MathsCBSE Class 10 · Basic · Board exam 2024, set 430/5/1 · 2 marks · Very short answer

CBSE Class 10 Mathematics Basic 2024: question 22(b)

Set in the CBSE Class 10 board exam of 2024, paper 430/5/1, section B, question 22(b), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q22(b) · 2 marks · Very short answerCoordinate Geometry

Find a point on y-axis which is equidistant from the points A(6, 5) and B( $- 4$ , 3).

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/5/1.

Hints

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

Hint 1

What is the x-coordinate of every point on the y-axis?

Hint 2

Equate the squares of the distances PA and PB; the $y^{2}$ 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.

Any point on the y-axis has x-coordinate 0; let it be $P (0, y)$ .
½ mark
$P A = P B \Rightarrow P A^{2} = P B^{2}$
$(0 - 6)^{2} + (y - 5)^{2} = (0 + 4)^{2} + (y - 3)^{2}$
1 mark
$36 + y^{2} - 10 y + 25 = 16 + y^{2} - 6 y + 9 \Rightarrow 61 - 10 y = 25 - 6 y \Rightarrow 4 y = 36 \Rightarrow y = 9$
Required point is $(0, 9)$ .
½ mark

Answer: (0, 9)

Where the marks are

Takes the point as $(0, y)$ (½ mark)
Uses distance formula with $P A^{2} = P B^{2}$ (1 mark)
Solves to get $y = 9$ , point $(0, 9)$ (½ mark)

Revise the chapter

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

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 430/5/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 22(b)

Hints

Step-by-step solution

Revise the chapter

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