# NCERT Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2 NCERT Solutions

## NCERT Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2 NCERT Solutions

The academic team at Neutronclasses has meticulously crafted NCERT solutions for Chapter 7 Exercise 7.2 of Class 10 Mathematics, which focuses on Coordinate Geometry. These solutions comprehensively cover all the exercises within this chapter. Presented below are the step-by-step solutions to all the questions found in the NCERT textbook for Chapter 7.

### 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.**

**Answer:** Let P(x, y) be the required point. Using the section formula

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).****Answer:**

Let P (x_{1},y_{1}) and Q (x_{2},y_{2}) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

Therefore P(x_{1},y_{1}) = (2, -5/3)

Point Q divides AB internally in the ratio 2:1.

Q (x_{2} ,y_{2}) = (0, -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 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th 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 flagexactly halfway between the line segment joining the two flags, where should she post her flag?**

**Answer:**

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

**Answer:**

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.**

**Answer:**

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.

To find the coordinates let’s substitute the value of k in equation(1)

Required point = [(- 4(1) + 1) / (1 + 1), (5(1) – 5) / (1 + 1)]

= [(- 4 + 1) / 2, (5 – 5) / 2]

= [- 3/2, 0]

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

**Answer:**

Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.

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

**Answer:**

Let (x,y) be the coordinate of A.

Since AB is the diameter of the circle, the centre will be the mid-point of AB.

now, as centre is the mid-point of AB.

x-coordinate of centre = (2x+1)/2

y-coordinate of centre = (2y+4)/2

But given that centre of circle is (2,−3).

Therefore,

(2x+1)/2=2⇒x=3

(2y+4)/2=−3⇒y=−10

Thus the coordinate of A is (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.**

**Answer:**

As given the coordinates of A(−2,−2) and B(2,−4) and P is a point lies on AB.

And AP = 3/7 AB

∴BP = 4/7

Then, ratio of AP and PB = m_{1}:m_{2 }= 3:4

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

∴ x = (m_{1}x_{2} + m_{2}x_{1}) / (m_{1} + m_{2})

⇒ x = (3 × 2 + 4 × (−2)) / (3 + 4) = (6 − 8) / 7 = −2 / 7

And y = (m_{1}y_{2} + m_{2}y_{1}) / (m_{1} + m_{2})

⇒ y = ((3 × (−4) + 4 × (−2)) / (3 + 4) = (−12−8) / 7 = −20 / 7

∴ Coordinates of P = −2 / 7, −20 / 7

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

**Answer:**

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.

Using Sectional Formula, we get,

Coordinates of X = ((1 × 2 + 3 × (−2)) / (1 + 3), (1 × 8 + 3 × 2) / (1 + 3))

= (−1, 7/2)

Coordinates of Y = (2 − 2) / 2, (2 + 8) / 2 = (0,5)

Coordinates of Z = ((3 × 2 + 1 × (−2)) / (1 + 3), (3 × 8 + 1 × 2) / (1 + 3)

= (1, 13/2)

**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)]**

**Answer:**

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

Length of the diagonal AC=

Length of the diagonal BD=

Area of rhombus ABCD = 1/2 X 4√2 X 6√2= 24 square units.

Therefore, the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is 24 square units.

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