MathsCBSE Class 10 · Standard · Board exam 2024, set 30/3/1 · 2 marks · Very short answer

CBSE Class 10 Mathematics Standard 2024: question 23(a)

Set in the CBSE Class 10 board exam of 2024, paper 30/3/1, section B, question 23(a), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q23(a) · 2 marks · Very short answerCoordinate Geometry

Find a relation between $x$ and $y$ such that the point P $(x, y)$ is equidistant from the points A(7, 1) and B(3, 5).

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/3/1.

Hints

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

Hint 1

Equidistant means PA = PB.

Hint 2

Square both distances; the $x^{2}$ and $y^{2}$ terms cancel.

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.

PA = PB, so PA² = PB²: $(x - 7)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2}$ .
1 mark
Expanding: $- 14 x - 2 y + 50 = - 6 x - 10 y + 34$ , so $- 8 x + 8 y + 16 = 0$ . The relation is $x - y - 2 = 0$ , i.e. $x - y = 2$ .
1 mark

Answer: $x - y - 2 = 0$ (i.e. $x - y = 2$ )

Where the marks are

Writes PA² = PB² using the distance formula (1 mark)
Simplifies to $x - y - 2 = 0$ (1 mark)

Revise the chapter

Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
Topic: Applications of the 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 paper2024 board paper, set 30/3/1 (Standard): 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 Standard 2024: question 23(a)

Hints

Step-by-step solution

Revise the chapter

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