Coordinate Geometry is the study of geometric figures using coordinates, axes and algebraic formulas. NCERT Solutions Class 10 Maths Chapter 7 connect this chapter with distance formula, section formula, midpoint formula and coordinate-based figure verification.
Chapter 7 Coordinate Geometry helps students use algebra to study points, line segments and figures on the coordinate plane. It begins with the x-axis, y-axis, abscissa and ordinate, then explains how to find the distance between two points and how to locate a point that divides a line segment in a given ratio.
NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry
Q.
Find the distance between following pairs of points:
(i) (2, 3), (4, 1)
(ii) (–5, 7), (–1, 3)
(iii) (a, b), (–a, –b)
Q.
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Q.
Find the point on y-axis which is equidistant form the points (5, –2) and (–3, 2).
[CBSE - 2019]
Q.
Find the are of triangle PQR formed by the points P(–5, 7), Q(–4, –5) and R(4, 5).
[CBSE - 2020]
Q.
The coordinates of the centre of a circle are (2a, a – 7). Find the value(s) of 'a' if the circle passes through the point (11, –9) and has diameter
102 units.
[CBSE - 2025]
Q.
If the distance between the points (3, –5) and (x, –5) is 15 units, then the value of x are:
[CBSE - 2024]
Q.
The mid-point of the line segment joining the points P(–4, 5) and Q(4, 6) lies on :
[CBSE - 2025]
Q.
Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).
Q.
Q.
Q.
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.
Q.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).
Q.
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.
Q.
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Q.
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Q.
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)
Q.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. 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.
Q.
Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
Q.
Determine if the points (1, 5), (2, 3) and (– 2, –11) are collinear.
Q.
Find the coordinates of the mid-point of the line segment joining the points P(–4, 5) and Q(4, 6). Justify on which axis this point lies.
NCERT Solutions For Class 10 Maths Chapter 7 cover Exercise 7.1 and Exercise 7.2 in textbook order. Students practise the distance formula through questions on triangles, quadrilaterals and collinearity, then use the section formula and midpoint formula to solve line-segment division questions. A quick formula section is also included for revision of all major coordinate geometry formulas Class 10 students need from this chapter.
Key Takeaways
- Distance Formula: Used to find the distance between two coordinate points.
- Collinearity: Three points are collinear when the sum of two smaller distances equals the largest distance.
- Section Formula: Used to find a point that divides a line segment in a given ratio.
- Midpoint Formula: Used when a line segment is divided in the ratio 1 : 1.
NCERT Solutions For Class 10 Maths Chapter 7 Structure 2026
| Exercise No. |
Main Topic |
Question Count |
| Exercise 7.1 |
Distance formula and coordinate figures |
10 |
| Exercise 7.2 |
Section formula and midpoint formula |
10 |
| Chapter 7 |
Coordinate geometry formulas |
Distance, section and midpoint |
NCERT Class 10 Maths Chapter 7 Exercise 7.1 Solutions
Exercise 7.1 focuses on the distance formula. Students find distances between pairs of points, check collinearity, verify triangles and quadrilaterals, and find points equidistant from two given points.
Q1. Find the distance between the following pairs of points.
Use:
Distance = √[(x2 - x1)² + (y2 - y1)²]
Q1(i). (2, 3), (4, 1)
Distance = √[(4 - 2)² + (1 - 3)²]
Distance = √[2² + (-2)²]
Distance = √(4 + 4)
Distance = √8
Distance = 2√2 units
Answer:
2√2 units
Q1(ii). (-5, 7), (-1, 3)
Distance = √[(-1 + 5)² + (3 - 7)²]
Distance = √[4² + (-4)²]
Distance = √(16 + 16)
Distance = √32
Distance = 4√2 units
Answer:
4√2 units
Q1(iii). (a, b), (-a, -b)
Distance = √[(-a - a)² + (-b - b)²]
Distance = √[(-2a)² + (-2b)²]
Distance = √(4a² + 4b²)
Distance = 2√(a² + b²)
Answer:
2√(a² + b²)
Q2. 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?
Distance = √[(36 - 0)² + (15 - 0)²]
Distance = √(36² + 15²)
Distance = √(1296 + 225)
Distance = √1521
Distance = 39
Answer:
The distance between the points is 39 units.
The distance between towns A and B is 39 km.
Q3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Let:
A = (1, 5)
B = (2, 3)
C = (-2, -11)
Find AB:
AB = √[(2 - 1)² + (3 - 5)²]
AB = √(1 + 4)
AB = √5
Find BC:
BC = √[(-2 - 2)² + (-11 - 3)²]
BC = √[(-4)² + (-14)²]
BC = √(16 + 196)
BC = √212
Find AC:
AC = √[(-2 - 1)² + (-11 - 5)²]
AC = √[(-3)² + (-16)²]
AC = √(9 + 256)
AC = √265
Since the sum of two distances does not equal the third distance, the points are not collinear.
Answer:
The given points are not collinear.
Q4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Let:
A = (5, -2)
B = (6, 4)
C = (7, -2)
Find AB:
AB = √[(6 - 5)² + (4 + 2)²]
AB = √(1² + 6²)
AB = √37
Find BC:
BC = √[(7 - 6)² + (-2 - 4)²]
BC = √(1² + (-6)²)
BC = √37
Find AC:
AC = √[(7 - 5)² + (-2 + 2)²]
AC = √(2² + 0²)
AC = 2
Since:
AB = BC
Answer:
Yes, the given points are the vertices of an isosceles triangle.
Q5. In a classroom, four friends are seated at A, B, C and D as shown in Fig. 7.8. Champa says ABCD is a square, while Chameli disagrees. Using distance formula, find who is correct.
From Fig. 7.8, take the points as:
A = (3, 4)
B = (6, 7)
C = (9, 4)
D = (6, 1)
Find sides:
AB = √[(6 - 3)² + (7 - 4)²]
AB = √(9 + 9)
AB = 3√2
BC = √[(9 - 6)² + (4 - 7)²]
BC = √(9 + 9)
BC = 3√2
CD = √[(6 - 9)² + (1 - 4)²]
CD = √(9 + 9)
CD = 3√2
DA = √[(3 - 6)² + (4 - 1)²]
DA = √(9 + 9)
DA = 3√2
Find diagonals:
AC = √[(9 - 3)² + (4 - 4)²]
AC = √36
AC = 6
BD = √[(6 - 6)² + (1 - 7)²]
BD = √36
BD = 6
Since all four sides are equal and both diagonals are equal, ABCD is a square.
Answer:
Champa is correct.
Class 10 Coordinate Geometry Solutions for Quadrilaterals
Distance formula helps identify quadrilaterals by checking side lengths, diagonals and collinearity.
Q6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
Q6(i). (-1, -2), (1, 0), (-1, 2), (-3, 0)
Let the points be A, B, C and D in order.
AB = √[(1 + 1)² + (0 + 2)²]
AB = √(4 + 4)
AB = 2√2
BC = √[(-1 - 1)² + (2 - 0)²]
BC = √(4 + 4)
BC = 2√2
CD = √[(-3 + 1)² + (0 - 2)²]
CD = √(4 + 4)
CD = 2√2
DA = √[(-1 + 3)² + (-2 - 0)²]
DA = √(4 + 4)
DA = 2√2
Diagonals:
AC = √[(-1 + 1)² + (2 + 2)²]
AC = 4
BD = √[(-3 - 1)² + (0 - 0)²]
BD = 4
All sides are equal and diagonals are equal.
Answer:
The quadrilateral is a square.
Q6(ii). (-3, 5), (3, 1), (0, 3), (-1, -4)
Let the points be A, B, C and D.
Check A, B and C:
Slope of AB = (1 - 5)/(3 + 3)
Slope of AB = -4/6
Slope of AB = -2/3
Slope of BC = (3 - 1)/(0 - 3)
Slope of BC = 2/(-3)
Slope of BC = -2/3
So, A, B and C are collinear.
Answer:
The given points do not form a quadrilateral.
Q6(iii). (4, 5), (7, 6), (4, 3), (1, 2)
Let the points be A, B, C and D.
AB = √[(7 - 4)² + (6 - 5)²]
AB = √(9 + 1)
AB = √10
BC = √[(4 - 7)² + (3 - 6)²]
BC = √(9 + 9)
BC = √18
CD = √[(1 - 4)² + (2 - 3)²]
CD = √(9 + 1)
CD = √10
DA = √[(4 - 1)² + (5 - 2)²]
DA = √(9 + 9)
DA = √18
Here:
AB = CD
BC = DA
Opposite sides are equal.
Answer:
The quadrilateral is a parallelogram.
Q7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Let the point on the x-axis be:
P = (x, 0)
Given points:
A = (2, -5)
B = (-2, 9)
Since P is equidistant:
PA = PB
So:
PA² = PB²
(x - 2)² + (0 + 5)² = (x + 2)² + (0 - 9)²
(x - 2)² + 25 = (x + 2)² + 81
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
x² - 4x + 29 = x² + 4x + 85
-8x = 56
x = -7
Answer:
The required point is (-7, 0).
Q8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Given:
P = (2, -3)
Q = (10, y)
Distance = 10
Use:
PQ² = 10²
(10 - 2)² + (y + 3)² = 100
8² + (y + 3)² = 100
64 + (y + 3)² = 100
(y + 3)² = 36
y + 3 = ±6
So:
y = 3 or y = -9
Answer:
y = 3 or y = -9
Q9. 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.
Given:
Q = (0, 1)
P = (5, -3)
R = (x, 6)
Since Q is equidistant from P and R:
QP = QR
Find QP²:
QP² = (5 - 0)² + (-3 - 1)²
QP² = 25 + 16
QP² = 41
Find QR²:
QR² = (x - 0)² + (6 - 1)²
QR² = x² + 25
Since QP² = QR²:
41 = x² + 25
x² = 16
x = ±4
Now:
QR = √41
If x = 4:
R = (4, 6)
PR = √[(4 - 5)² + (6 + 3)²]
PR = √(1 + 81)
PR = √82
If x = -4:
R = (-4, 6)
PR = √[(-4 - 5)² + (6 + 3)²]
PR = √(81 + 81)
PR = √162
PR = 9√2
Answer:
x = 4 or x = -4
QR = √41
If x = 4, PR = √82
If x = -4, PR = 9√2
Q10. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
Let:
P = (x, y)
A = (3, 6)
B = (-3, 4)
Since P is equidistant from A and B:
PA = PB
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
3x + y = 5
Answer:
The required relation is:
3x + y = 5
NCERT Class 10 Maths Chapter 7 Exercise 7.2 Solutions
Exercise 7.2 focuses on the section formula and midpoint formula. The section formula gives the coordinates of the point that divides a line segment joining two given points in a given ratio.
Q1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Let:
A = (-1, 7)
B = (4, -3)
Ratio = 2 : 3
Use section formula:
Point = [(m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)]
x = [2(4) + 3(-1)]/(2 + 3)
x = (8 - 3)/5
x = 1
y = [2(-3) + 3(7)]/(2 + 3)
y = (-6 + 21)/5
y = 3
Answer:
The required point is (1, 3).
Q2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Let:
A = (4, -1)
B = (-2, -3)
The points of trisection divide AB in ratios 1 : 2 and 2 : 1.
For ratio 1 : 2:
x = [1(-2) + 2(4)]/3
x = 6/3
x = 2
y = [1(-3) + 2(-1)]/3
y = -5/3
First point:
(2, -5/3)
For ratio 2 : 1:
x = [2(-2) + 1(4)]/3
x = 0
y = [2(-3) + 1(-1)]/3
y = -7/3
Second point:
(0, -7/3)
Answer:
The points of trisection are:
(2, -5/3) and (0, -7/3)
Q3. To conduct Sports Day activities, lines have been drawn with chalk powder at a distance of 1 m each. Find the distance between the green and red flags and the position of the blue flag.
From the question:
Niharika runs 1/4 of AD on the 2nd line.
Preet runs 1/5 of AD on the 8th line.
Since 100 flower pots are placed 1 m apart along AD, take:
AD = 100 m
Green flag position:
G = (2, 25)
Red flag position:
R = (8, 20)
Distance between flags:
GR = √[(8 - 2)² + (20 - 25)²]
GR = √[6² + (-5)²]
GR = √(36 + 25)
GR = √61 m
Rashmi posts the blue flag at the midpoint of G and R.
Midpoint = [(2 + 8)/2, (25 + 20)/2]
Midpoint = (5, 45/2)
Midpoint = (5, 22.5)
Answer:
Distance between the green and red flags = √61 m
Rashmi should post the blue flag at (5, 22.5).
Q4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Let:
A = (-3, 10)
B = (6, -8)
P = (-1, 6)
Let P divide AB in the ratio m : n.
Using x-coordinate:
-1 = [m(6) + n(-3)]/(m + n)
-1(m + n) = 6m - 3n
-m - n = 6m - 3n
7m = 2n
m/n = 2/7
Answer:
The ratio is 2 : 7.
Q5. 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.
Let the point of division be on the x-axis.
So its y-coordinate is 0.
Let the ratio be m : n.
Using y-coordinate:
0 = [m(5) + n(-5)]/(m + n)
5m - 5n = 0
m = n
Ratio = 1 : 1
Since the point divides AB in the ratio 1 : 1, it is the midpoint.
Point = [(1 - 4)/2, (-5 + 5)/2]
Point = (-3/2, 0)
Answer:
The x-axis divides the line segment in the ratio 1 : 1.
Point of division = (-3/2, 0)
Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Let:
A = (1, 2)
B = (4, y)
C = (x, 6)
D = (3, 5)
In a parallelogram, diagonals bisect each other.
Midpoint of AC = midpoint of BD
[(1 + x)/2, (2 + 6)/2] = [(4 + 3)/2, (y + 5)/2]
[(1 + x)/2, 4] = [7/2, (y + 5)/2]
Compare x-coordinates:
(1 + x)/2 = 7/2
1 + x = 7
x = 6
Compare y-coordinates:
4 = (y + 5)/2
8 = y + 5
y = 3
Answer:
x = 6 and y = 3
Q7. 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).
Let:
A = (x, y)
B = (1, 4)
Centre = midpoint of AB = (2, -3)
Using midpoint formula:
[(x + 1)/2, (y + 4)/2] = (2, -3)
Compare x-coordinates:
(x + 1)/2 = 2
x + 1 = 4
x = 3
Compare y-coordinates:
(y + 4)/2 = -3
y + 4 = -6
y = -10
Answer:
A = (3, -10)
Q8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3AB/7 and P lies on the line segment AB.
Given:
AP = 3AB/7
So:
AP : PB = 3 : 4
Let:
A = (-2, -2)
B = (2, -4)
Using section formula with ratio 3 : 4:
x = [3(2) + 4(-2)]/7
x = (6 - 8)/7
x = -2/7
y = [3(-4) + 4(-2)]/7
y = (-12 - 8)/7
y = -20/7
Answer:
P = (-2/7, -20/7)
Q9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
To divide AB into four equal parts, find three points.
Point 1 divides AB in ratio 1 : 3.
Point 2 divides AB in ratio 2 : 2.
Point 3 divides AB in ratio 3 : 1.
For ratio 1 : 3:
x = [1(2) + 3(-2)]/4
x = -1
y = [1(8) + 3(2)]/4
y = 14/4
y = 7/2
Point 1 = (-1, 7/2)
For ratio 2 : 2:
This is the midpoint.
x = (-2 + 2)/2
x = 0
y = (2 + 8)/2
y = 5
Point 2 = (0, 5)
For ratio 3 : 1:
x = [3(2) + 1(-2)]/4
x = 1
y = [3(8) + 1(2)]/4
y = 26/4
y = 13/2
Point 3 = (1, 13/2)
Answer:
The required points are:
(-1, 7/2), (0, 5), and (1, 13/2)
Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
Let:
A = (3, 0)
B = (4, 5)
C = (-1, 4)
D = (-2, -1)
Area of rhombus = 1/2 × product of diagonals
Find diagonal AC:
AC = √[(-1 - 3)² + (4 - 0)²]
AC = √[(-4)² + 4²]
AC = √(16 + 16)
AC = 4√2
Find diagonal BD:
BD = √[(-2 - 4)² + (-1 - 5)²]
BD = √[(-6)² + (-6)²]
BD = √(36 + 36)
BD = 6√2
Area = 1/2 × AC × BD
Area = 1/2 × 4√2 × 6√2
Area = 1/2 × 24 × 2
Area = 24 square units
Answer:
The area of the rhombus is 24 square units.
Coordinate Geometry Formulas Class 10
Coordinate geometry formulas Class 10 students need in Chapter 7 are distance formula, section formula and midpoint formula. These formulas help solve questions on distances, collinearity, line segment division, quadrilaterals and coordinates of unknown points.
Distance Formula Class 10
Distance between P(x1, y1) and Q(x2, y2):
Distance = √[(x2 - x1)² + (y2 - y1)²]
Distance of point P(x, y) from origin:
OP = √(x² + y²)
Section Formula Class 10
If P divides A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2, then:
P = [(m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)]
Midpoint Formula Class 10
If P is the midpoint of A(x1, y1) and B(x2, y2), then:
P = [(x1 + x2)/2, (y1 + y2)/2]
Quick Formula Table for NCERT Solutions For Class 10 Maths Chapter 7
| Concept |
Copy-Friendly Formula |
Used In |
| Distance formula |
√[(x2 - x1)² + (y2 - y1)²] |
Exercise 7.1 |
| Section formula |
[(m1x2 + m2x1)/(m1 + m2), (m1y2 + m2y1)/(m1 + m2)] |
Exercise 7.2 |
| Midpoint formula |
[(x1 + x2)/2, (y1 + y2)/2] |
Exercise 7.2 |
Useful Links for Class 10 Maths NCERT Solutions