Show that the points $(-2, 3)$, $(8, 3)$ and $(6, 7)$ are the vertices of a right-angled triangle.

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Accepted Answer

Step 1: Let $A(-2, 3)$, $B(8, 3)$, $C(6, 7)$. $AB = 10$, $BC = \sqrt{4 + 16} = \sqrt{20}$
Step 2: $AC = \sqrt{64 + 16} = \sqrt{80}$
Step 3: $AB^2 = 100 = 20 + 80 = BC^2 + AC^2$. $	herefore$ the given points are the vertices of a right angled triangle.

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