P $(x, y)$, Q $(-2, -3)$ and R $(2, 3)$ are the vertices of a right triangle PQR right angled at P. Find the relationship between $x$ and $y$. Hence, find all possible values of $x$ for which $y = 2$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Since ∠P = 90°, PQ ⊥ PR, so PQ² + PR² = QR².
$$QR^2 = (2-(-2))^2 + (3-(-3))^2 = 16 + 36 = 52$$
$$PQ^2 = (x+2)^2 + (y+3)^2$$
$$PR^2 = (x-2)^2 + (y-3)^2$$
Using PQ² + PR² = QR²:
$$(x+2)^2 + (y+3)^2 + (x-2)^2 + (y-3)^2 = 52$$
$$2x^2 + 8 + 2y^2 + 18 = 52$$
$$x^2 + y^2 = 13 \quad \text{...(required relationship)}$$
For y = 2:
$$x^2 + 4 = 13 \implies x^2 = 9 \implies x = \pm 3$$
So the possible values of $x$ are 3 and −3.
Source: Chapter 7, Section 7.2 (Distance Formula)
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Explanation
- The key condition is ∠P = 90°, which means PQ ⊥ PR → by Pythagoras: PQ² + PR² = QR².
- Expand and simplify both sides — the $x^2$, $y^2$ cross terms cancel neatly, leaving $x^2 + y^2 = 13$.
- Examiners award 1 mark for setting up the Pythagoras condition, 1 mark for the correct relationship, and 1 mark for finding both values of $x$ when $y = 2$. Don't forget the ± sign.