AP 10th Class Maths 7th Chapter Coordinate Geometry Exercise 7.2 Solutions

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 m₁ : m₂ = 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 m₁ : m₂ = 2 : 1
Section formula

ii) If m₁ : m₂ = 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 m₁ : m₂.
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
6m₁ – 3m₂ = -1 (m₁ + m₂)
6m₁ – 3m₂ = -m₁ – m₂
6m₁ + m₁ = -m₂ + 3m₂
7m₁ = 2m₂
\(\frac{\mathrm{m}_1}{\mathrm{~m}_2}\) = \(\frac{2}{7}\)
Therefore, m₁ : m₂ = 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 m₁ : m₂.
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(x₁, y₁), Q(x₂, y₂) 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.