Question

Prove that the points $(0,0),(5,5)$ and $-5,5)$ are the vertices of a right isosceles triangle.

Answer 1

Let the vertices of the right isosceles triangle be A(0,0),B(5,5),C(-5,5)

$AB=\sqrt{25+25}=\sqrt{50}$

$BC=\sqrt{(5+5{)}^2+(5-5{)}^2}=\sqrt{100}=10$

$AC=\sqrt{(-5-0{)}^2+(5-0{)}^2}=\sqrt{50}$

$\therefore AB=AC=\sqrt{50}$

$\therefore △ABC\text{is an isosceles right angled}$

Also, $A{B}^2+A{C}^2=B{C}^2$

$\Rightarrow 50+50=100$

$△ABC$ is an isosceles right-angled triangle.