Mathematics — Standard (041) — CBSE Class 10 board question

Given: Q(0, 1) is equidistant from P(5, –3) and R(x, 6), so QP = QR.

$$QP = \sqrt{(5-0)^2 + (-3-1)^2} = \sqrt{25 + 16} = \sqrt{41}$$

$$QR = \sqrt{(x-0)^2 + (6-1)^2} = \sqrt{x^2 + 25}$$

Setting QP = QR:

$$\sqrt{41} = \sqrt{x^2 + 25}$$

$$41 = x^2 + 25$$

$$x^2 = 16 \implies x = \pm 4$$

Therefore, x = 4 or x = –4.

Source: Exercise 7.1, Q.9, Chapter 7 — Coordinate Geometry

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• The key step is applying the distance formula and equating QP = QR, then squaring both sides to remove surds.
• The question only asks for values of x (the full question also asks for QR and PR, but this extract stops at finding x — answer those parts if your question includes them).
• Remember: squaring gives $x^2 = 16$, yielding two values ±4. Students often miss the negative value — examiners check for both.