Well-designed AP 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Exercise 7.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Coordinate Geometry Class 10 Exercise 7.2 Solutions – 10th Class Maths 7.2 Exercise Solutions

Question 1.

Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.

Solution:

Let P(x, y) can divide the line joining by the points A(-1, 7) and B(4, -3) in the ratio m_{1} : m_{2} = 2 : 3

Section formula

Question 2.

Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Solution:

Given points are P(4, -1) and Q(-2, -3).

A point which divides the line segment in the ratio 2: 1 (or) 1: 2 is called the Trisectional points.

i) If m_{1} : m_{2} = 2 : 1

Section formula

ii) If m_{1} : m_{2} = 1 : 2

Trisectional point B

Question 3.

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 1oo flower pots have been placed at a distance of 1m from each other along AD, as shown In given figure. Niharika runs \(\frac{1}{4}\)th the distance AD on the 2^{nd} line and posts a green flag. Preet runs \(\frac{1}{5}\) th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Solution:

According to the given instructions, Niharika posted the green flag at \(\frac{1}{4}\)th of the distance AD.

That is \(\frac{1}{4}\) th of 100

So, Niharika is from the starting point of the 2^{nd} line.

Therefore, point is (2, 25)

Now, Preet posted a red flag at \(\frac{1}{5}\) th of the distance AD.

That is \(\frac{1}{5}\) th of 100

So, Preet is from the starting point at 8^{th} line.

Therefore, point is (8, 20).

Rashmi post a blue flag between them at an half distance. That is mid point of (2, 25) and (8, 20).

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

= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)

= \(\left(\frac{10}{2}, \frac{45}{2}\right)\)

Therefore, Rashmi post the blue flag at = (5, 22.5)

Distance between Niharika flag (2, 25) and Preet flag (8, 20)

Therefore, distance between Niharika’s green flag and Preet’s red flag is \(\sqrt{61}\) m.

Question 4.

Find the ratio in which the line seg-ment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6).

Solution:

Let P(-1, 6) can divide the line joining by the points A(-3, 10) and B(6, -8) in the ratio m_{1} : m_{2}.

Section formula

= \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)

P(x, y)

= \(\left(\frac{m_1(6)+m_2(-3)}{m_1+m_2}, \frac{m_1(-8)+m_2(10)}{m_1+m_2}\right)\)

= (-1, 6)

By comparing x and y – coordinates

\(\frac{6 m_1-3 m_2}{m_1+m_2}\) = -1 and \(\frac{-8 m_1+10 m_2}{m_1+m_2}\) = 6

6m_{1} – 3m_{2} = -1 (m_{1} + m_{2})

6m_{1} – 3m_{2} = -m_{1} – m_{2}

6m_{1} + m_{1} = -m_{2} + 3m_{2}

7m_{1} = 2m_{2}

\(\frac{\mathrm{m}_1}{\mathrm{~m}_2}\) = \(\frac{2}{7}\)

Therefore, m_{1} : m_{2} = 2 : 7

Question 5.

Find the ratio in which the line segment joining A(1, – 5) and B(- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:

Let P(x, 0) is a point on the x-axis which divides the line joining by the points A(1, -5) and B(-4, 5) in the ratio m_{1} : m_{2}.

Section formula

Question 6.

If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution:

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

In parallelogram diagonals bisect each other.

Question 7.

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is(1, 4).

Solution:

Let A(x, y) and B(1, 4) are the two end points of the diameter AB whose centre O is (2, -3).

We know that, O is the midpoint of AB.

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

Midpoint of AB = \(\left(\frac{x+1}{2}, \frac{y+4}{2}\right)\) = (2, -3)

By comparing x, y-coordinates.

Therefore A(x, y) = A(3, -10)

Question 8.

If A and B are (-2, -2) and (2, -4), re-spectively, find the coordinates of P such that AP = \(\frac{3}{7}\)AB and P lies on the line segment AB.

Solution:

Let P(x, y) divides the line joining by the points A(-2, -2) and B(2, -4) such that AP = \(\frac{3}{7}\)AB

That is \(\frac{\mathrm{AP}}{\mathrm{AB}}\) = \(\frac{3}{7}\) but \(\frac{\mathrm{AP}}{\mathrm{PB}}\) = \(\frac{3}{4}\)

Section formula

Question 9.

Find the coordinates of the points which divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts.

Solution:

Let A(-2, 2) and P(x_{1}, y_{1}), Q(x_{2}, y_{2}) and B(2, 8) are the points on a line which divides the AB into 4 equal parts.

Therefore, C(-1, \(\frac{7}{2}\)), D(0, 5) and E(1, \(\frac{13}{2}\)) are equal distance.

Question 10.

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. [Hint: Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals)]

Solution:

Let A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) are the vertices of ABCD.

In rhombus diagonals AC and BD per-pendicularly bisect each other.