---
title: "NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1"
url: https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions/exercise-7-1
dateModified: 2026-10-07T15:54:16+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

Chapter 7: Coordinate Geometry. Every question from Exercise 7.1, with full working and the final answer.

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## EXERCISE 7.1

### Question 1

*3 marks · Short answer*

Find the distance between the following pairs of points :
(i) $(2, 3), (4, 1)$ 
(ii) $(-5, 7), (-1, 3)$ 
(iii) $(a, b), (-a, -b)$

**Part (i)**

1. Let $P(2, 3)$ and $Q(4, 1)$ be the given points.
2. Distance $PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$ units.

Answer (i): $2\sqrt{2}$ units

**Part (ii)**

1. Let $P(-5, 7)$ and $Q(-1, 3)$ be the given points.
2. Distance $PQ = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}$ units.

Answer (ii): $4\sqrt{2}$ units

**Part (iii)**

1. Let $P(a, b)$ and $Q(-a, -b)$ be the given points.
2. Distance $PQ = \sqrt{(-a - a)^2 + (-b - b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2}$ units.

Answer (iii): $2\sqrt{a^2 + b^2}$ units

**Answer:** (i) $2\sqrt{2}$, (ii) $4\sqrt{2}$, (iii) $2\sqrt{a^2 + b^2}$

> Common mistake: Errors in handling negative signs inside the distance formula brackets.

### Question 2

*3 marks · Short answer*

Find the distance between the points $(0, 0)$ and $(36, 15)$. Can you now find the distance between the two towns A and B discussed in Section 7.2.

**Solution**

1. Let $P(0, 0)$ and $Q(36, 15)$ be the given points.
2. Distance $PQ = \sqrt{(36 - 0)^2 + (15 - 0)^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39$ units.
3. Yes, we can now find the distance between the two towns A and B discussed in Section 7.2, which is $39\text{ km}$.

**Answer:** $39$ units and $39\text{ km}$

> Common mistake: Calculation error in squaring large numbers like $36$.

### Question 3

*3 marks · Short answer*

Determine if the points $(1, 5)$, $(2, 3)$ and $(-2, -11)$ are collinear.

**Solution**

1. Let $A(1, 5)$, $B(2, 3)$ and $C(-2, -11)$ be the given points.
2. Find the distances between the points: $AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{1 + 4} = \sqrt{5}$.
3. Find $BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53}$ and $AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} = \sqrt{9 + 256} = \sqrt{265}$.
4. Since $AB + BC \neq AC$, the points are not collinear.

**Answer:** The points are not collinear.

> Common mistake: Assuming points are collinear without checking the triangle inequality sum condition properly.

### Question 4

*3 marks · Short answer*

Check whether $(5, -2)$, $(6, 4)$ and $(7, -2)$ are the vertices of an isosceles triangle.

**Solution**

1. Let $A(5, -2)$, $B(6, 4)$ and $C(7, -2)$ be the given points.
2. Calculate the lengths of the sides: $AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2} = \sqrt{1 + 36} = \sqrt{37}$.
3. $BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{1 + 36} = \sqrt{37}$ and $AC = \sqrt{(7 - 5)^2 + (-2 - (-2))^2} = \sqrt{4} = 2$.
4. Since $AB = BC = \sqrt{37}$, two sides are equal, so ABC is an isosceles triangle.

**Answer:** Yes, they are the vertices of an isosceles triangle.

> Common mistake: Incorrect subtraction of negative coordinates.

### Question 5

*3 marks · Short answer*

In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, "Don't you think ABCD is a square?" Chameli disagrees. Using distance formula, find which of them is correct.

**Solution**

1. From Fig. 7.8, the coordinates of the points are $A(3, 4)$, $B(6, 7)$, $C(9, 4)$ and $D(6, 1)$.
2. Calculate the lengths of all four sides: $AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{9 + 9} = \sqrt{18}$, $BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{9 + 9} = \sqrt{18}$, $CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = \sqrt{9 + 9} = \sqrt{18}$, and $DA = \sqrt{(3 - 6)^2 + (4 - 1)^2} = \sqrt{9 + 9} = \sqrt{18}$.
3. Calculate the diagonals: $AC = \sqrt{(9 - 3)^2 + (4 - 4)^2} = \sqrt{36} = 6$ and $BD = \sqrt{(6 - 6)^2 + (1 - 7)^2} = \sqrt{36} = 6$.
4. Since all four sides are equal and both diagonals are equal, ABCD is a square. Therefore, Champa is correct.

**Answer:** Champa is correct.

> Common mistake: Forgetting to check the diagonals while proving a square.

### Question 6

*3 marks · Short answer*

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) $(-1, -2), (1, 0), (-1, 2), (-3, 0)$
(ii) $(-3, 5), (3, 1), (0, 3), (-1, -4)$
(iii) $(4, 5), (7, 6), (4, 3), (1, 2)$

**Part (i)**

1. Let $A(-1, -2)$, $B(1, 0)$, $C(-1, 2)$, and $D(-3, 0)$ be the points.
2. Finding side lengths: $AB = \sqrt{8}$, $BC = \sqrt{8}$, $CD = \sqrt{8}$, $DA = \sqrt{8}$.
3. Finding diagonals: $AC = 4$ and $BD = 4$.
4. Since all sides are equal and diagonals are equal, it is a square.

Answer (i): Square

**Part (ii)**

1. Let $A(-3, 5)$, $B(3, 1)$, $C(0, 3)$, and $D(-1, -4)$ be the points.
2. Check collinearity for points A, C, B: $AC = \sqrt{13}$, $CB = \sqrt{13}$, $AB = \sqrt{52} = 2\sqrt{13}$.
3. Since $AC + CB = AB$, the points A, C, B are collinear.
4. Thus, the given points do not form a proper quadrilateral.

Answer (ii): No quadrilateral is formed as three points are collinear.

**Part (iii)**

1. Let $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, and $D(1, 2)$ be the points.
2. Finding side lengths: $AB = \sqrt{10}$, $BC = \sqrt{18}$, $CD = \sqrt{10}$, $DA = \sqrt{18}$.
3. Finding diagonals: $AC = \sqrt{4} = 2$ and $BD = \sqrt{40} = 2\sqrt{10}$.
4. Since opposite sides are equal ($AB = CD$ and $BC = DA$) but diagonals are not equal, it is a parallelogram.

Answer (iii): Parallelogram

**Answer:** (i) Square, (ii) General quadrilateral, (iii) Parallelogram

> Common mistake: Failing to check for collinearity in part (ii).

### Question 7

*3 marks · Short answer*

Find the point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.

**Solution**

1. Let the point on the $x$-axis be $P(x, 0)$.
2. Let $A(2, -5)$ and $B(-2, 9)$ be the given points.
3. Since $P$ is equidistant from $A$ and $B$, $PA = PB$, so $PA^2 = PB^2$.
4. $(x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2$.
5. $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$.
6. $-8x = 56$, which gives $x = -7$.
7. Therefore, the required point on the $x$-axis is $(-7, 0).$

**Answer:** $(-7, 0)$

> Common mistake: Writing a point on the $x$-axis as $(0, y)$ instead of $(x, 0)$.

### Question 8

*3 marks · Short answer*

Find the values of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is $10$ units.

**Solution**

1. We are given the points $P(2, -3)$ and $Q(10, y)$, and the distance $PQ = 10$.
2. Using the distance formula, $PQ = \sqrt{(10 - 2)^2 + (y - (-3))^2} = 10$.
3. Squaring both sides, $(8)^2 + (y + 3)^2 = 100$
4. $64 + y^2 + 6y + 9 = 100$
5. $y^2 + 6y - 27 = 0$
6. $(y + 9)(y - 3) = 0$, which gives $y = 3$ or $y = -9$.

**Answer:** $y = 3$ or $y = -9$

> Common mistake: Forgetting to consider both positive and negative values of $y$ after taking the square root.

### Question 9

*3 marks · Short answer*

If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$, find the values of $x$. Also find the distances $QR$ and $PR$.

**Solution**

1. We are given that $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$, so $QP = QR$.
2. $QP^2 = (5 - 0)^2 + (-3 - 1)^2 = 25 + 16 = 41$.
3. $QR^2 = (x - 0)^2 + (6 - 1)^2 = x^2 + 25$.
4. Since $QP^2 = QR^2$, we have $x^2 + 25 = 41$, which gives $x^2 = 16$, so $x = \pm 4$.
5. For $x = 4$, $R$ is $(4, 6)$, so $QR = \sqrt{41}$ and $PR = \sqrt{(4 - 5)^2 + (6 - (-3))^2} = \sqrt{1 + 81} = \sqrt{82}$.
6. For $x = -4$, $R$ is $(-4, 6)$, so $QR = \sqrt{41}$ and $PR = \sqrt{(-4 - 5)^2 + (6 - (-3))^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$.

**Answer:** $x = \pm 4$; for $x = 4$, $QR = \sqrt{41}, PR = \sqrt{82}$; for $x = -4$, $QR = \sqrt{41}, PR = 9\sqrt{2}$

> Common mistake: Not finding both cases for $x$ when solving $x^2 = 16$.

### Question 10

*3 marks · Short answer*

Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the point $(3, 6)$ and $(-3, 4)$.

**Solution**

1. Let $P(x, y)$ be equidistant from $A(3, 6)$ and $B(-3, 4)$.
2. Therefore, $PA = PB$, which implies $PA^2 = PB^2$.
3. $(x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2$
4. $x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$
5. $-6x - 12y + 45 = 6x - 8y + 25$
6. $12x + 4y - 20 = 0$, which simplifies to $3x + y - 5 = 0$.

**Answer:** $3x + y - 5 = 0$

> Common mistake: Sign errors while expanding terms with negative coordinates.

## Related pages

- [All Chapter 7 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions)
- [Exercise 7.2](https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions/exercise-7-2)

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