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

# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

This chapter's questions cover the concepts of coordinate geometry, including the distance formula, section formula, and their applications to geometric figures. Students practice finding distances between points, determining collinearity, finding ratios of division, and locating coordinates of points dividing line segments.

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 7.1

### 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.

## 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.

## Frequently asked questions

### How many exercises and questions are there in NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry for the 2026-27 session?

This chapter has a total of 2 exercises with 20 questions in the NCERT textbook. Exercise 7.1 contains 10 questions, and Exercise 7.2 also contains 10 questions. SwaVid provides the free PDF and step-by-step solutions for all these questions on this page only.

### Which important mathematical concepts and topics are covered in these exercises?

Exercise 7.1 covers concepts like the Distance Formula, collinearity of points, equidistant points, properties of triangles, and types of quadrilaterals. Exercise 7.2 covers the Section Formula, Mid-Point Formula, trisection points, division into four parts, and properties of parallelograms and rhombuses.

### What is the hardest question type in this chapter and how should we approach it?

Multi-step problems involving the Section Formula for trisection, finding unknown coordinates using the area of a rhombus, or proving geometric properties of quadrilaterals are usually found challenging. To approach these, first write down the given coordinates, clearly identify the required formula such as $\left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$, and substitute the values carefully.

### How can students write answers in board exams to score full marks in Coordinate Geometry?

To score full marks, always start by writing the relevant formula at the beginning of each step and state the given values clearly. Drawing a rough schematic diagram for problems involving triangles, quadrilaterals, or the section formula helps avoid calculation errors. You can refer to SwaVid's free PDF and step-by-step solutions on this page only to practice the proper presentation format.

### Is the free PDF for these NCERT solutions available for download?

Yes, the complete chapter solutions are fully accessible. SwaVid's free PDF and step-by-step solutions are on this page only to help students prepare effectively for their Class 10 board examinations.

## Related pages

- [Exercise 7.1 solutions](https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions/exercise-7-1)
- [Exercise 7.2 solutions](https://www.swavid.com/maths/class/10/chapter/coordinate-geometry/ncert-solutions/exercise-7-2)
- [Coordinate Geometry: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/coordinate-geometry)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

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