Question

Prove that the points $A(1,-3),B(-3,0)$ and $C(4,1)$ are the vertices of a right angled Isosceles triangle

Answer 1

$AB=\sqrt{(1-(-3){)}^2+(-3-0{)}^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

$BC=\sqrt{(-3-4{)}^2+(0-1{)}^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$

$AC=\sqrt{(1-4{)}^2+(-3-1{)}^2}=\sqrt{9+16}=\sqrt{25}=5$

$A{B}^2+A{C}^2=5^2+5^2=50=B{C}^2$

By Pythagoras theorem

$\angle A={90}^0$

and $AC=AB$

$\therefore ABC$ is right isosceles $\Delta$