MathsCBSE Class 10 · Basic · Board exam 2024, set 430/3/1 · 2 marks · Very short answer

CBSE Class 10 Mathematics Basic 2024: question 21(a)

Set in the CBSE Class 10 board exam of 2024, paper 430/3/1, section B, question 21(a), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q21(a) · 2 marks · Very short answerCoordinate Geometry

If $Q (0, 2)$ is equidistant from $P (5, -3)$ and $R (x, 7)$ , find the value(s) of $x$ .

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/3/1.

Hints

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

Hint 1

Equidistant means the distances QP and QR are equal, so their squares are also equal.

Hint 2

Write $Q P^{2} = (0 - 5)^{2} + (2 + 3)^{2}$ and $Q R^{2} = (0 - x)^{2} + (2 - 7)^{2}$ , then equate them.

Hint 3

Remember that $x^{2} = 25$ has two solutions.

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.

Since $Q$ is equidistant from $P$ and $R$ , we have $Q P^{2} = Q R^{2}$ . By the distance formula:
$(0 - 5)^{2} + (2 + 3)^{2} = (0 - x)^{2} + (2 - 7)^{2}$
1 mark
$25 + 25 = x^{2} + 25 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$
1 mark

Answer: $x = 5$ or $x = - 5$

Where the marks are

Writes $Q P^{2} = Q R^{2}$ correctly using the distance formula (1 mark)
Simplifies to $x^{2} = 25$ and gets $x = \pm 5$ (1 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 paper2024 board paper, set 430/3/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 21(a)

Hints

Step-by-step solution

Revise the chapter

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