the ratio $1:2$, and point Q divides the line segment AB internally in the ratio $2:1$.

---

Formula:

The coordinates $(x, y)$ of a point that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are given by the section formula:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

---

Calculation for the first point of trisection P (dividing in ratio 1:2):

Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, $m_1 = 1$, and $m_2 = 2$.

The x-coordinate of P is:

$x_P = \frac{(1)(-2) + (2)(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2$

The y-coordinate of P is:

$y_P = \frac{(1)(-3) + (2)(-1)}{1 + 2} = \frac{-3 - 2}{3} = \frac{-5}{3}$

So, the coordinates of the first point of trisection P are $(2, -\frac{5}{3})$.

---

Calculation for the second point of trisection Q (dividing in ratio 2:1):

Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, $m_1 = 2$, and $m_2 = 1$.

The x-coordinate of Q is:

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

The y-coordinate of Q is:

$y_Q = \frac{(2)(-3) + (1)(-1)}{2 + 1} = \frac{-6 - 1}{3} = \frac{-7}{3}$

So, the coordinates of the second point of trisection Q are $(0, -\frac{7}{3})$.

---

Answer:

The coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$ are $(2, -\frac{5}{3})$ and $(0, -\frac{7}{3})$.

Given points:

Let the vertices of the quadrilateral be A$(1, 7)$, B$(4, 2)$, C$(-1, -1)$, and D$(-4, 4)$.

---

To Determine:

The type of quadrilateral ABCD by finding the lengths of its sides and diagonals.

---

Formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

---

Calculation of Side Lengths:

Length of AB: $(x_1, y_1) = (1, 7)$, $(x_2, y_2) = (4, 2)$

$AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$

Length of BC: $(x_1, y_1) = (4, 2)$, $(x_2, y_2) = (-1, -1)$

$BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$

Length of CD: $(x_1, y_1) = (-1, -1)$, $(x_2, y_2) = (-4, 4)$

$CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-4 + 1)^2 + (4 + 1)^2} = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34}$

Length of DA: $(x_1, y_1) = (-4, 4)$, $(x_2, y_2) = (1, 7)$

$DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{(1 + 4)^2 + (3)^2} = \sqrt{(5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34}$

---

Observation: AB = BC = CD = DA = $\sqrt{34}$. This means all four sides are equal. The quadrilateral is either a rhombus or a square.

---

Calculation of Diagonal Lengths:

Length of AC (Diagonal 1): $(x_1, y_1) = (1, 7)$, $(x_2, y_2) = (-1, -1)$

$AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$

Length of BD (Diagonal 2): $(x_1, y_1) = (4, 2)$, $(x_2, y_2) = (-4, 4)$

$BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68}$

---

Observation: AC = BD = $\sqrt{68}$. This means the diagonals are equal in length.

---

Conclusion:

Since all four sides are equal (AB = BC = CD = DA) and the diagonals are equal (AC = BD), the quadrilateral formed by the given points is a Square.

Given vertices:

Let the vertices of the rhombus be A$(3, 0)$, B$(4, 5)$, C$(-1, 4)$, and D$(-2, -1)$ taken in order.

---

To Find:

The area of the rhombus ABCD.

---

Formula:

The area of a rhombus is given by half the product of its diagonals ($d_1$ and $d_2$):

Area $= \frac{1}{2} \times d_1 \times d_2$

We need to find the lengths of the diagonals AC and BD using the distance formula:

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

---

Calculation of Diagonals:

Length of Diagonal AC ($d_1$):

Using points A$(3, 0)$ and C$(-1, 4)$:

$d_1 = \sqrt{(-1 - 3)^2 + (4 - 0)^2}$

$d_1 = \sqrt{(-4)^2 + (4)^2}$

$d_1 = \sqrt{16 + 16}$

$d_1 = \sqrt{32}$

$d_1 = \sqrt{16 \times 2}$

$d_1 = 4\sqrt{2}$ units.

---

Length of Diagonal BD ($d_2$):

Using points B$(4, 5)$ and D$(-2, -1)$:

$d_2 = \sqrt{(-2 - 4)^2 + (-1 - 5)^2}$

$d_2 = \sqrt{(-6)^2 + (-6)^2}$

$d_2 = \sqrt{36 + 36}$

$d_2 = \sqrt{72}$

$d_2 = \sqrt{36 \times 2}$

$d_2 = 6\sqrt{2}$ units.

---

Calculation of Area:

Now, using the area formula for a rhombus:

Area $= \frac{1}{2} \times d_1 \times d_2$

Area $= \frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2})$

Area $= \frac{1}{2} \times 4 \times 6 \times \sqrt{2} \times \sqrt{2}$

Area $= \frac{1}{2} \times 24 \times 2$

Area $= 12 \times 2$

Area $= 24$ square units.

---

Answer:

The area of the rhombus with the given vertices is $24$ square units.

Given:

Center of the circle C: $(h, k) = (2, -3)$

Point on the circle P: $(x, y) = (5, 1)$

---

To Find:

The radius of the circle, denoted by $r$.

---

Concept:

The radius of a circle is the distance from its center to any point on its circumference. In this case, the radius is the distance between the center C$(2, -3)$ and the point P$(5, 1)$ on the circle.

---

Formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

The radius $r$ is the distance between C$(2, -3)$ and P$(5, 1)$, so $r = CP$.

---

Calculation:

Substitute the coordinates of the center C and the point P into the distance formula:

$r = \sqrt{(5 - 2)^2 + (1 - (-3))^2}$

$r = \sqrt{(3)^2 + (1 + 3)^2}$

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

$r = \sqrt{9 + 16}$

$r = \sqrt{25}$

$r = 5$

---

Answer:

The radius of the circle with center $(2, -3)$ that passes through the point $(5, 1)$ is $5$ units.

Given:

Point P$(x, y)$ is equidistant from point A$(5, 1)$ and point B$(-1, 5)$.

---

To Prove:

$3x = 2y$

---

Solution:

The point P$(x, y)$ being equidistant from points A$(5, 1)$ and B$(-1, 5)$ means that the distance PA is equal to the distance PB.

We use the distance formula to find the lengths of PA and PB. The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

Using this formula, the distance PA between P$(x, y)$ and A$(5, 1)$ is:

$PA = \sqrt{(x - 5)^2 + (y - 1)^2}$

The distance PB between P$(x, y)$ and B$(-1, 5)$ is:

$PB = \sqrt{(x - (-1))^2 + (y - 5)^2} = \sqrt{(x + 1)^2 + (y - 5)^2}$

From equation (i), we have $PA = PB$. Squaring both sides eliminates the square roots:

$PA^2 = PB^2$

$(x - 5)^2 + (y - 1)^2 = (x + 1)^2 + (y - 5)^2$

Expand the squared terms using the algebraic identities $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$:

$(x^2 - 2 \times x \times 5 + 5^2) + (y^2 - 2 \times y \times 1 + 1^2) = (x^2 + 2 \times x \times 1 + 1^2) + (y^2 - 2 \times y \times 5 + 5^2)$

$(x^2 - 10x + 25) + (y^2 - 2y + 1) = (x^2 + 2x + 1) + (y^2 - 10y + 25)$

Remove the parentheses and combine terms on each side:

$x^2 - 10x + y^2 - 2y + 26 = x^2 + 2x + y^2 - 10y + 26$

Subtract $x^2$, $y^2$, and $26$ from both sides of the equation:

$x^2 - 10x + y^2 - 2y + 26 - x^2 - y^2 - 26 = x^2 + 2x + y^2 - 10y + 26 - x^2 - y^2 - 26$

This simplifies to:

$-10x - 2y = 2x - 10y$

Now, rearrange the terms to group $x$ terms on one side and $y$ terms on the other side. Add $10y$ to both sides and add $10x$ to both sides, or equivalently, move $2x$ to the left and $-2y$ to the right:

$-10x - 2x = -10y + 2y$

Combine the terms:

$-12x = -8y$

To get the desired form, divide both sides by the greatest common divisor of $12$ and $8$, which is $4$. Let's divide by $-4$ to make both sides positive:

$\frac{-12x}{-4} = \frac{-8y}{-4}$

$3x = 2y$

---

Thus, it is proved that if the point P$(x, y)$ is equidistant from the points A$(5, 1)$ and B$(-1, 5)$, then $3x = 2y$.

Given points:

Point A: $(x_1, y_1) = (1, -5)$

Point B: $(x_2, y_2) = (-4, 5)$

The line dividing the segment is the x-axis.

---

To Find:

The ratio in which the x-axis divides the line segment AB and the coordinates of the point of division.

---

Let:

Let the x-axis divide the line segment joining A$(1, -5)$ and B$(-4, 5)$ at the point P$(x, y)$ in the ratio $k : 1$ internally.

Since the point P lies on the x-axis, its y-coordinate is $0$. So, P$(x, 0)$.

---

Formula:

Using the section formula, the coordinates of the point P$(x, y)$ that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1 : m_2$ internally are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

---

Calculation of Ratio:

Here, $m_1 = k$, $m_2 = 1$, $(x_1, y_1) = (1, -5)$, and $(x_2, y_2) = (-4, 5)$. The point of intersection is P$(x, 0)$, so $y = 0$.

Using the y-coordinate part of the section formula:

$0 = \frac{k(5) + 1(-5)}{k + 1}$

$0 = \frac{5k - 5}{k + 1}$

For this equation to be true, the numerator must be $0$ (assuming $k+1 \neq 0$, which is true for a finite ratio):

$5k - 5 = 0$

$5k = 5$

$k = 1$

The ratio is $k : 1 = 1 : 1$. Since $k$ is positive, the x-axis divides the line segment internally in the ratio $1:1$. This means the point of division is the midpoint of the line segment.

---

Calculation of Point of Division:

Now, we use the x-coordinate part of the section formula with the ratio $m_1 : m_2 = 1 : 1$ ($m_1 = 1$, $m_2 = 1$):

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

$x = \frac{-4 + 1}{2}$

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

The coordinates of the point of division P are $(x, 0)$.

P$(-\frac{3}{2}, 0)$

---

Answer:

The x-axis divides the line segment joining the points $(1, -5)$ and $(-4, 5)$ in the ratio $1:1$ internally.

The coordinates of the point of division are $(-\frac{3}{2}, 0)$.

Given vertices:

Let the vertices of the parallelogram be A$(1, 2)$, B$(4, y)$, C$(x, 6)$, and D$(3, 5)$ taken in order.

---

Property of Parallelogram:

In a parallelogram, the diagonals bisect each other. This means that the midpoint of one diagonal is the same as the midpoint of the other diagonal.

Let the diagonals be AC and BD. Their midpoints must coincide.

---

Formula:

The coordinates of the midpoint of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ are given by the midpoint formula:

Midpoint $(x_m, y_m) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

---

Calculation of Midpoint of AC:

Using points A$(1, 2)$ and C$(x, 6)$: $(x_1, y_1) = (1, 2)$, $(x_2, y_2) = (x, 6)$.

Midpoint of AC $= (\frac{1 + x}{2}, \frac{2 + 6}{2})$

Midpoint of AC $= (\frac{1 + x}{2}, \frac{8}{2})$

Midpoint of AC $= (\frac{1 + x}{2}, 4)$

---

Calculation of Midpoint of BD:

Using points B$(4, y)$ and D$(3, 5)$: $(x_1, y_1) = (4, y)$, $(x_2, y_2) = (3, 5)$.

Midpoint of BD $= (\frac{4 + 3}{2}, \frac{y + 5}{2})$

Midpoint of BD $= (\frac{7}{2}, \frac{y + 5}{2})$

---

Equating the Midpoints:

Since the midpoints of AC and BD are the same, we equate their respective coordinates:

---

Solving for $x$:

$\frac{1 + x}{2} = \frac{7}{2}$

Multiply both sides by 2:

$1 + x = 7$

Subtract 1 from both sides:

$x = 7 - 1$

$x = 6$

---

Solving for $y$:

$\frac{8}{2} = \frac{y + 5}{2}$

Simplify the left side:

$4 = \frac{y + 5}{2}$

Multiply both sides by 2:

$4 \times 2 = y + 5$

$8 = y + 5$

Subtract 5 from both sides:

$8 - 5 = y$

$y = 3$

---

Answer:

The values of $x$ and $y$ are $6$ and $3$ respectively.

Thus, the vertices of the parallelogram are A$(1, 2)$, B$(4, 3)$, C$(6, 6)$, and D$(3, 5)$.

Given:

Center of the circle (let's call it O): $(2, -3)$

One endpoint of the diameter AB (Point B): $(1, 4)$

AB is the diameter of the circle.

---

To Find:

The coordinates of the other endpoint of the diameter (Point A).

---

Concept:

The center of a circle is always the midpoint of its diameter. Therefore, the center O$(2, -3)$ is the midpoint of the line segment AB.

---

Formula:

The coordinates of the midpoint $(x_m, y_m)$ of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ are given by the midpoint formula:

Midpoint $(x_m, y_m) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

---

Calculation:

Let the coordinates of point A be $(x, y)$. We are given that the midpoint of AB is O$(2, -3)$ and the coordinates of B are $(1, 4)$.

Using the midpoint formula for the x-coordinate:

$2 = \frac{x + 1}{2}$

Multiply both sides by 2:

$2 \times 2 = x + 1$

$4 = x + 1$

Subtract 1 from both sides:

$x = 4 - 1$

$x = 3$

Using the midpoint formula for the y-coordinate:

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

Multiply both sides by 2:

$-3 \times 2 = y + 4$

$-6 = y + 4$

Subtract 4 from both sides:

$y = -6 - 4$

$y = -10$

---

Coordinates of Point A:

The coordinates of point A are $(x, y) = (3, -10)$.

---

Answer:

The coordinates of point A are $(3, -10)$.

Given vertices:

Let the vertices of the triangle be A$(0, 0)$, B$(2, 0)$, and C$(0, 3)$.

$(x_1, y_1) = (0, 0)$

$(x_2, y_2) = (2, 0)$

$(x_3, y_3) = (0, 3)$

---

To Find:

The area of the triangle formed by these points.

---

Formula:

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

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

---

Calculation:

Substitute the given coordinates into the formula:

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

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

Area $= \frac{1}{2} |0 + 6 + 0|$

Area $= \frac{1}{2} |6|$

Area $= \frac{1}{2} \times 6$

Area $= 3$

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Answer:

The area of the triangle formed by the points $(0, 0)$, $(2, 0),$ and $(0, 3)$ is $3$ square units.

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Alternate Method (Geometric Approach):

The vertices are O$(0, 0)$, A$(2, 0)$ on the x-axis, and C$(0, 3)$ on the y-axis. This forms a right-angled triangle with the right angle at the origin O$(0, 0)$.

The base of the triangle can be taken as the distance OA along the x-axis.

Base = OA = Distance between $(0, 0)$ and $(2, 0) = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{2^2} = 2$ units.

The height of the triangle can be taken as the distance OC along the y-axis.

Height = OC = Distance between $(0, 0)$ and $(0, 3) = \sqrt{(0-0)^2 + (3-0)^2} = \sqrt{3^2} = 3$ units.

The area of a right-angled triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

Area $= \frac{1}{2} \times OA \times OC$

Area $= \frac{1}{2} \times 2 \times 3$

Area $= \frac{1}{2} \times 6$

Area $= 3$ square units.

The distance formula is a formula used to find the distance between two points in a coordinate plane. It is derived from the Pythagorean theorem and allows us to calculate the length of a line segment connecting two points with given coordinates.

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Derivation using the Pythagorean Theorem:

Let P be a point with coordinates $(x_1, y_1)$ and Q be a point with coordinates $(x_2, y_2)$ in the Cartesian plane. We want to find the distance between P and Q.

Draw a line segment connecting P and Q.

Draw a horizontal line through P and a vertical line through Q. These two lines intersect at a point R. The coordinates of point R will be $(x_2, y_1)$.

Now, consider the triangle PRQ. This triangle is a right-angled triangle with the right angle at R.

The length of the horizontal side PR is the absolute difference between the x-coordinates of P and R (or P and Q, as they have the same x-coordinate difference):

$PR = |x_2 - x_1|$

The length of the vertical side RQ is the absolute difference between the y-coordinates of R and Q (or P and Q, as they have the same y-coordinate difference):

$RQ = |y_2 - y_1|$

The distance between P and Q, which is the length of the segment PQ, is the hypotenuse of the right-angled triangle PRQ.

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In triangle PRQ:

$PQ^2 = PR^2 + RQ^2$

Substitute the lengths of PR and RQ:

$PQ^2 = |x_2 - x_1|^2 + |y_2 - y_1|^2$

Since the square of an absolute value is the same as the square of the number itself (e.g., $|-a|^2 = (-a)^2 = a^2$), we can remove the absolute value signs:

$PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

To find the distance PQ, take the square root of both sides:

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

This is the distance formula.

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Distance between P$(x_1, y_1)$ and Q$(x_2, y_2)$:

The distance between the points P$(x_1, y_1)$ and Q$(x_2, y_2)$ is given by the formula:

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

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Showing the points form a square:

Given points:

Let the vertices of the quadrilateral be A$(1, 7)$, B$(4, 2)$, C$(-1, -1)$, and D$(-4, 4)$ taken in order.

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To Show:

The points A, B, C, and D are the vertices of a square.

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Solution:

To show that the quadrilateral ABCD is a square, we need to prove that all four sides are equal in length and that the two diagonals are also equal in length. We use the distance formula to calculate the lengths of the sides and diagonals.

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Calculating Side Lengths:

Length of AB: Using A$(1, 7)$ and B$(4, 2)$

$AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$

Length of BC: Using B$(4, 2)$ and C$(-1, -1)$

$BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$

Length of CD: Using C$(-1, -1)$ and D$(-4, 4)$

$CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-4 + 1)^2 + (4 + 1)^2} = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34}$

Length of DA: Using D$(-4, 4)$ and A$(1, 7)$

$DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{(1 + 4)^2 + (3)^2} = \sqrt{(5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34}$

Since $AB = BC = CD = DA = \sqrt{34}$, all four sides of the quadrilateral are equal. This indicates that the quadrilateral is either a rhombus or a square.

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Calculating Diagonal Lengths:

Length of AC (Diagonal 1): Using A$(1, 7)$ and C$(-1, -1)$

$AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$

Length of BD (Diagonal 2): Using B$(4, 2)$ and D$(-4, 4)$

$BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68}$

Since $AC = BD = \sqrt{68}$, the two diagonals of the quadrilateral are equal in length.

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Conclusion:

A quadrilateral with all four sides equal and diagonals equal is a square.

Since we have shown that $AB = BC = CD = DA$ and $AC = BD$, the points $(1, 7), (4, 2), (-1, -1),$ and $(-4, 4)$ are indeed the vertices of a square.

The section formula for internal division is used to find the coordinates of a point that lies on the line segment joining two given points and divides it into two segments with a specific ratio of lengths. 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 $m_1 : m_2$, its coordinates are given by the formula:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

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Derivation of the Section Formula (Internal Division):

Let the two given points be A$(x_1, y_1)$ and B$(x_2, y_2)$. Let P$(x, y)$ be the point that divides the line segment AB internally in the ratio $m_1 : m_2$. This means $\frac{AP}{PB} = \frac{m_1}{m_2}$.

Draw perpendiculars from A, P, and B to the x-axis, meeting the x-axis at A', P', and B' respectively. Their x-coordinates are $x_1, x,$ and $x_2$.

Draw perpendiculars from A, P, and B to the y-axis, meeting the y-axis at A'', P'', and B'' respectively. Their y-coordinates are $y_1, y,$ and $y_2$.

From the property of similar triangles (consider the triangles formed by A, P, and B and the perpendiculars to the axes), we have:

$\frac{AP}{PB} = \frac{A'P'}{P'B'} = \frac{A''P''}{P''B''}$

Consider the equation involving x-coordinates:

$m_1 (x_2 - x) = m_2 (x - x_1)$

$m_1 x_2 - m_1 x = m_2 x - m_2 x_1$

Rearrange terms to solve for x:

$m_1 x_2 + m_2 x_1 = m_2 x + m_1 x$

$m_1 x_2 + m_2 x_1 = x (m_1 + m_2)$

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

Similarly, consider the equation involving y-coordinates:

$m_1 (y_2 - y) = m_2 (y - y_1)$

$m_1 y_2 - m_1 y = m_2 y - m_2 y_1$

Rearrange terms to solve for y:

$m_1 y_2 + m_2 y_1 = m_2 y + m_1 y$

$m_1 y_2 + m_2 y_1 = y (m_1 + m_2)$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

These are the coordinates of the point P$(x, y)$ that divides the line segment AB internally in the ratio $m_1 : m_2$.

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Finding the coordinates of the point dividing in ratio 3:2:

Given points:

Point A: $(x_1, y_1) = (-1, 3)$

Point B: $(x_2, y_2) = (4, -7)$

Ratio of internal division: $m_1 : m_2 = 3 : 2$, so $m_1 = 3$ and $m_2 = 2$.

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Calculation:

Using the section formula:

For the x-coordinate:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{(3)(4) + (2)(-1)}{3 + 2} = \frac{12 - 2}{5} = \frac{10}{5} = 2$

For the y-coordinate:

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{(3)(-7) + (2)(3)}{3 + 2} = \frac{-21 + 6}{5} = \frac{-15}{5} = -3$

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Answer:

The coordinates of the point which divides the line segment joining A$(-1, 3)$ and B$(4, -7)$ internally in the ratio $3:2$ are $(2, -3)$.

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Finding the coordinates of the midpoint:

Given points:

Point A: $(x_1, y_1) = (-1, 3)$

Point B: $(x_2, y_2) = (4, -7)$

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Concept:

The midpoint is the point that divides the line segment in the ratio $1:1$. So, we can use the section formula with $m_1 = 1$ and $m_2 = 1$, or simply use the midpoint formula, which is a simplified version.

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Formula (Midpoint):

The coordinates of the midpoint $(x_m, y_m)$ of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ are:

$x_m = \frac{x_1 + x_2}{2}$

$y_m = \frac{y_1 + y_2}{2}$

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Calculation:

For the x-coordinate of the midpoint:

$x_m = \frac{-1 + 4}{2} = \frac{3}{2}$

For the y-coordinate of the midpoint:

$y_m = \frac{3 + (-7)}{2} = \frac{3 - 7}{2} = \frac{-4}{2} = -2$

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Answer:

The coordinates of the midpoint of the line segment joining A$(-1, 3)$ and B$(4, -7)$ are $(\frac{3}{2}, -2)$.

The area of a triangle in coordinate geometry can be calculated if the coordinates of its three vertices are known. This formula is derived using the coordinates of the vertices and is related to the concept of determinants or by dividing the triangle into trapezoids and calculating their areas.

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

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

The absolute value sign is used because the area of a geometrical figure is always non-negative. If the calculated value inside the absolute value is negative, we take its positive counterpart.

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Given vertices:

Let the vertices of the triangle be A$(-5, -1)$, B$(3, -5)$, and C$(5, 2)$.

So, $(x_1, y_1) = (-5, -1)$

$(x_2, y_2) = (3, -5)$

$(x_3, y_3) = (5, 2)$

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To Find:

The area of $\triangle ABC$.

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Formula:

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

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Calculation:

Substitute the given coordinates into the area formula:

Area $= \frac{1}{2} |(-5)((-5) - 2) + 3(2 - (-1)) + 5((-1) - (-5))|$

Area $= \frac{1}{2} |(-5)(-5 - 2) + 3(2 + 1) + 5(-1 + 5)|$

Area $= \frac{1}{2} |(-5)(-7) + 3(3) + 5(4)|$

Area $= \frac{1}{2} |35 + 9 + 20|$

Area $= \frac{1}{2} |64|$

Area $= \frac{1}{2} \times 64$

Area $= 32$

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Answer:

The area of the triangle formed by the points $(-5, -1)$, $(3, -5)$, and $(5, 2)$ is $32$ square units.

Given:

Points of the line segment: A$(x_1, y_1) = (2, -2)$ and B$(x_2, y_2) = (3, 7)$.

Equation of the line: $2x + y - 4 = 0$.

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To Find:

The ratio in which the line divides the line segment AB and the coordinates of the point of division.

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Let:

Let the line $2x + y - 4 = 0$ divide the line segment AB at point P$(x, y)$ in the ratio $k : 1$ internally.

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Formula:

Using the section formula, the coordinates of the point P$(x, y)$ that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1 : m_2$ internally are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Here, $m_1 = k$ and $m_2 = 1$.

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Calculation of Ratio:

The coordinates of P in terms of $k$ are:

$x = \frac{k(3) + 1(2)}{k + 1} = \frac{3k + 2}{k + 1}$

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

Since the point P$(x, y)$ lies on the line $2x + y - 4 = 0$, its coordinates must satisfy the equation of the line. Substitute the expressions for $x$ and $y$ into the line equation:

$2\left(\frac{3k + 2}{k + 1}\right) + \left(\frac{7k - 2}{k + 1}\right) - 4 = 0$

Multiply the entire equation by $(k + 1)$ to clear the denominators (assuming $k+1 \neq 0$):

$2(3k + 2) + (7k - 2) - 4(k + 1) = 0$

$6k + 4 + 7k - 2 - 4k - 4 = 0$

Combine the terms:

$(6k + 7k - 4k) + (4 - 2 - 4) = 0$

$9k - 2 = 0$

$9k = 2$

$k = \frac{2}{9}$

The ratio is $k : 1 = \frac{2}{9} : 1$. To express this in its simplest form, multiply both parts of the ratio by $9$:

Ratio $= 2 : 9$

Since the value of $k = \frac{2}{9}$ is positive, the line divides the line segment internally.

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Calculation of Point of Division:

Now substitute the value of $k = \frac{2}{9}$ back into the expressions for $x$ and $y$ to find the coordinates of P.

$x = \frac{3(\frac{2}{9}) + 2}{\frac{2}{9} + 1} = \frac{\frac{6}{9} + 2}{\frac{2 + 9}{9}} = \frac{\frac{2}{3} + 2}{\frac{11}{9}} = \frac{\frac{2 + 6}{3}}{\frac{11}{9}} = \frac{\frac{8}{3}}{\frac{11}{9}}$

$x = \frac{8}{3} \times \frac{9}{11} = \frac{8 \times \cancel{9}^3}{\cancel{3}_1 \times 11} = \frac{24}{11}$

$y = \frac{7(\frac{2}{9}) - 2}{\frac{2}{9} + 1} = \frac{\frac{14}{9} - 2}{\frac{11}{9}} = \frac{\frac{14 - 18}{9}}{\frac{11}{9}} = \frac{\frac{-4}{9}}{\frac{11}{9}}$

$y = \frac{-4}{\cancel{9}} \times \frac{\cancel{9}}{11} = \frac{-4}{11}$

The coordinates of the point of division P are $(\frac{24}{11}, -\frac{4}{11})$.

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Answer:

The line $2x + y - 4 = 0$ divides the line segment joining A$(2, -2)$ and B$(3, 7)$ in the ratio $2:9$ internally.

The coordinates of the point of division are $(\frac{24}{11}, -\frac{4}{11})$.

Given:

Let the vertices of the triangle be A$(x_1, y_1)$, B$(x_2, y_2)$, and C$(x_3, y_3)$.

Let the midpoints of the sides BC, CA, and AB be D$(1, 2)$, E$(0, -1)$, and F$(2, -1)$ respectively.

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To Find:

The coordinates of the vertices A$(x_1, y_1)$, B$(x_2, y_2)$, and C$(x_3, y_3)$.

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Formula:

The coordinates of the midpoint of a line segment joining points $(x_a, y_a)$ and $(x_b, y_b)$ are given by the midpoint formula:

Midpoint $(x_m, y_m) = (\frac{x_a + x_b}{2}, \frac{y_a + y_b}{2})$

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Setting up Equations using Midpoint Formula:

Since D$(1, 2)$ is the midpoint of BC:

Since E$(0, -1)$ is the midpoint of CA:

Since F$(2, -1)$ is the midpoint of AB:

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Solving for x-coordinates:

We have a system of linear equations for $x_1, x_2, x_3$:

(1) $x_2 + x_3 = 2$

(2) $x_3 + x_1 = 0$

(3) $x_1 + x_2 = 4$

Add equations (1), (2), and (3):

$(x_2 + x_3) + (x_3 + x_1) + (x_1 + x_2) = 2 + 0 + 4$

$2x_1 + 2x_2 + 2x_3 = 6$

Divide by 2:

Now subtract equations (1), (2), and (3) from equation (7) to find $x_1, x_2, x_3$:

Subtract (1) from (7): $(x_1 + x_2 + x_3) - (x_2 + x_3) = 3 - 2 \implies x_1 = 1$

Subtract (2) from (7): $(x_1 + x_2 + x_3) - (x_3 + x_1) = 3 - 0 \implies x_2 = 3$

Subtract (3) from (7): $(x_1 + x_2 + x_3) - (x_1 + x_2) = 3 - 4 \implies x_3 = -1$

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Solving for y-coordinates:

We have a system of linear equations for $y_1, y_2, y_3$:

(4) $y_2 + y_3 = 4$

(5) $y_3 + y_1 = -2$

(6) $y_1 + y_2 = -2$

Add equations (4), (5), and (6):

$(y_2 + y_3) + (y_3 + y_1) + (y_1 + y_2) = 4 + (-2) + (-2)$

$2y_1 + 2y_2 + 2y_3 = 0$

Divide by 2:

Now subtract equations (4), (5), and (6) from equation (8) to find $y_1, y_2, y_3$:

Subtract (4) from (8): $(y_1 + y_2 + y_3) - (y_2 + y_3) = 0 - 4 \implies y_1 = -4$

Subtract (5) from (8): $(y_1 + y_2 + y_3) - (y_3 + y_1) = 0 - (-2) \implies y_2 = 2$

Subtract (6) from (8): $(y_1 + y_2 + y_3) - (y_1 + y_2) = 0 - (-2) \implies y_3 = 2$

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Coordinates of the Vertices:

A$(x_1, y_1) = (1, -4)$

B$(x_2, y_2) = (3, 2)$

C$(x_3, y_3) = (-1, 2)$

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Answer:

The coordinates of the vertices of the triangle are A$(1, -4)$, B$(3, 2)$, and C$(-1, 2)$.

Given vertices:

The vertices of the parallelogram ABCD are given in order as A$(6, 1)$, B$(8, 2)$, C$(9, 4)$, and D$(p, 3)$.

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To Find:

The value of $p$.

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Method Explanation:

A key property of a parallelogram is that its diagonals bisect each other. This means the point where the two diagonals intersect is the midpoint of both diagonals. If the vertices are taken in order A, B, C, and D, the diagonals are AC and BD.

Therefore, the midpoint of diagonal AC must be the same as the midpoint of diagonal BD.

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Formula:

The coordinates of the midpoint of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ are given by the midpoint formula:

Midpoint $(x_m, y_m) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

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Calculation of Midpoints:

Midpoint of diagonal AC:

Using points A$(6, 1)$ and C$(9, 4)$: $(x_1, y_1) = (6, 1)$, $(x_2, y_2) = (9, 4)$.

Midpoint of AC $= (\frac{6 + 9}{2}, \frac{1 + 4}{2})$

Midpoint of AC $= (\frac{15}{2}, \frac{5}{2})$

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Midpoint of diagonal BD:

Using points B$(8, 2)$ and D$(p, 3)$: $(x_1, y_1) = (8, 2)$, $(x_2, y_2) = (p, 3)$.

Midpoint of BD $= (\frac{8 + p}{2}, \frac{2 + 3}{2})$

Midpoint of BD $= (\frac{8 + p}{2}, \frac{5}{2})$

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Equating the Midpoints:

Since the midpoints of AC and BD are the same, we equate their respective coordinates:

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Solving for $p$:

From the equation of the x-coordinates:

$\frac{15}{2} = \frac{8 + p}{2}$

Multiply both sides by 2:

$15 = 8 + p$

Subtract 8 from both sides:

$p = 15 - 8$

$p = 7$

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Answer:

The value of $p$ is $7$.

Given vertices:

Let the vertices of the quadrilateral be A$(-4, -2)$, B$(-3, -5)$, C$(3, -2)$, and D$(2, 3)$ taken in order.

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To Find:

The area of the quadrilateral ABCD.

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Method:

We can divide the quadrilateral into two triangles by joining any two non-adjacent vertices (drawing a diagonal). Let's draw the diagonal AC. This divides the quadrilateral ABCD into two triangles, $\triangle ABC$ and $\triangle ADC$. The area of the quadrilateral will be the sum of the areas of these two triangles.

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Formula for Area of a Triangle:

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

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

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Calculating the Area of $\triangle ABC$:

The vertices are A$(-4, -2)$, B$(-3, -5)$, and C$(3, -2)$.

Let $(x_1, y_1) = (-4, -2)$, $(x_2, y_2) = (-3, -5)$, $(x_3, y_3) = (3, -2)$.

Area($\triangle ABC$) $= \frac{1}{2} |(-4)((-5) - (-2)) + (-3)((-2) - (-2)) + 3((-2) - (-5))|$

Area($\triangle ABC$) $= \frac{1}{2} |(-4)(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)|$

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

Area($\triangle ABC$) $= \frac{1}{2} |12 + 0 + 9|$

Area($\triangle ABC$) $= \frac{1}{2} |21|$

Area($\triangle ABC$) $= \frac{21}{2}$ square units.

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Calculating the Area of $\triangle ADC$:

The vertices are A$(-4, -2)$, D$(2, 3)$, and C$(3, -2)$. Note the order of vertices to maintain consistency when using the formula (A, D, C). Alternatively, use A, C, D.

Let's use A$(-4, -2)$, C$(3, -2)$, D$(2, 3)$. So, $(x_1, y_1) = (-4, -2)$, $(x_2, y_2) = (3, -2)$, $(x_3, y_3) = (2, 3)$.

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

Area($\triangle ADC$) $= \frac{1}{2} |(-4)(-5) + 3(3 + 2) + 2(-2 + 2)|$

Area($\triangle ADC$) $= \frac{1}{2} |20 + 3(5) + 2(0)|$

Area($\triangle ADC$) $= \frac{1}{2} |20 + 15 + 0|$

Area($\triangle ADC$) $= \frac{1}{2} |35|$

Area($\triangle ADC$) $= \frac{35}{2}$ square units.

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Calculating the Area of Quadrilateral ABCD:

Area(ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

Area(ABCD) $= \frac{21}{2} + \frac{35}{2}$

Area(ABCD) $= \frac{21 + 35}{2}$

Area(ABCD) $= \frac{56}{2}$

Area(ABCD) $= 28$ square units.

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Answer:

The area of the quadrilateral whose vertices are $(-4, -2)$, $(-3, -5)$, $(3, -2)$, and $(2, 3)$ taken in order is $28$ square units.

Given vertices:

The vertices of the triangle are A$(2, 3)$, B$(-1, 0)$, and C$(2, -4)$.

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Part 1: Find the area of $\triangle ABC$.

To Find:

Area of $\triangle ABC$.

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Formula:

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

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

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Calculation:

Using the vertices A$(2, 3)$, B$(-1, 0)$, C$(2, -4)$:

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

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

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

Substitute these values 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 $= \frac{1}{2} \times 21$

Area $= \frac{21}{2}$ square units.

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Answer (Area):

The area of the triangle formed by the points A$(2, 3)$, B$(-1, 0)$, C$(2, -4)$ is $\frac{21}{2}$ square units.

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Part 2: Find the length of the median from vertex A to the side BC.

To Find:

The length of the median AM, where M is the midpoint of BC.

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Concept:

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The median from vertex A connects A to the midpoint of side BC.

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Finding the Midpoint of BC:

Let M be the midpoint of the side BC. Using the midpoint formula for points B$(-1, 0)$ and C$(2, -4)$:

Midpoint M$(x_m, y_m) = (\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2})$

$x_m = \frac{-1 + 2}{2} = \frac{1}{2}$

$y_m = \frac{0 + (-4)}{2} = \frac{-4}{2} = -2$

So, the coordinates of the midpoint M are $(\frac{1}{2}, -2)$.

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Finding the Length of the Median AM:

The length of the median AM is the distance between point A$(2, 3)$ and point M$(\frac{1}{2}, -2)$. Using the distance formula:

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

Using A$(2, 3)$ as $(x_1, y_1)$ and M$(\frac{1}{2}, -2)$ as $(x_2, y_2)$:

$AM = \sqrt{(\frac{1}{2} - 2)^2 + (-2 - 3)^2}$

$AM = \sqrt{(\frac{1 - 4}{2})^2 + (-5)^2}$

$AM = \sqrt{(\frac{-3}{2})^2 + (-5)^2}$

$AM = \sqrt{\frac{9}{4} + 25}$

$AM = \sqrt{\frac{9}{4} + \frac{100}{4}}$

$AM = \sqrt{\frac{9 + 100}{4}}$

$AM = \sqrt{\frac{109}{4}}$

$AM = \frac{\sqrt{109}}{\sqrt{4}}$

$AM = \frac{\sqrt{109}}{2}$ units.

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Answer (Median Length):

The length of the median from vertex A to the side BC is $\frac{\sqrt{109}}{2}$ units.

Given points:

Point A: $(x_1, y_1) = (1, -5)$

Point B: $(x_2, y_2) = (-4, 5)$

The line dividing the segment is the x-axis.

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To Find:

The ratio in which the x-axis divides the line segment AB and the coordinates of the point of division.

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Let:

Let the x-axis divide the line segment joining A$(1, -5)$ and B$(-4, 5)$ at the point P$(x, y)$ in the ratio $k : 1$ internally.

Since the point P lies on the x-axis, its y-coordinate is $0$. So, P$(x, 0)$.

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Formula:

Using the section formula, the coordinates of the point P$(x, y)$ that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1 : m_2$ internally are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Here, $m_1 = k$ and $m_2 = 1$.

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Calculation of Ratio:

Using the y-coordinate part of the section formula, and knowing that the y-coordinate of the point on the x-axis is $0$:

$y = \frac{k y_2 + 1 y_1}{k + 1}$

$0 = \frac{k(5) + 1(-5)}{k + 1}$

$0 = \frac{5k - 5}{k + 1}$

For this equation to be true, the numerator must be $0$ (assuming $k+1 \neq 0$):

$5k - 5 = 0$

$5k = 5$

$k = 1$

The ratio is $k : 1 = 1 : 1$. Since $k = 1$ is positive, the division is internal.

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Explanation based on y-coordinates:

The y-coordinate of point A is $-5$ (negative) and the y-coordinate of point B is $5$ (positive). This means that point A is below the x-axis and point B is above the x-axis.

When a line segment connects two points that lie on opposite sides of a line (in this case, the x-axis), the line must intersect the segment somewhere between the two endpoints. This intersection point represents internal division.

The ratio $k = \frac{m_1}{m_2} = -\frac{y_1}{y_2}$ is a shortcut derived from the section formula when the dividing line is the x-axis (setting the y-coordinate of the point of division to 0):

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = 0 \implies m_1 y_2 + m_2 y_1 = 0 \implies m_1 y_2 = -m_2 y_1 \implies \frac{m_1}{m_2} = -\frac{y_1}{y_2}$.

Using the y-coordinates of A$(1, -5)$ and B$(-4, 5)$: $y_1 = -5$ and $y_2 = 5$.

Ratio $\frac{m_1}{m_2} = -\frac{-5}{5} = \frac{5}{5} = 1$.

So the ratio is $1:1$. The positive value of the ratio confirms internal division, which is consistent with the y-coordinates having opposite signs.

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Calculation of Point of Division:

Now substitute the value of $k = 1$ (or $m_1 = 1, m_2 = 1$) back into the expressions for $x$ and $y$ to find the coordinates of P. Since the ratio is $1:1$, P is the midpoint.

Using the section formula or midpoint formula:

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

$y = \frac{1(5) + 1(-5)}{1 + 1} = \frac{5 - 5}{2} = \frac{0}{2} = 0$

The coordinates of the point of division P are $(-\frac{3}{2}, 0)$.

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Answer:

The x-axis divides the line segment joining the points $(1, -5)$ and $(-4, 5)$ in the ratio $1:1$ internally.

The coordinates of the point of division are $(-\frac{3}{2}, 0)$.

Given points:

Let the given points be A$(1, 5)$, B$(2, 3)$, and C$(-2, -11)$.

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Condition for Collinearity using Distance Formula:

Three points A, B, and C are collinear if the sum of the lengths of any two line segments formed by these points is equal to the length of the third line segment. That is, if $AB + BC = AC$ or $AB + AC = BC$ or $AC + BC = AB$. We need to calculate the distances between all three pairs of points (AB, BC, and AC) and check this condition.

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Formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

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

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Calculation of Distances:

Distance AB: Using A$(1, 5)$ and B$(2, 3)$

$AB = \sqrt{(2 - 1)^2 + (3 - 5)^2}$

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

$AB = \sqrt{1 + 4}$

$AB = \sqrt{5}$ units.

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Distance BC: Using B$(2, 3)$ and C$(-2, -11)$

$BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2}$

$BC = \sqrt{(-4)^2 + (-14)^2}$

$BC = \sqrt{16 + 196}$

$BC = \sqrt{212}$ units.

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Distance AC: Using A$(1, 5)$ and C$(-2, -11)$

$AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2}$

$AC = \sqrt{(-3)^2 + (-16)^2}$

$AC = \sqrt{9 + 256}$

$AC = \sqrt{265}$ units.

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Checking for Collinearity:

We need to check if the sum of the lengths of the two smaller segments equals the length of the longest segment. The lengths are $\sqrt{5}$, $\sqrt{212}$, and $\sqrt{265}$.

Approximate values: $\sqrt{5} \approx 2.236$, $\sqrt{212} \approx 14.560$, $\sqrt{265} \approx 16.279$.

Let's check the sums:

$AB + BC = \sqrt{5} + \sqrt{212} \approx 2.236 + 14.560 = 16.796$

$AC = \sqrt{265} \approx 16.279$

Since $AB + BC \neq AC$ ($16.796 \neq 16.279$), the points are not collinear.

Let's double-check other combinations:

$AB + AC = \sqrt{5} + \sqrt{265} \approx 2.236 + 16.279 = 18.515$ (Not equal to $\sqrt{212}$)

$AC + BC = \sqrt{265} + \sqrt{212} \approx 16.279 + 14.560 = 30.839$ (Not equal to $\sqrt{5}$)

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Conclusion:

Since the sum of the lengths of any two segments formed by the points is not equal to the length of the third segment, the points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are not collinear.

Given points:

Let the points be A$(x_1, y_1) = (2, -2)$ and B$(x_2, y_2) = (-7, 4)$.

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To Find:

1. The coordinates of the point that divides AB internally in the ratio $1:2$.

2. The coordinates of the point that divides AB internally in the ratio $2:1$.

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Formula:

The coordinates $(x, y)$ of a point that divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are given by the section formula:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

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Part 1: Dividing in the ratio 1:2 internally.

Here, $m_1 = 1$ and $m_2 = 2$.

Using the section formula:

For the x-coordinate:

$x = \frac{(1)(-7) + (2)(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1$

For the y-coordinate:

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

The coordinates of the point are $(-1, 0)$.

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Part 2: Dividing in the ratio 2:1 internally.

Here, $m_1 = 2$ and $m_2 = 1$.

Using the section formula:

For the x-coordinate:

$x = \frac{(2)(-7) + (1)(2)}{2 + 1} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4$

For the y-coordinate:

$y = \frac{(2)(4) + (1)(-2)}{2 + 1} = \frac{8 - 2}{3} = \frac{6}{3} = 2$

The coordinates of the point are $(-4, 2)$.

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Answer:

The coordinates of the point that divides the line segment joining $(2, -2)$ and $(-7, 4)$ in the ratio $1:2$ internally are $(-1, 0)$.

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

Given vertices:

Let the two given opposite vertices of the square be A$(-1, 2)$ and C$(3, 2)$. Let the other two vertices be B$(x, y)$ and D$(x', y')$.

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To Find:

The coordinates of the vertices B and D.

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Method Explanation:

We use the key properties of a square:

1. The diagonals of a square bisect each other. This means their intersection point is the midpoint of both diagonals.

2. The diagonals of a square are perpendicular to each other.

3. The diagonals of a square are equal in length.

4. All sides of a square are equal in length.

Let M be the midpoint of the diagonal AC. M is also the midpoint of the diagonal BD. Since the diagonals are perpendicular, the line segment BD lies on a line perpendicular to AC and passing through M. Also, the distance from the center M to each vertex (A, B, C, D) is the same, which is half the length of a diagonal.

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Calculation:

First, find the coordinates of the midpoint M of the diagonal AC using the midpoint formula $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$:

$M = \left(\frac{-1 + 3}{2}, \frac{2 + 2}{2}\right) = \left(\frac{2}{2}, \frac{4}{2}\right) = (1, 2)$

So, the center of the square is M$(1, 2)$. M is also the midpoint of the other diagonal BD.

Next, examine the line segment AC. Since the y-coordinates of A and C are the same ($y_A = y_C = 2$), the line segment AC is horizontal.

Because the diagonals of a square are perpendicular, the diagonal BD must be vertical, passing through the midpoint M$(1, 2)$. A vertical line has a constant x-coordinate. Thus, the x-coordinates of the other two vertices B and D must be equal to the x-coordinate of M, which is 1.

So, the coordinates of the other two vertices are B$(1, y)$ and D$(1, y')$.

The distance from the center M to any vertex is half the length of the diagonal. Let's find the distance MA (half the length of AC):

$MA = \sqrt{(-1 - 1)^2 + (2 - 2)^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2$

The distance MB (or MD) must also be equal to MA, which is 2. Using the distance formula for MB, where M$(1, 2)$ and B$(1, y)$:

$MB = \sqrt{(1 - 1)^2 + (y - 2)^2}$

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

$2 = \sqrt{(y - 2)^2}$

Squaring both sides:

$2^2 = (y - 2)^2$

$4 = (y - 2)^2$

Taking the square root of both sides:

$y - 2 = \pm \sqrt{4}$

$y - 2 = \pm 2$

This gives two possible values for $y$:

Case 1: $y - 2 = 2 \implies y = 2 + 2 = 4$. One vertex is $(1, 4)$.

Case 2: $y - 2 = -2 \implies y = 2 - 2 = 0$. The other vertex is $(1, 0)$.

So, the coordinates of the other two vertices are $(1, 4)$ and $(1, 0)$.

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Answer:

The coordinates of the other two vertices of the square are $(1, 4)$ and $(1, 0)$.