---
title: "NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2"
url: https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions/exercise-7-2
dateModified: 2026-10-07T15:54:16+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2

Chapter 7: Coordinate Geometry. Every question from Exercise 7.2, with full working and the final answer.

Free PDF (9 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-64ea47ba88.pdf

## EXERCISE 7.2

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Let P(x, y) be the required point dividing the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$.
2. Using the section formula, $x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = 1$.
3. Similarly, $y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = 3$.

**Answer:** (1, 3)

> Common mistake: Interchanging the values of $m_1$ and $m_2$ or using the mid-point formula instead of the section formula.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Let P and Q be the points of trisection of the line segment joining A(4, -1) and B(-2, -3) such that AP = PQ = QB.
2. Point P divides AB internally in the ratio $1 : 2$, so its coordinates are $\left(\frac{1(-2) + 2(4)}{1 + 2}, \frac{1(-3) + 2(-1)}{1 + 2}\right) = (2, -5/3)$.
3. Point Q divides AB internally in the ratio $2 : 1$, so its coordinates are $\left(\frac{2(-2) + 1(4)}{2 + 1}, \frac{2(-3) + 1(-1)}{2 + 1}\right) = (0, -7/3)$.

**Answer:** $(2, -\frac{5}{3})$ and $(0, -\frac{7}{3})$

> Common mistake: Taking incorrect ratios for the points P and Q.

### Question 3

*3 marks · Short answer*

To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of $1\text{m}$ each. $100$ flower pots have been placed at a distance of $1\text{m}$ from each other along AD, as shown in Fig. 7.12. Niharika runs $\frac{1}{4}\text{th}$ the distance AD on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}\text{th}$ 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**

1. The green flag is posted by Niharika at $\left(2, \frac{1}{4} \times 100\right) = (2, 25)$ and the red flag is posted by Preet at $\left(8, \frac{1}{5} \times 100\right) = (8, 20)$.
2. The distance between the two flags is given by $\sqrt{(8 - 2)^2 + (20 - 25)^2} = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61}\text{ m}$.
3. Rashmi's blue flag is posted halfway between the two flags, so its coordinates are $\left(\frac{2 + 8}{2}, \frac{25 + 20}{2}\right) = (5, 22.5)$.

**Answer:** Distance is $\sqrt{61}\text{ m}$ and the blue flag is at $(5, 22.5)$ on the 5th line.

> Common mistake: Calculating coordinates as just the fractions without multiplying by the total distance of 100.

### Question 4

*3 marks · Short answer*

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

**Solution**

1. Let the point $(-1, 6)$ divide the line segment joining $(-3, 10)$ and $(6, -8)$ in the ratio $k : 1$.
2. Using the section formula for the x-coordinate, $-1 = \frac{6k - 3}{k + 1}$.
3. Solving for $k$, $-k - 1 = 6k - 3 \implies 7k = 2 \implies k : 1 = 2 : 7$.

**Answer:** $2 : 7$

> Common mistake: Forgetting to write the ratio in the form $k : 1$ or making algebraic sign errors.

### Question 5

*3 marks · Short answer*

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**

1. Let the ratio in which the x-axis divides the line segment be $k : 1$. The point of division on the x-axis is of the form $(x, 0)$.
2. Using the section formula for the y-coordinate, $0 = \frac{k(5) + 1(-5)}{k + 1}$.
3. Solving for $k$, $5k - 5 = 0 \implies k = 1$, so the ratio is $1 : 1$, and the point of division is $\left(\frac{1(-4) + 1(1)}{1 + 1}, 0\right) = \left(-\frac{3}{2}, 0\right)$.

**Answer:** Ratio is $1 : 1$ and the point is $(-\frac{3}{2}, 0)$

> Common mistake: Equating the x-coordinate to zero instead of the y-coordinate for a point on the x-axis.

### Question 6

*3 marks · Short answer*

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

**Solution**

1. We know that the diagonals of a parallelogram bisect each other.
2. Equating the mid-point of diagonal joining $(1, 2)$ and $(x, 6)$ with the mid-point of diagonal joining $(4, y)$ and $(3, 5)$, we get $\left(\frac{1 + x}{2}, \frac{2 + 6}{2}\right) = \left(\frac{4 + 3}{2}, \frac{y + 5}{2}\right)$.
3. Comparing the coordinates, $\frac{1 + x}{2} = \frac{7}{2} \implies x = 6$ and $\frac{8}{2} = \frac{y + 5}{2} \implies y = 3$.

**Answer:** $x = 6$, $y = 3$

> Common mistake: Taking the vertices in a non-sequential order instead of taking them in order around the perimeter.

### Question 7

*3 marks · Short answer*

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**

1. Let the coordinates of point A be $(x, y)$.
2. Since AB is the diameter of the circle with centre $(2, -3)$ and B$(1, 4)$, the centre is the mid-point of AB.
3. Using the mid-point formula, we have $\frac{x + 1}{2} = 2$ and $\frac{y + 4}{2} = -3$.
4. Solving for $x$, we get $x + 1 = 4$, so $x = 3$.
5. Solving for $y$, we get $y + 4 = -6$, so $y = -10$.
6. Therefore, the coordinates of A are $(3, -10)$.

**Answer:** $(3, -10)$

> Common mistake: Confusing the centre as one of the endpoints instead of using it as the mid-point.

### Question 8

*3 marks · Short answer*

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

**Solution**

1. We are given $AP = \frac{3}{7}AB$, which means $\frac{AP}{AB} = \frac{3}{7}$.
2. Thus, $PB = AB - AP = AB - \frac{3}{7}AB = \frac{4}{7}AB$.
3. The ratio in which P divides AB is $AP : PB = \frac{3}{7} : \frac{4}{7} = 3 : 4$.
4. Using the section formula for points A$(-2, -2)$ and B$(2, -4)$, the coordinates of P are $\left(\frac{3(2) + 4(-2)}{3 + 4}, \frac{3(-4) + 4(-2)}{3 + 4}\right)$.
5. Simplifying the coordinates, we get $\left(\frac{6 - 8}{7}, \frac{-12 - 8}{7}\right) = \left(\frac{-2}{7}, \frac{-20}{7}\right)$.
6. Therefore, the coordinates of P are $\left(-\frac{2}{7}, -\frac{20}{7}\right)$.

**Answer:** $\left(-\frac{2}{7}, -\frac{20}{7}\right)$

> Common mistake: Taking the ratio as $3 : 7$ instead of finding $PB$ to get the internal ratio $3 : 4$.

### Question 9

*3 marks · Short answer*

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

**Solution**

1. Let the points dividing the line segment AB into four equal parts be P, Q, and R such that $AP = PQ = QR = RB$.
2. Thus, Q is the mid-point of AB, P is the mid-point of AQ, and R is the mid-point of QB.
3. Using the mid-point formula for A$(-2, 2)$ and B$(2, 8)$, the coordinates of Q are $\left(\frac{-2 + 2}{2}, \frac{2 + 8}{2}\right) = (0, 5)$.
4. For point P, which is the mid-point of A$(-2, 2)$ and Q$(0, 5)$, the coordinates are $\left(\frac{-2 + 0}{2}, \frac{2 + 5}{2}\right) = \left(-1, \frac{7}{2}\right)$.
5. For point R, which is the mid-point of Q$(0, 5)$ and B$(2, 8)$, the coordinates are $\left(\frac{0 + 2}{2}, \frac{5 + 8}{2}\right) = \left(1, \frac{13}{2}\right)$.
6. Therefore, the required points are $\left(-1, \frac{7}{2}\right)$, $(0, 5)$, and $\left(1, \frac{13}{2}\right)$.

**Answer:** $\left(-1, \frac{7}{2}\right), (0, 5), \left(1, \frac{13}{2}\right)$

> Common mistake: Applying incorrect ratios like trisection instead of dividing into four equal parts using mid-points.

### Question 10

*3 marks · Short answer*

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**

1. Let the vertices be A$(3, 0)$, B$(4, 5)$, C$(-1, 4)$, and D$(-2, -1)$.
2. Using the distance formula, find the length of diagonal AC: $AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.
3. Find the length of diagonal BD: $BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$.
4. Using the given hint, Area of rhombus = $\frac{1}{2} \times (\text{product of its diagonals})$.
5. Area = $\frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24$ square units.

**Answer:** $24$ square units

> Common mistake: Calculating side lengths instead of diagonal lengths when using the rhombus area formula.

## Related pages

- [All Chapter 7 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions)
- [Exercise 7.1](https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions/exercise-7-1)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student