If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$, find the values of $x$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:29 · grounding rag
Model Answer
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
---
Explanation
- 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.