Question

$A(x, y)$, $B(-3, -2)$ and $C(3, 2)$ are the vertices of a right triangle $ABC$, right-angled at $A$. Find the relationship between $x$ and $y$. Hence, find all the possible values of $y$ for which $x = 3$.

Accepted Answer

Since $	riangle ABC$ is right-angled at $A$, by Pythagoras theorem: $$AB^2 + AC^2 = BC^2$$ $$AB^2 = (x + 3)^2 + (y + 2)^2$$ $$AC^2 = (x - 3)^2 + (y - 2)^2$$ $$BC^2 = (3 - (-3))^2 + (2 - (-2))^2 = 6^2 + 4^2 = 52$$ Adding $AB^2 + AC^2$: $$[(x+3)^2 + (x-3)^2] + [(y+2)^2 + (y-2)^2] = 52$$ $$2(x^2 + 9) + 2(y^2 + 4) = 52 \implies 2x^2 + 18 + 2y^2 + 8 = 52$$ $$2x^2 + 2y^2 = 26 \implies x^2 + y^2 = 13$$ When $x = 3$: $$3^2 + y^2 = 13 \implies 9 + y^2 = 13 \implies y^2 = 4 \implies y = \pm 2$$ When $y = 2$, $A(3, 2) = C(3, 2)$, which is not a triangle. Thus $y = -2$.

Right triangle vertices and coordinate relation

Explanation

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