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By Ankit Gupta
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Updated on 15 Jul 2026, 16:29 IST
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry help students understand how points, lines, and shapes are studied on a coordinate plane. This chapter connects algebra with geometry and teaches students how to solve practical problems by using coordinates. The solutions are written in simple language and explain every step clearly, so students can learn the method instead of only memorising the final answer.
In Chapter 7, students mainly study the distance formula, section formula, and area of a triangle. The distance formula is used to find the distance between two points. It also helps students check whether three points are collinear and identify shapes such as triangles, squares, rectangles, rhombuses, and parallelograms. The section formula is used to find the coordinates of a point that divides a line segment in a given ratio. Students also learn how to find the midpoint of a line segment. The area formula helps them calculate the area of a triangle when the coordinates of its three vertices are known.
These NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry cover all questions from Exercise 7.1 and Exercise 7.2. Each answer follows the correct formula, substitutes the given values, simplifies the calculation, and presents the final result clearly. This step-by-step method makes difficult questions easier to understand and reduces common calculation mistakes.
The solutions are useful for completing homework, preparing class notes, revising important formulas, and studying for school examinations. They can also help students prepare for the CBSE Class 10 board exam because NCERT questions build the basic concepts required for many exam problems. Students should first try each question on their own and then use the solution to check their method and answer.
Coordinate Geometry becomes easier with regular practice. Students should carefully note the signs of coordinates, especially negative values, and write each calculation in an organised way. They should also remember which formula is suitable for a particular question. By studying these detailed solutions, learners can improve their problem-solving skills, strengthen their understanding of the chapter, and gain confidence in mathematics. This resource provides a clear and reliable way to learn Class 10 Coordinate Geometry and revise the complete chapter effectively.
Download the Class 10 Maths Chapter 7 Coordinate Geometry NCERT Solutions PDF given below for clear and easy-to-understand answers. It contains step-by-step solutions to all the exercise questions, including the distance formula, section formula, midpoint formula, collinear points, and coordinate-based geometry problems. The solutions are written in simple words and presented in a clean format to help students understand the methods, revise the chapter, complete homework, and prepare for exams.
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Question 1: Find the distance between each pair of points.
(i) (2, 3) and (4, 1)
Let A = (2, 3) and B = (4, 1).
AB = √[(4 − 2)2 + (1 − 3)2]
AB = √[(2)2 + (−2)2] = √(4 + 4) = √8
√8 = √(4 × 2) = 2√2
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Answer: The distance is 2√2 units.
(ii) (−5, 7) and (−1, 3)
Let A = (−5, 7) and B = (−1, 3).
AB = √[((−1) − (−5))2 + (3 − 7)2]
AB = √[(4)2 + (−4)2] = √(16 + 16) = √32
√32 = √(16 × 2) = 4√2
Answer: The distance is 4√2 units.
(iii) (a, b) and (−a, −b)
Let A = (a, b) and B = (−a, −b).
AB = √[((−a) − a)2 + ((−b) − b)2]
AB = √[(−2a)2 + (−2b)2]
AB = √(4a2 + 4b2) = 2√(a2 + b2)
Answer: The distance is 2√(a2 + b2) units.
Question 2: Find the distance between (0, 0) and (36, 15). Also find the distance between towns A and B.
The coordinates of the towns are A = (0, 0) and B = (36, 15).
AB = √[(36 − 0)2 + (15 − 0)2]
AB = √(362 + 152) = √(1296 + 225)
AB = √1521 = 39
Therefore, the same calculation gives the distance between the two towns.
Answer: The distance between A and B is 39 km.
Question 3: Check whether (1, 5), (2, 3) and (−2, −11) are collinear.
Let A = (1, 5), B = (2, 3) and C = (−2, −11). Three points are collinear when the area of the triangle formed by them is zero.
Area of triangle = 1/2 × |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|
Area = 1/2 × |1(3 − (−11)) + 2((−11) − 5) + (−2)(5 − 3)|
Area = 1/2 × |14 − 32 − 4| = 1/2 × 22 = 11 square units
The area is not zero, so the three points do not lie on one straight line.
Answer: The points are not collinear.
Question 4: Check whether (5, −2), (6, 4) and (7, −2) form an isosceles triangle.
Let A = (5, −2), B = (6, 4) and C = (7, −2).
| Side | Calculation | Length |
| AB | √[(6 − 5)2 + (4 − (−2))2] | √(1 + 36) = √37 |
| BC | √[(7 − 6)2 + ((−2) − 4)2] | √(1 + 36) = √37 |
| AC | √[(7 − 5)2 + ((−2) − (−2))2] | √4 = 2 |
Since AB = BC, two sides have equal length.
Answer: Yes. The points form an isosceles triangle.
Question 5: The four students are at A(3, 4), B(6, 7), C(9, 4) and D(6, 1). Is ABCD a square?
To check whether ABCD is a square, compare its four sides and two diagonals.
| Line segment | Calculation | Length |
| AB | √[(6 − 3)2 + (7 − 4)2] | √18 = 3√2 |
| BC | √[(9 − 6)2 + (4 − 7)2] | √18 = 3√2 |
| CD | √[(6 − 9)2 + (1 − 4)2] | √18 = 3√2 |
| DA | √[(3 − 6)2 + (4 − 1)2] | √18 = 3√2 |
| AC | √[(9 − 3)2 + (4 − 4)2] | 6 |
| BD | √[(6 − 6)2 + (1 − 7)2] | 6 |
All four sides are equal, and both diagonals are also equal. Therefore, ABCD satisfies the required conditions for a square.
Answer: Champa is correct. ABCD is a square.
Question 6: Name the quadrilateral formed by each set of points, if possible. Give reasons.
(i) A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0)
| Segment | Length |
| AB | √[22 + 22] = 2√2 |
| BC | √[(−2)2 + 22] = 2√2 |
| CD | √[(−2)2 + (−2)2] = 2√2 |
| DA | √[22 + (−2)2] = 2√2 |
| AC | 4 |
| BD | 4 |
All four sides are equal, and the diagonals are equal.
Answer: The points form a square.
(ii) A(−3, 5), B(3, 1), C(0, 3), D(−1, −4)
First check whether A, B and C lie on one straight line.
Slope of AB = (1 − 5)/(3 − (−3)) = −4/6 = −2/3
Slope of AC = (3 − 5)/(0 − (−3)) = −2/3
The slopes are equal, so A, B and C are collinear. Three vertices of a quadrilateral cannot lie on the same straight line.
Answer: No quadrilateral is formed.
(iii) A(4, 5), B(7, 6), C(4, 3), D(1, 2)
| Segment | Length |
| AB | √[(3)2 + (1)2] = √10 |
| BC | √[(−3)2 + (−3)2] = 3√2 |
| CD | √[(−3)2 + (−1)2] = √10 |
| DA | √[(3)2 + (3)2] = 3√2 |
| AC | 2 |
| BD | √[(6)2 + (4)2] = 2√13 |
AB = CD and BC = DA. Thus, both pairs of opposite sides are equal. The diagonals are not equal, so it is not a rectangle or square.
Answer: The points form a parallelogram.
Question 7: Find the point on the x-axis that is equidistant from (2, −5) and (−2, 9).
A point on the x-axis has y-coordinate 0. Let the required point be P = (x, 0).
Because P is equally far from A = (2, −5) and B = (−2, 9), PA = PB.
PA2 = (x − 2)2 + (0 − (−5))2 = (x − 2)2 + 25
PB2 = (x − (−2))2 + (0 − 9)2 = (x + 2)2 + 81
(x − 2)2 + 25 = (x + 2)2 + 81
x2 − 4x + 4 + 25 = x2 + 4x + 4 + 81
−4x + 29 = 4x + 85
−8x = 56, so x = −7
Answer: The required point is (−7, 0).
Question 8: Find the values of y if the distance between P(2, −3) and Q(10, y) is 10 units.
PQ = √[(10 − 2)2 + (y − (−3))2]
10 = √[82 + (y + 3)2]
Squaring both sides:
100 = 64 + (y + 3)2
(y + 3)2 = 36
y + 3 = ±6
y = 3 or y = −9
Answer: y = 3 or y = −9.
Question 9: Q(0, 1) is equidistant from P(5, −3) and R(x, 6). Find x, QR and PR.
Since Q is equally far from P and R, QP = QR.
QP2 = (5 − 0)2 + ((−3) − 1)2 = 25 + 16 = 41
QR2 = (x − 0)2 + (6 − 1)2 = x2 + 25
x2 + 25 = 41
x2 = 16, so x = 4 or x = −4
In both cases, QR = √41 units.
Case 1: R = (4, 6)
PR = √[(4 − 5)2 + (6 − (−3))2] = √(1 + 81) = √82
Case 2: R = (−4, 6)
PR = √[((−4) − 5)2 + (6 − (−3))2] = √(81 + 81) = √162 = 9√2
Answer: x = 4 or −4. In both cases QR = √41 units. PR = √82 units when x = 4, and PR = 9√2 units when x = −4.
Question 10: Find the relation between x and y if (x, y) is equidistant from (3, 6) and (−3, 4).
Let P = (x, y), A = (3, 6) and B = (−3, 4). Since PA = PB, their squared distances are equal.
(x − 3)2 + (y − 6)2 = (x + 3)2 + (y − 4)2
x2 − 6x + 9 + y2 − 12y + 36 = x2 + 6x + 9 + y2 − 8y + 16
Cancel x2, y2 and 9 from both sides.
−6x − 12y + 36 = 6x − 8y + 16
20 = 12x + 4y
5 = 3x + y
Answer: The required relation is 3x + y − 5 = 0.
Question 1: Find the point that divides the line joining (−1, 7) and (4, −3) in the ratio 2 : 3.
Let A = (−1, 7), B = (4, −3), and let P divide AB internally in the ratio 2 : 3.
P = ((2 × 4 + 3 × (−1))/(2 + 3), (2 × (−3) + 3 × 7)/(2 + 3))
P = ((8 − 3)/5, (−6 + 21)/5)
P = (5/5, 15/5) = (1, 3)
Answer: The required point is (1, 3).
Question 2: Find the trisection points of the line segment joining (4, −1) and (−2, −3).
Let A = (4, −1) and B = (−2, −3). The two trisection points divide AB into three equal parts.
First trisection point P
P divides AB in the ratio 1 : 2.
P = ((1 × (−2) + 2 × 4)/3, (1 × (−3) + 2 × (−1))/3)
P = ((−2 + 8)/3, (−3 − 2)/3) = (2, −5/3)
Second trisection point Q
Q divides AB in the ratio 2 : 1.
Q = ((2 × (−2) + 1 × 4)/3, (2 × (−3) + 1 × (−1))/3)
Q = ((−4 + 4)/3, (−6 − 1)/3) = (0, −7/3)
Answer: The trisection points are (2, −5/3) and (0, −7/3).
Question 3: Find the distance between the green and red flags. Also find the position of the blue flag placed halfway between them.
Niharika runs one-fourth of 100 m on the second line.
One-fourth of 100 m = 100/4 = 25 m
Therefore, the green flag is at P = (2, 25).
Preet runs one-fifth of 100 m on the eighth line.
One-fifth of 100 m = 100/5 = 20 m
Therefore, the red flag is at Q = (8, 20).
Distance between the flags
PQ = √[(8 − 2)2 + (20 − 25)2]
PQ = √(62 + (−5)2) = √(36 + 25) = √61 m
Position of the blue flag
The blue flag is placed at the midpoint of P and Q.
Midpoint = ((2 + 8)/2, (25 + 20)/2)
Midpoint = (10/2, 45/2) = (5, 22.5)
Answer: The distance between the green and red flags is √61 m. Rashmi should place the blue flag on the fifth line, 22.5 m from the starting side.
Question 4: Find the ratio in which (−1, 6) divides the line segment joining (−3, 10) and (6, −8).
Let A = (−3, 10), B = (6, −8), and P = (−1, 6). Suppose P divides AB in the ratio k : 1.
Using the x-coordinate in the section formula:
−1 = (6k + 1 × (−3))/(k + 1)
−1 = (6k − 3)/(k + 1)
−k − 1 = 6k − 3
2 = 7k, so k = 2/7
Therefore, k : 1 = 2/7 : 1 = 2 : 7.
Answer: The point divides the line segment in the ratio 2 : 7.
Question 5: Find the ratio in which the x-axis divides the line segment joining A(1, −5) and B(−4, 5). Also find the point of division.
Let P divide AB in the ratio k : 1. Since P lies on the x-axis, its y-coordinate is 0.
0 = (k × 5 + 1 × (−5))/(k + 1)
0 = (5k − 5)/(k + 1)
5k − 5 = 0, so k = 1
Thus, the ratio is 1 : 1, so P is the midpoint of A and B.
P = ((1 + (−4))/2, ((−5) + 5)/2)
P = (−3/2, 0)
Answer: The x-axis divides the line segment in the ratio 1 : 1 at the point (−3/2, 0).
Question 6: If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram in order, find x and y.
Let the vertices be A(1, 2), B(4, y), C(x, 6) and D(3, 5). The diagonals of a parallelogram bisect each other, so the midpoint of AC equals the midpoint of BD.
Midpoint of AC = ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4)
Midpoint of BD = ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2)
Equate the x-coordinates:
(1 + x)/2 = 7/2
1 + x = 7, so x = 6
Equate the y-coordinates:
4 = (y + 5)/2
8 = y + 5, so y = 3
Answer: x = 6 and y = 3.
Question 7: AB is a diameter of a circle. The centre is (2, −3) and B is (1, 4). Find A.
Let A = (x, y). Since the centre of a circle is the midpoint of a diameter, the midpoint of A and B is (2, −3).
((x + 1)/2, (y + 4)/2) = (2, −3)
Equate the x-coordinates:
(x + 1)/2 = 2, so x + 1 = 4 and x = 3
Equate the y-coordinates:
(y + 4)/2 = −3, so y + 4 = −6 and y = −10
Answer: A = (3, −10).
Question 8: A(−2, −2) and B(2, −4) are given. Find P if AP = (3/7)AB and P lies on AB.
AP is 3/7 of the full line segment AB. Therefore, the remaining part PB is 4/7 of AB.
AP : PB = 3 : 4
Use the section formula with m = 3 and n = 4.
P = ((3 × 2 + 4 × (−2))/7, (3 × (−4) + 4 × (−2))/7)
P = ((6 − 8)/7, (−12 − 8)/7)
P = (−2/7, −20/7)
Answer: P = (−2/7, −20/7).
Question 9: Find the three points that divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.
Call the required points P1, P2 and P3.
Point P1
P1 divides AB in the ratio 1 : 3.
P1 = ((1 × 2 + 3 × (−2))/4, (1 × 8 + 3 × 2)/4)
P1 = ((2 − 6)/4, (8 + 6)/4) = (−1, 7/2)
Point P2
P2 is the midpoint, so it divides AB in the ratio 1 : 1.
P2 = (((−2) + 2)/2, (2 + 8)/2) = (0, 5)
Point P3
P3 divides AB in the ratio 3 : 1.
P3 = ((3 × 2 + 1 × (−2))/4, (3 × 8 + 1 × 2)/4)
P3 = ((6 − 2)/4, (24 + 2)/4) = (1, 13/2)
Answer: The three points are (−1, 7/2), (0, 5) and (1, 13/2).
Question 10: Find the area of the rhombus with vertices (3, 0), (4, 5), (−1, 4) and (−2, −1), taken in order.
Let the vertices be A(3, 0), B(4, 5), C(−1, 4) and D(−2, −1). The diagonals are AC and BD.
Length of diagonal AC
AC = √[((−1) − 3)2 + (4 − 0)2]
AC = √[(−4)2 + 42] = √32 = 4√2
Length of diagonal BD
BD = √[((−2) − 4)2 + ((−1) − 5)2]
BD = √[(−6)2 + (−6)2] = √72 = 6√2
Area of the rhombus
Area = 1/2 × AC × BD
Area = 1/2 × 4√2 × 6√2
Area = 1/2 × 24 × 2 = 24
Answer: The area of the rhombus is 24 square units.
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The solutions cover important topics such as the distance formula, section formula, midpoint formula, division of a line segment, area of a triangle, collinear points, and coordinate-based questions on geometrical shapes.
These solutions explain every question in a simple, step-by-step manner. They help students understand how to choose the correct formula, substitute values, complete calculations, and write the final answer properly.
Yes, the solutions include the questions from Exercise 7.1 and Exercise 7.2 of NCERT Class 10 Maths Chapter 7. Students can use them to check answers and revise the complete chapter.
Yes. NCERT questions are important for CBSE board exam preparation. Practising these solutions can help students understand the main concepts, avoid common mistakes, and answer coordinate geometry questions confidently.
Students should mainly learn the distance formula, section formula, midpoint formula, and the formula for finding the area of a triangle using coordinates. They should also understand how these formulas are applied in different questions.
Students should first read the question and try to solve it independently. After completing their attempt, they can compare their method with the given solution. They should note their mistakes, revise the related formula, and practise similar questions again.