Aspire Faculty ID #15296 · Topic: JAMIA MCA 2016 · Just now
JAMIA MCA 2016

The points $A(12,8)$, $B(-2,6)$ and $C(6,0)$ are the vertices of …

Solution

$AB^{2}=14^{2}+2^{2}=200,\ BC^{2}=(-8)^{2}+6^{2}=100,\ CA^{2}=(-6)^{2}+(-8)^{2}=100$. Since $BC=CA$ the triangle is isosceles, and $BC^{2}+CA^{2}=AB^{2}$ ⇒ right-angled at $C$.
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