Coordinate Geometry: CBSE Class 10 Previous Year Questions

209 different questions from Coordinate Geometry (NCERT Chapter 7) asked in CBSE Class 10 Maths board exams 2022–2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions97 questions 2 marks questions57 questions 3 marks questions44 questions 4 marks questions11 questions

Most asked Coordinate Geometry questions

Q81 markMCQCoordinate Geometry CBSE 2026 · Standard 30/1/1

The mid-point of the line segment joining the points $(5, - 4)$ and $(6, 4)$ lies on :

(A) $x$ -axis
(B) $y$ -axis
(C)origin
(D)neither $x$ -axis nor $y$ -axis

Show answer & solution

Answer: (A)

x

-axis

Mid-point $= (\frac{5 + 6}{2}, \frac{- 4 + 4}{2}) = (\frac{11}{2}, 0)$ .
Its $y$ -coordinate is 0, so it lies on the $x$ -axis.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q171 markMCQCoordinate Geometry CBSE 2026 · Standard 30/3/1

In the given figure, $Δ$ ABC is an equilateral triangle. AD is a median of the triangle joining the points A $(0, \frac{5 3}{2})$ , D(0, 0). Points B and C are (in same order) :

Diagram for CBSE 2026 Class 10 Maths question 17

(A)(– 5, 0), (5, 0)
(B) $(- \frac{5}{2}, 0), (\frac{5}{2}, 0)$
(C)(– 10, 0), (10, 0)
(D) $(- 53, 0), (53, 0)$

Show answer & solution

Answer: (B)

(- \frac{5}{2}, 0), (\frac{5}{2}, 0)

AD $= \frac{5 3}{2}$ is the height of the equilateral triangle
Height $= \frac{3}{2} \times$ side, so side $= 5$
D is the midpoint of BC on the x-axis, so BD = DC $= \frac{5}{2}$
B $(- \frac{5}{2}, 0)$ , C $(\frac{5}{2}, 0)$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q71 markMCQCoordinate Geometry CBSE 2026 · Standard 30/4/1

In the given figure, a circle is centred at (1, 2). The diameter of the circle is

Diagram for CBSE 2026 Class 10 Maths question 7

(A)4
(B) $22$
(C) $5$
(D) $25$

Show answer & solution

Answer: (D)

25

The circle passes through the origin O(0, 0).
Radius $= (1 - 0)^{2} + (2 - 0)^{2} = 5$
Diameter $= 25$

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q21 markMCQCoordinate Geometry CBSE 2026 · Standard 30/5/1

The line segment joining the points $P (- 4, - 2)$ and $Q (10, 4)$ is divided by y-axis in the ratio

(A)2 : 5
(B)1 : 2
(C)2 : 1
(D)5 : 2

Show answer & solution

Answer: (A) 2 : 5

Let the y-axis divide PQ in the ratio $k : 1$ .
The point of division has x-coordinate $\frac{10 k - 4}{k + 1}$ , which is 0 on the y-axis.
$10 k - 4 = 0$ , so $k = \frac{2}{5}$ .
Ratio = 2 : 5

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q232 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/3/1

Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are (9, – 2) and (1, 6) respectively.
(i) Find the co-ordinates of point P.
(ii) Find the length of the side of the square.

Diagram for CBSE 2026 Class 10 Maths question 23

Show answer & solution

Answer: (i) P(5, 2) (ii) 8 units

(i) Diagonals of a square bisect each other, so P is the midpoint of BD
P $= (\frac{9 + 1}{2}, \frac{- 2 + 6}{2}) = (5, 2)$
(ii) BD $= (9 - 1)^{2} + (- 2 - 6)^{2} = 128 = 82$
Diagonal $= 2 \times$ side, so side $= \frac{8 2}{2} = 8$ units

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q23 (OR) (OR)2 marksVery Short AnswerCoordinate Geometry CBSE 2026 · Standard 30/3/1

Find the coordinates of a point on the line $x + y = 5$ which is equidistant from (6, 4) and (5, 2).

Show answer & solution

Answer:

(- \frac{3}{2}, \frac{13}{2})

Let the point be $(x, y)$
$(x - 6)^{2} + (y - 4)^{2} = (x - 5)^{2} + (y - 2)^{2}$
$- 12 x + 36 - 8 y + 16 = - 10 x + 25 - 4 y + 4$ , so $2 x + 4 y = 23$
With $y = 5 - x$ : $2 x + 20 - 4 x = 23$ , so $x = - \frac{3}{2}$ , $y = \frac{13}{2}$
Point: $(- \frac{3}{2}, \frac{13}{2})$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q313 marksShort AnswerCoordinate Geometry CBSE 2026 · Basic 430/4/1

The vertices of a rhombus ABCD are A(– 3, – 4), B(5, – 3), C(1, 4) and D(– 7, 3). Find the length of both the diagonals. Hence, find area of the rhombus ABCD.

Show answer & solution

Answer: AC =

45

units, BD =

65

units, area = 60 sq. units

AC = $(1 + 3)^{2} + (4 + 4)^{2} = 16 + 64 = 80 = 45$ units.
BD = $(- 7 - 5)^{2} + (3 + 3)^{2} = 144 + 36 = 180 = 65$ units.
Area = $\frac{1}{2} \times A C \times B D = \frac{1}{2} \times 45 \times 65 = 60$ sq. units.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q31 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2026 · Basic 430/4/1

The line segment joining the points A (– 5, 1) and B (7, 6) is trisected at the points P and Q such that P is nearer to A. If P lies on the line $x$ + y = k, then find the value of k.

Show answer & solution

Answer: k =

\frac{5}{3}

P divides AB in the ratio 1 : 2.
P = $(\frac{1 ( 7 ) + 2 ( - 5 )}{3}, \frac{1 ( 6 ) + 2 ( 1 )}{3}) = (- 1, \frac{8}{3})$ .
P lies on $x + y = k$ : $k = - 1 + \frac{8}{3} = \frac{5}{3}$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q293 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/5/1

A circle centered at $(2, 1)$ passes through the points $A (5, 6)$ and $B (- 3, K)$ . Find the value(s) of K. Hence find length of chord AB.

Show answer & solution

Answer: K = 4 or K = −2; AB =

217

units (K = 4) or

82

units (K = −2)

Let centre $O (2, 1)$ . $O A^{2} = (5 - 2)^{2} + (6 - 1)^{2} = 9 + 25 = 34$
$O B^{2} = O A^{2}$ : $(- 3 - 2)^{2} + (K - 1)^{2} = 34$
$25 + (K - 1)^{2} = 34$ , so $(K - 1)^{2} = 9$ and $K - 1 = \pm 3$
$K = 4$ or $K = - 2$
For $K = 4$ : $A B = (5 + 3)^{2} + (6 - 4)^{2} = 68 = 217$
For $K = - 2$ : $A B = (5 + 3)^{2} + (6 + 2)^{2} = 128 = 82$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q29 (OR) (OR)3 marksShort AnswerCoordinate Geometry CBSE 2026 · Standard 30/5/1

Prove that the point P dividing the line segment joining the points $A (- 1, 7)$ and $B (4, - 3)$ in the ratio 3 : 2, lies on the line $x - 3 y = - 1$ . Also find length of PA and PB.

Show answer & solution

Answer: P(2, 1) lies on the line; PA =

35

units, PB =

25

units

$P = (\frac{3 ( 4 ) + 2 ( - 1 )}{3 + 2}, \frac{3 ( - 3 ) + 2 ( 7 )}{3 + 2}) = (2, 1)$
Check: $x - 3 y = 2 - 3 = - 1$ . So P lies on the line $x - 3 y = - 1$ .
$P A = (2 + 1)^{2} + (1 - 7)^{2} = 9 + 36 = 35$
$P B = (4 - 2)^{2} + (- 3 - 1)^{2} = 4 + 16 = 25$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

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