Question

Show that $(3,2),(2,-3),(0,0)$ are the vertices of a right angled triangle using distance formula.

Answer 1

Let $A(x_1,y_1)=(3,2),B(x_2,y_2)=(2,-3)$ and $C(x_3,y_3)=(0,0)$

$\therefore$ Distance between two points $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{{(2-3)}^2+{(-3-2)}^2}=\sqrt{1+25}=\sqrt{26}$units

$BC=\sqrt{{(0-2)}^2+{(0+3)}^2}=\sqrt{4+9}=\sqrt{13}$units

$CA=\sqrt{{(0-3)}^2+{(0-2)}^2}=\sqrt{9+4}=\sqrt{13}$units

${\left(\sqrt{13}\right)}^2+{\left(\sqrt{13}\right)}^2={\left(\sqrt{26}\right)}^2$

$\therefore {BC}^2+{CA}^2={AB}^2$

Hence the given points form a right-angled triangle.