Question

Let the vertices of a triangle ABC be (4, 4), (3, 5), $(-1,-1)$, then show that $\Delta ABC$ is right angled.

Answer 1

Let $A\equiv (4,4),B\equiv (3,5),C\equiv (-1,-1)$

$AB=\sqrt{(3-4{)}^2+(5-4{)}^2}=\sqrt{2}$
$BC=\sqrt{(-1-3{)}^2+(-1-5{)}^2}=\sqrt{52}$
$AC=\sqrt{(-1-4{)}^2+(-1-4{)}^2}=\sqrt{50}$
$\therefore A{B}^2+A{C}^2=B{C}^2$
Hence by the Pythagoras Theorem, ABC is a right angled triangle.