Question

Prove that the points $(-3,0),(1,-3)$ and $(4,1)$ are the vertices of a right-angled isosceles triangle.

Answer 1

Vertices of the triangle are $A(-3,0),$ $B(1,-3),$ $C(4,1).$

Distance between two points = $\sqrt{(}{x}_2-x_1{)}^2+(y_2-y_1{)}^2$
$AB=\sqrt{(1+3{)}^2+(-3-0{)}^2}=5$
$BC=\sqrt{(4-1{)}^2+(1+3{)}^2}=5$
$AC=\sqrt{(4+3{)}^2+(1-0{)}^2}=5\sqrt{2}$
$AB=BC$
Therefore, $\Delta ABC$ is an isosceles triangle.

$(AB{)}^2+(BC{)}^2=5^2+5^2=50$
and $(AC{)}^2=(5\sqrt{2}{)}^2=50$

$\therefore (AB{)}^2+(BC{)}^2=(AC{)}^2$

So, the triangle satisfies the Pythagoras theorem and hence it is a right angled triangle.