MathsCBSE Class 10 · Basic · Sample paper, session 2026-27 · 2 marks · Very short answer

CBSE Class 10 Mathematics Basic sample paper 2026-27: question 25(B)

From the official CBSE sample question paper for 2026-27, section B, question 25(B), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q25(B) · 2 marks · Very short answerCoordinate Geometry

Find a linear relation between $x$ and $y$ such that P $(x, y)$ is equidistant from the points A (1, 4) and B ( $- 1$ , 2).

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

Equidistant means PA = PB. Use the distance formula.

Hint 2

Square both sides; the $x^{2}$ and $y^{2}$ terms cancel, leaving a linear equation.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

$P A = P B \Rightarrow (x - 1)^{2} + (y - 4)^{2} = (x + 1)^{2} + (y - 2)^{2}$
½ mark
Squaring both sides: $x^{2} - 2 x + 1 + y^{2} - 8 y + 16 = x^{2} + 2 x + 1 + y^{2} - 4 y + 4$ , which simplifies to $4 x + 4 y = 12$
1 mark
$x + y = 3$
½ mark

Answer: $x + y = 3$

Where the marks are

Writes PA = PB using the distance formula (½ mark)
Squares and simplifies correctly (1 mark)
$x + y = 3$ (½ mark)

Revise the chapter

Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
Topic: Distance Formula

Practise this chapterBoard-style questions from Coordinate Geometry, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic sample paper 2026-27: question 25(B)

Hints

Step-by-step solution

Revise the chapter

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