CBSE Class 10 · Chapter 7 · Coordinate Geometry · 2026-27

Coordinate Geometry Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceDistance formula

How far apart are the points $(1, - 2)$ and $(6, 10)$ ?

Show answer

Answer: (a)

13

Why: d = √(5² + 12²) = √169.

$d = \sqrt{(6 - 1)^{2} + (10 + 2)^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$ . (Adding the differences, $5 + 12 = 17$ , is a common slip.)

Q2

·1 mark·Multiple choiceDistance formula

The distance of the point $(- 8, 15)$ from the origin is

Show answer

Answer: (c)

17

Why: √(64 + 225) = √289.

$\sqrt{(- 8)^{2} + 15^{2}} = \sqrt{64 + 225} = \sqrt{289} = 17$ .

Q3

·1 mark·Multiple choiceMid-point formula

The mid-point of the segment joining $(- 3, 8)$ and $(7, - 2)$ is

Show answer

Answer: (c)

(2, 3)

Why: Average the coordinates: ((−3 + 7)/2, (8 − 2)/2).

$(\frac{- 3 + 7}{2}, \frac{8 - 2}{2}) = (2, 3)$ .

Q4

·1 mark·Multiple choiceSection formula

The point dividing the segment joining $(1, 4)$ and $(10, - 2)$ internally in the ratio $2 : 1$ is

Show answer

Answer: (b)

(7, 0)

Why: Section formula with m : n = 2 : 1: ((2·10 + 1·1)/3, (2·(−2) + 1·4)/3).

$(\frac{2 (10) + 1 (1)}{3}, \frac{2 (- 2) + 1 (4)}{3}) = (7, 0)$ .

Q5

·1 mark·Multiple choiceDistance formula

The point on the $y$ -axis that is equidistant from $(2, 3)$ and $(- 4, 1)$ is

Show answer

Answer: (d)

(0, - 1)

Why: Take (0, y): 4 + (y − 3)² = 16 + (y − 1)² gives y = −1.

$2^{2} + (y - 3)^{2} = 4^{2} + (y - 1)^{2} \Rightarrow 13 - 6 y = 17 - 2 y \Rightarrow y = - 1$ . The point is $(0, - 1)$ .

Q6

·1 mark·Multiple choiceSection formula

The $x$ -axis divides the segment joining $(3, - 6)$ and $(7, 4)$ in the ratio

Show answer

Answer: (a)

3 : 2

Why: On the x-axis y = 0: (4k − 6)/(k + 1) = 0 gives k = 3/2.

Let the ratio be $k : 1$ . $\frac{4 k - 6}{k + 1} = 0 \Rightarrow k = \frac{3}{2}$ , i.e. $3 : 2$ .

Q7

·1 mark·Multiple choiceMid-point formula

If $(3, b)$ is the mid-point of the segment joining $(a, 5)$ and $(4, - 1)$ , then $a + b$ equals

Show answer

Answer: (d)

4

Why: (a + 4)/2 = 3 gives a = 2; b = (5 − 1)/2 = 2.

$\frac{a + 4}{2} = 3 \Rightarrow a = 2$ ; $b = \frac{5 + (- 1)}{2} = 2$ . So $a + b = 4$ .

Q8

·1 mark·Multiple choiceDistance formula

The points $A (1, 1)$ , $B (4, 5)$ and $C (7, 1)$ are the vertices of

Show answer

Answer: (b) an isosceles triangle that is not equilateral

Why: AB = BC = 5 but AC = 6.

$A B = \sqrt{9 + 16} = 5$ , $B C = \sqrt{9 + 16} = 5$ , $A C = 6$ . Two equal sides, not three: isosceles.

Q9

·1 mark·Multiple choiceDistance formula

If the distance between $(p, 3)$ and $(5, - 1)$ is $5$ units, then $p$ is

Show answer

Answer: (b)

2

8

Why: (p − 5)² + 16 = 25 gives p − 5 = ±3.

$(p - 5)^{2} + 4^{2} = 25 \Rightarrow (p - 5)^{2} = 9 \Rightarrow p = 8$ or $p = 2$ .

Q10

·1 mark·Multiple choiceMid-point formula

The centre of a circle is $(2, - 3)$ and one end of a diameter is $(- 1, 1)$ . The other end of that diameter is

Show answer

Answer: (c)

(5, - 7)

Why: The centre is the mid-point of the diameter: other end = 2 × centre − known end.

$(x, y) = (2 \times 2 - (- 1), 2 \times (- 3) - 1) = (5, - 7)$ .

Q11

·1 mark·Multiple choiceDistance formula

The distance of the point $(6, - 9)$ from the $x$ -axis is

Show answer

Answer: (d)

9

Why: Distance from the x-axis is |y|.

The foot of the perpendicular is $(6, 0)$ , so the distance is $| - 9 | = 9$ .

Q12

·1 mark·Multiple choiceMid-point formula

For the points $A (- 1, - 2)$ , $B (1, 0)$ and $C (3, 2)$ , which statement is true?

Show answer

Answer: (a)

B

is the mid-point of

A C

Why: AB = BC = 2√2 and AC = 4√2 = AB + BC, and ((−1 + 3)/2, (−2 + 2)/2) = (1, 0).

Mid-point of $A C$ $= (\frac{- 1 + 3}{2}, \frac{- 2 + 2}{2}) = (1, 0) = B$ . (Also $A B + B C = 2 \sqrt{2} + 2 \sqrt{2} = 4 \sqrt{2} = A C$ , so the points are collinear: no triangle.)

Q13

·1 mark·Multiple choiceSection formula

$P$ lies on the segment joining $A (- 2, 5)$ and $B (4, - 1)$ with $\frac{A P}{P B} = \frac{1}{2}$ . The coordinates of $P$ are

Show answer

Answer: (d)

(0, 3)

Why: Section formula with m : n = 1 : 2.

$P = (\frac{1 (4) + 2 (- 2)}{3}, \frac{1 (- 1) + 2 (5)}{3}) = (0, 3)$ .

Q14

·1 mark·Multiple choiceDistance formula

A right-angled triangle has its corners at the origin, at $(5, 0)$ and at $(0, 12)$ . How long is its boundary?

Show answer

Answer: (c)

30

Why: Sides 5, 12 and √(25 + 144) = 13.

Sides: $5$ , $12$ and $\sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13$ . Perimeter $= 30$ .

Q15

·1 mark·Multiple choiceSection formula

The points $P$ and $Q$ trisect the segment joining $A (2, 3)$ and $B (6, 7)$ , with $P$ nearer to $A$ . Then $P$ is

Show answer

Answer: (b)

(\frac{10}{3}, \frac{13}{3})

Why: P divides AB in the ratio 1 : 2.

$P = (\frac{1 (6) + 2 (2)}{3}, \frac{1 (7) + 2 (3)}{3}) = (\frac{10}{3}, \frac{13}{3})$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Drone deliveries (4 marks)

A delivery company plots its service area on a grid in which $1$ unit is $1$ km. Its drone warehouse is at $W (- 2, 1)$ , and two customers live at $H_{1} (6, 7)$ and $H_{2} (4, - 7)$ . Drones fly in straight lines.

(i) How far does a drone fly from the warehouse to

H_{1}

? [1 mark]

Show answer

Answer:

10

W H_{1} = \sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10

km. A1

(ii) A charging pad is placed at the mid-point of

H_{1} H_{2}

. Find its coordinates. [1 mark]

Show answer

Answer:

(5, 0)

(\frac{6 + 4}{2}, \frac{7 - 7}{2}) = (5, 0)

. A1

(iii) A relay mast

R

stands on

W H_{1}

with

W R : R H_{1} = 1 : 3

. Find the coordinates of

R

. [2 marks]

Show answer

Answer:

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

R = (\frac{1 (6) + 3 (- 2)}{4}, \frac{1 (7) + 3 (1)}{4})

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

. A1

OR Show that the warehouse is equally far from both customers. [2 marks]

Show answer

Answer:

W H_{1} = W H_{2} = 10

W H_{2} = \sqrt{(4 + 2)^{2} + (- 7 - 1)^{2}} = \sqrt{36 + 64}

= 10 = W H_{1}

. A1

Case study 2: Park fountain (4 marks)

A rectangular park is drawn on a grid in which $1$ unit is $10$ m. Its corners are $P (1, 1)$ , $Q (9, 1)$ , $R (9, 7)$ and $S (1, 7)$ . A fountain is to be built where the diagonals $P R$ and $Q S$ cross.

(i) Find the length of the diagonal

P R

in grid units. [1 mark]

Show answer

Answer:

10

units

P R = \sqrt{8^{2} + 6^{2}} = 10

units (i.e.

100

m). A1

(ii) Find the coordinates of the fountain. [1 mark]

Show answer

Answer:

(5, 4)

The diagonals bisect each other: mid-point of

P R = (\frac{1 + 9}{2}, \frac{1 + 7}{2}) = (5, 4)

. A1

(iii) A lamp

L

P R

divides it in the ratio

P L : L R = 3 : 1

. Find the coordinates of

L

. [2 marks]

Show answer

Answer:

(7, \frac{11}{2})

L = (\frac{3 (9) + 1 (1)}{4}, \frac{3 (7) + 1 (1)}{4})

= (7, \frac{11}{2})

. A1

OR A bench is to be placed on side

Q R

at the point equidistant from

P

and

S

. Find its coordinates. [2 marks]

Show answer

Answer:

(9, 4)

Take

(9, y)

64 + (y - 1)^{2} = 64 + (y - 7)^{2}

\Rightarrow y = 4

; the bench is at

(9, 4)

. A1

Next steps for Coordinate Geometry

This free set is separate from the chapter's question bank. On the Coordinate Geometry chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 40 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Coordinate Geometry: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.