NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2

NCERT Solutions Class 10 Maths
Chapter – 7 (Coordinate Geometry)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 7 Coordinate Geometry Exercise 7.2 has been provided here to help the students in solving the questions from this exercise.

Chapter : 7 Coordinate Geometry

Exercise – 7.2

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) be the required point. Using the section formula, we get
x = $\frac{(2 \times 4 + 3 \times (-1))}{(2 + 3)}$ = $\frac{8-3}{5}$ = 1

y = $\frac{(2 \times -3 + 3 \times 7)}{(2 + 3)}$ = $\frac{-6+21}{5}$ = 3
Therefore, the point is (1, 3).

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

Solution – Let P (x₁, y₁) and Q (x₂, y2) be the points of trisection of the line segment joining the given points,
i.e. AP = PQ = QB
NCERT Class 10 Maths Solution
Therefore, point P divides AB internally in the ratio 1 : 2.
x₁= $\frac{(1 \times (-2) + 2 \times4)}{3}$ = $\frac{-2 +8}{3}$ = $\frac{6}{3}$ = 2

y₁ = $\frac{(1\times(-3)+2\times(-1))}{1+2}$ = $\frac{-3-2}{3}$ = $\frac{-5}{3}$

Therefore : P (x₁, y₁) = P(2, $\frac{-5}{3}$ )
Point Q divides AB internally in the ratio 2 : 1.
x₂ = $\frac{(2\times (-2) +1\times 4)}{2+1}$ = $\frac{-4+4}{3}$ = 0

y₂ = $\frac{(2\times (-3)+1\times(-1))}{2+1}$ = $\frac{-6-1}{3}$ = $\frac{-7}{3}$
The coordinates of the point Q are (0, $\frac{-7}{3}$ )

3. To conduct sports day activities in your rectangular-shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4^th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5^th the distance AD on the eighth line and posts a red flag. What is the distance between both 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 – From the given instruction, we observed that Niharika posted the green flag at 1/4^th of the distance AD, i.e. (1/4 × 100) m = 25 m from the starting point of the 2nd line. Therefore, the coordinates of this point are (2, 25).
Similarly, Preet posted a red flag at 1/5 of the distance AD, i.e. (1/5 × 100) m = 20 m from the starting point of the 8th line. Therefore, the coordinates of this point are (8, 20).
Distance between these flags can be calculated by using the distance formula,
= $\sqrt{(8-2)^2 +(20-25)^2}$
= $\sqrt{36+25}$
= $\sqrt{61}$

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let’s say this point is P(x, y).
x = (2 + 8)/2 = 10/2 = 5
y = (20 + 25)/2 = 45/2
Hence, P( x, y) = (5, 45/2)
Therefore, Rashmi should post her blue flag at 45/2 = 22.5m on the 5th line.

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

Solution – Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k : 1.
Therefore, -1 = 6k-3/k+1
–k – 1 = 6k -3
7k = 2
k = 2/7
Therefore, the required ratio is 2:7.

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 the ratio in which the line segment joining A (1, – 5) and B ( – 4, 5) is divided by x-axis be k : 1.
Therefore, the coordinates of the point of division is (-4k+1/k+1, 5k-5/k+1).
We know that y-coordinate of any point on x-axis is 0.
∴ 5k-5/k+1 = 0
Therefore, x-axis divides it in the ratio 1 : 1.

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 –
NCERT Class 10 Maths Solution
Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.
Mid point of diagonal AC is O = ((1 + x)/2 , (2 + 6)/2) = ((1 + x)/2 , 4)
Midpoint of diagonal BD is O = ((4 + 3)/2 , (5 + y)/2 ) = (7/2 , (5 + y)/2)
Since the diagonals of a parallelogram bisect each other, the mid point of AC and BD are same.
∴ (x + 1)/2 = 7/2 and 4 = (5 + y)/2
⇒ x + 1 = 7 and 5 + y = 8
⇒ x = 6 and y = 3

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

Solution – Let the coordinates of point A be (x, y).
Midpoint of AB is (2, – 3), which is the centre of the circle.
Coordinate of B = (1, 4)
(2, -3) =((x + 1)/2 , (y + 4)/2)
(x + 1)/2 = 2 and (y + 4)/2 = -3
x + 1 = 4 and y + 4 = -6
x = 3 and y = -10
The coordinates of A(3, -10).

8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

Solution –

The coordinates of points A and B are (-2,-2) and (2,-4), respectively.
Since AP = 3/7 AB
Therefore,
AP : PB = 3 : 4
Point P divides the line segment AB in the ratio 3 : 4.
Coordinate of P = $\left ( \frac{3 \times 2 +4}{3+4}, \frac{3\times(-4)+4\times(-2)}{3+4} \right )$
= $\left ( \frac{6-8}{7}, \frac{-12-8}{7} \right )$
= $\left ( \frac{-2}{7}, \frac{-20}{7} \right )$

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 – Draw a figure, line dividing by 4 points.

From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1 : 3, 1 : 1, 3 : 1 respectively.

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 = 1/2 (product of its diagonals)

Solution –

Let (3, 0), (4, 5), ( – 1, 4) and ( – 2, – 1) are the vertices A, B, C, D of a rhombus ABCD.
NCERT Class 10 Maths Solution

Go Back To Chapters

NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2