NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry (NCERT 2026–27)

These Class 10 Maths Chapter 7 solutions cover Coordinate Geometry from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 7.1 and Exercise 7.2 is solved step by step using the distance formula and the section formula, with full working, so you can master the chapter and revise it quickly before exams.

Class: 10 Subject: Mathematics Chapter: 7 – Coordinate Geometry Exercises: 7.1, 7.2 Topics: Distance & Section Formula Session: 2026–27
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Chapter 7 Overview

Chapter 7 of Class 10 Maths, Coordinate Geometry, extends the work on the coordinate plane begun in Class IX. It develops coordinate geometry as an algebraic tool for studying geometry — the idea that points, lines and figures on a plane can be analysed using numbers. The chapter introduces two central tools: the distance formula, used to find the length of the line segment joining two points (and hence to test for collinearity, isosceles/equilateral triangles, squares, rectangles and rhombuses), and the section formula, used to find the coordinates of a point that divides a line segment in a given ratio, including the special cases of the mid-point and points of trisection. These ideas are applied in physics, engineering, navigation and design.

Key Concepts & Definitions

Coordinate axes: the position of a point on a plane is fixed by a pair of perpendicular axes. The distance from the y-axis is the x-coordinate (abscissa); the distance from the x-axis is the y-coordinate (ordinate).

Points on the axes: a point on the x-axis has the form (x, 0); a point on the y-axis has the form (0, y).

Distance between two points: the length of the straight line segment joining them, always a non-negative value (positive square root).

Collinear points: three points are collinear if they lie on one straight line; equivalently, the sum of two of the pairwise distances equals the third.

Internal division: a point P lies on segment AB and divides it so that PA : PB = m1 : m2.

Mid-point: the point dividing a segment in the ratio 1 : 1.

Points of trisection: the two points that divide a segment into three equal parts (ratios 1 : 2 and 2 : 1).

Important Formulas (Chapter 7)

Distance formula: the distance between P(x1, y1) and Q(x2, y2) is √[(x2 − x1)2 + (y2 − y1)2].

Distance from the origin: the distance of P(x, y) from O(0, 0) is √(x2 + y2).

Section formula (internal): the point dividing the join of A(x1, y1) and B(x2, y2) in the ratio m1 : m2 is  [ (m1x2 + m2x1)/(m1 + m2) , (m1y2 + m2y1)/(m1 + m2) ].

Ratio k : 1 form: P = [ (kx2 + x1)/(k + 1) , (ky2 + y1)/(k + 1) ].

Mid-point formula: the mid-point of A(x1, y1) and B(x2, y2) is  [ (x1 + x2)/2 , (y1 + y2)/2 ].

Diagonals of a parallelogram bisect each other — so the mid-points of the two diagonals coincide.

Exercise 7.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given in the book.

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

SOLUTION Use the distance formula d = √[(x2 − x1)2 + (y2 − y1)2]. (i) d = √[(4 − 2)2 + (1 − 3)2] = √(4 + 4) = √8 = 2√2 units. (ii) d = √[(−1 − (−5))2 + (3 − 7)2] = √(42 + (−4)2) = √(16 + 16) = √32 = 4√2 units. (iii) d = √[(−a − a)2 + (−b − b)2] = √(4a2 + 4b2) = 2√(a2 + b2) units.

2. 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 Distance = √[(36 − 0)2 + (15 − 0)2] = √(1296 + 225) = √1521 = 39 units. Town B is 36 km east and 15 km north of town A, so taking A as the origin, B is at (36, 15). Hence the distance between the two towns is 39 km.

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

SOLUTION Let A(1, 5), B(2, 3), C(−2, −11). AB = √[(2 − 1)2 + (3 − 5)2] = √(1 + 4) = √5 ≈ 2.24. BC = √[(−2 − 2)2 + (−11 − 3)2] = √(16 + 196) = √212 ≈ 14.56. AC = √[(−2 − 1)2 + (−11 − 5)2] = √(9 + 256) = √265 ≈ 16.28. Since AB + BC ≈ 16.80 ≠ AC ≈ 16.28, the three points do not lie on a line. ∴ the points are not collinear.

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

SOLUTION Let A(5, −2), B(6, 4), C(7, −2). AB = √[(6 − 5)2 + (4 − (−2))2] = √(1 + 36) = √37. BC = √[(7 − 6)2 + (−2 − 4)2] = √(1 + 36) = √37. CA = √[(5 − 7)2 + (−2 − (−2))2] = √(4 + 0) = 2. Since AB = BC = √37, two sides are equal. ∴ the points are the vertices of an isosceles triangle.

5. 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 From the grid in Fig. 7.8 the four seats are A(3, 4), B(6, 7), C(9, 4) and D(6, 1). AB = √[(6 − 3)2 + (7 − 4)2] = √(9 + 9) = √18 = 3√2. BC = √[(9 − 6)2 + (4 − 7)2] = √(9 + 9) = √18 = 3√2. CD = √[(6 − 9)2 + (1 − 4)2] = √(9 + 9) = √18 = 3√2. DA = √[(3 − 6)2 + (4 − 1)2] = √(9 + 9) = √18 = 3√2. Diagonals: AC = √[(9 − 3)2 + 0] = 6 and BD = √[0 + (1 − 7)2] = 6. All four sides are equal (3√2) and both diagonals are equal (6), so ABCD is a square. ∴ Champa is correct.

6. 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)

SOLUTION (i) Let A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0). AB = √(4 + 4) = 2√2; BC = √(4 + 4) = 2√2; CD = √(4 + 4) = 2√2; DA = √(4 + 4) = 2√2. Diagonals AC = √(0 + 16) = 4 and BD = √(16 + 0) = 4. All sides equal and both diagonals equal → a square. (ii) Let A(−3, 5), B(3, 1), C(0, 3), D(−1, −4). AB = √(36 + 16) = √52 = 2√13; BC = √(9 + 4) = √13; AC = √(9 + 4) = √13. Here AC + CB = √13 + √13 = 2√13 = AB, so A, C, B are collinear. Since three of the points lie on one line, they do not form a quadrilateral. (iii) Let A(4, 5), B(7, 6), C(4, 3), D(1, 2). AB = √(9 + 1) = √10; BC = √(9 + 9) = 3√2; CD = √(9 + 1) = √10; DA = √(9 + 9) = 3√2. Opposite sides are equal (AB = CD, BC = DA). Diagonals AC = √(0 + 4) = 2 and BD = √(36 + 16) = √52 = 2√13 are unequal. Equal opposite sides with unequal diagonals → a parallelogram.

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

SOLUTION A point on the x-axis is of the form P(x, 0). Let it be equidistant from A(2, −5) and B(−2, 9), so PA2 = PB2. (x − 2)2 + (0 + 5)2 = (x + 2)2 + (0 − 9)2. x2 − 4x + 4 + 25 = x2 + 4x + 4 + 81. −4x + 29 = 4x + 85 ⇒ −8x = 56 ⇒ x = −7. ∴ the required point is (−7, 0).

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

SOLUTION PQ = 10 ⇒ PQ2 = 100. (10 − 2)2 + (y − (−3))2 = 100 ⇒ 64 + (y + 3)2 = 100. (y + 3)2 = 36 ⇒ y + 3 = ±6. So y = 6 − 3 = 3 or y = −6 − 3 = −9. ∴ y = 3 or y = −9.

9. 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 QP = QR ⇒ QP2 = QR2. (5 − 0)2 + (−3 − 1)2 = (x − 0)2 + (6 − 1)2 ⇒ 25 + 16 = x2 + 25. 41 = x2 + 25 ⇒ x2 = 16 ⇒ x = 4 or x = −4. When x = 4: QR = √[(4)2 + (6 − 1)2] = √(16 + 25) = √41 units; PR = √[(4 − 5)2 + (6 − (−3))2] = √(1 + 81) = √82 units. When x = −4: QR = √[(−4)2 + 25] = √(16 + 25) = √41 units; PR = √[(−4 − 5)2 + (9)2] = √(81 + 81) = √162 = 9√2 units.

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

SOLUTION Let P(x, y) be equidistant from A(3, 6) and B(−3, 4), so PA2 = PB2. (x − 3)2 + (y − 6)2 = (x + 3)2 + (y − 4)2. x2 − 6x + 9 + y2 − 12y + 36 = x2 + 6x + 9 + y2 − 8y + 16. −6x − 12y + 45 = 6x − 8y + 25 ⇒ −12x − 4y + 20 = 0. Dividing by −4: 3x + y − 5 = 0 is the required relation.

Exercise 7.2 Solutions

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

SOLUTION Section formula with A(−1, 7), B(4, −3), m1 : m2 = 2 : 3. x = (2×4 + 3×(−1)) / (2 + 3) = (8 − 3)/5 = 5/5 = 1. y = (2×(−3) + 3×7) / 5 = (−6 + 21)/5 = 15/5 = 3. ∴ the required 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 A(4, −1), B(−2, −3) and let P, Q trisect AB, so P divides AB in 1 : 2 and Q divides AB in 2 : 1. P: x = (1×(−2) + 2×4)/(1 + 2) = (−2 + 8)/3 = 6/3 = 2; y = (1×(−3) + 2×(−1))/3 = (−3 − 2)/3 = −5/3. So P = (2, −5/3). Q: x = (2×(−2) + 1×4)/3 = (−4 + 4)/3 = 0; y = (2×(−3) + 1×(−1))/3 = (−6 − 1)/3 = −7/3. So Q = (0, −7/3). ∴ the points of trisection are (2, −5/3) and (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 Fig. 7.12. Niharika runs ¼th the distance AD on the 2nd line and posts a green flag. Preet runs ⅕th 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 Take AD along the y-axis with each line as the x-coordinate. There are 100 m along AD. Green flag (Niharika): 2nd line, ¼ of 100 = 25 m up → G(2, 25). Red flag (Preet): 8th line, ⅕ of 100 = 20 m up → R(8, 20). Distance GR = √[(8 − 2)2 + (20 − 25)2] = √(36 + 25) = √61 m. Blue flag is the mid-point of GR = ((2 + 8)/2, (25 + 20)/2) = (5, 22.5), i.e. on the 5th line, 22.5 m along AD.

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

SOLUTION Let (−1, 6) divide A(−3, 10) and B(6, −8) in the ratio k : 1. x-coordinate: (k×6 + 1×(−3))/(k + 1) = −1 ⇒ 6k − 3 = −(k + 1) ⇒ 6k − 3 = −k − 1. 7k = 2 ⇒ k = 2/7, so the ratio is 2 : 7. Check with y: (2×(−8) + 7×10)/(2 + 7) = (−16 + 70)/9 = 54/9 = 6. ✓

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 On the x-axis the y-coordinate is 0. Let the ratio be k : 1. y = (k×5 + 1×(−5))/(k + 1) = 0 ⇒ 5k − 5 = 0 ⇒ k = 1. So the ratio is 1 : 1. Point of division: x = (1×(−4) + 1×1)/(1 + 1) = (−4 + 1)/2 = −3/2; y = 0. ∴ the x-axis divides AB in 1 : 1 at the point (−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.

SOLUTION Let A(1, 2), B(4, y), C(x, 6), D(3, 5). The diagonals AC and BD bisect each other, so their mid-points are equal. Mid-point of AC = ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4). Mid-point of BD = ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2). Equating x-parts: (1 + x)/2 = 7/2 ⇒ 1 + x = 7 ⇒ x = 6. Equating y-parts: 4 = (y + 5)/2 ⇒ y + 5 = 8 ⇒ y = 3.

7. 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 The centre is the mid-point of the diameter AB. Let A = (x, y). ((x + 1)/2, (y + 4)/2) = (2, −3). (x + 1)/2 = 2 ⇒ x + 1 = 4 ⇒ x = 3. (y + 4)/2 = −3 ⇒ y + 4 = −6 ⇒ y = −10. ∴ A = (3, −10).

8. If A and B are (−2, −2) and (2, −4), respectively, find the coordinates of P such that AP = &frac37; AB and P lies on the line segment AB.

SOLUTION AP = (3/7) AB, so PB = AB − AP = (4/7) AB. Hence AP : PB = 3 : 4, and P divides AB in the ratio 3 : 4. x = (3×2 + 4×(−2))/(3 + 4) = (6 − 8)/7 = −2/7. y = (3×(−4) + 4×(−2))/7 = (−12 − 8)/7 = −20/7. ∴ 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.

SOLUTION Let P, Q, R divide AB into four equal parts, so P divides AB in 1 : 3, Q is the mid-point (2 : 2), and R divides AB in 3 : 1. P: x = (1×2 + 3×(−2))/4 = (2 − 6)/4 = −1; y = (1×8 + 3×2)/4 = (8 + 6)/4 = 7/2. So P = (−1, 7/2). Q (mid-point of AB): ((−2 + 2)/2, (2 + 8)/2) = (0, 5). R: x = (3×2 + 1×(−2))/4 = (6 − 2)/4 = 1; y = (3×8 + 1×2)/4 = (24 + 2)/4 = 13/2. So R = (1, 13/2). ∴ the three points are (−1, 7/2), (0, 5) and (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 = ½(product of its diagonals)]

SOLUTION Let A(3, 0), B(4, 5), C(−1, 4), D(−2, −1). The diagonals are AC and BD. d1 = AC = √[(−1 − 3)2 + (4 − 0)2] = √(16 + 16) = √32 = 4√2. d2 = BD = √[(−2 − 4)2 + (−1 − 5)2] = √(36 + 36) = √72 = 6√2. Area = ½ × d1 × d2 = ½ × 4√2 × 6√2 = ½ × 48 = 24 square units.

Common Mistakes to Avoid

Watch out for these

  • Subtracting coordinates inconsistently in the distance formula — always do (x2 − x1) and (y2 − y1); since they are squared, the order does not matter, but mismatching x with y does.
  • Forgetting the ± when you take a square root in “find the value” questions (e.g. y = 3 or −9), and dropping the second answer.
  • Swapping the cross-multiplication in the section formula: it is m1x2 + m2x1, not m1x1 + m2x2.
  • For trisection, using the same ratio for both points — the first point is 1 : 2 and the second is 2 : 1.
  • To prove a square, checking only that the four sides are equal — a rhombus also has four equal sides, so you must also show the diagonals are equal.
  • Mixing up which diagonals bisect in a parallelogram — pair the opposite vertices (A with C, B with D) when the vertices are given “in order”.

Practice MCQs & Assertion–Reason

1. The distance between the points (0, 0) and (36, 15) is:

(a) 21    (b) 39    (c) 51    (d) 41

2. The distance of the point (−6, 8) from the origin is:

(a) 8    (b) 10    (c) 14    (d) 100

3. The mid-point of the segment joining (4, −1) and (−2, −3) is:

(a) (1, −2)    (b) (3, −2)    (c) (1, 2)    (d) (2, −1)

4. The point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3 is:

(a) (2, 4)    (b) (1, 3)    (c) (3, 1)    (d) (0, 5)

5. The point on the x-axis equidistant from (2, −5) and (−2, 9) is:

(a) (7, 0)    (b) (0, −7)    (c) (−7, 0)    (d) (0, 7)

6. The points (5, −2), (6, 4) and (7, −2) are the vertices of:

(a) a scalene triangle    (b) an isosceles triangle    (c) an equilateral triangle    (d) a right triangle

7. The x-axis divides the segment joining A(1, −5) and B(−4, 5) in the ratio:

(a) 1 : 2    (b) 2 : 1    (c) 1 : 1    (d) 3 : 2

8. If the distance between P(2, −3) and Q(10, y) is 10 units, then y equals:

(a) 3 or −9    (b) −3 or 9    (c) 6 or −6    (d) 3 or 9

9. The coordinates of the centre of a circle whose diameter has endpoints (1, 4) and (3, −10) are:

(a) (2, −3)    (b) (2, 3)    (c) (4, −6)    (d) (1, −3)

10. The area of a rhombus with diagonals 4√2 and 6√2 units is:

(a) 12 sq units    (b) 24 sq units    (c) 48 sq units    (d) 96 sq units

Answer key: 1-(b), 2-(b), 3-(a), 4-(b), 5-(c), 6-(b), 7-(c), 8-(a), 9-(a), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The distance of the point (3, 4) from the origin is 5 units.

Reason: The distance of a point P(x, y) from the origin is √(x2 + y2).

A-R 2. Assertion: The mid-point of the segment joining (2, −3) and (1, 4) is (3/2, 1/2).

Reason: The mid-point of A(x1, y1) and B(x2, y2) is ((x1 + x2)/2, (y1 + y2)/2).

A-R 3. Assertion: The points (1, 5), (2, 3) and (−2, −11) are collinear.

Reason: Three points are collinear when the sum of two of their pairwise distances equals the third.

A-R 4. Assertion: The diagonals of a parallelogram bisect each other.

Reason: For (1, 2), (4, 3), (6, 6) and (3, 5) taken in order, the mid-points of the two diagonals are equal.

A-R 5. Assertion: A quadrilateral with all four sides equal must be a square.

Reason: In a square the two diagonals are equal in length.

Answer key: 1-(A), 2-(A), 3-(D), 4-(B), 5-(D).

Quick Revision Summary

  • Distance between P(x1, y1) and Q(x2, y2) = √[(x2 − x1)2 + (y2 − y1)2].
  • Distance of a point from the origin = √(x2 + y2).
  • Section formula (ratio m1 : m2): [ (m1x2 + m2x1)/(m1 + m2) , (m1y2 + m2y1)/(m1 + m2) ].
  • Mid-point = [ (x1 + x2)/2 , (y1 + y2)/2 ]; trisection uses ratios 1 : 2 and 2 : 1.
  • Collinearity: distances satisfy AB + BC = AC (no area / equal sums fail to add up).
  • Shapes: equal sides ⇒ rhombus; equal sides + equal diagonals ⇒ square; equal opposite sides + unequal diagonals ⇒ parallelogram.
  • Diagonals of a parallelogram bisect each other (used to find a missing vertex).

How to score full marks in this chapter

Write the distance or section formula first, then substitute carefully with brackets so signs are not lost. For ratio questions, take the ratio as k : 1, solve from one coordinate and verify with the other. To classify a quadrilateral, list all four side lengths and both diagonals and state the deciding property. Keep surds in simplest form (e.g. √32 = 4√2), and always end with a clear concluding statement so each step earns its mark.

Frequently Asked Questions

What is Class 10 Maths Chapter 7 Coordinate Geometry about?

Chapter 7, Coordinate Geometry, teaches how to find the distance between two points using the distance formula and how to find the coordinates of a point dividing a line segment in a given ratio using the section formula, including the mid-point and points of trisection. These tools are used to test collinearity and to identify triangles and quadrilaterals.

How many exercises are there in Class 10 Maths Chapter 7?

The latest NCERT textbook (2026-27) has two exercises in Chapter 7 – Exercise 7.1 on the distance formula and Exercise 7.2 on the section formula – and both are fully solved step by step on this page.

What is the difference between the distance formula and the section formula?

The distance formula gives the length of the segment joining two points, √[(x2 – x1)^2 + (y2 – y1)^2]. The section formula gives the coordinates of a point that divides that segment in a given ratio m1 : m2; the mid-point formula is its special case for the ratio 1 : 1.

Are these Class 10 Maths Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Class 10 Mathematics textbook for the 2026-27 session, with every answer worked out and verified.

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