Question

Prove that the points $(3,0),(6,4)$ and $(-1,3)$ are the vertices of a right angled isosceles triangle.

Answer 1

Let $A(3,0),B(6,4)$ and $C(-1,3)$ be the vertices of a triangle ABC.




Length of $AB=\sqrt{(6-3{)}^2+(4-0{)}^2}$

=$\sqrt{(3{)}^2+(4{)}^2}$

= $\sqrt{9+16}=\sqrt{25}=5$ units

Length of $BC=\sqrt{(-1-6{)}^2+(3-4{)}^2}$

= $\sqrt{(-7{)}^2+(-1{)}^2}$

= $\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$ units.

And Length of $AC=\sqrt{(-1-3{)}^2+(3-0{)}^2}$

= $\sqrt{(-4{)}^2+(3{)}^2}$

= $\sqrt{16+9}=\sqrt{25}=5$ units

$\therefore AB=AC$

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

Hence, $\Delta$ ABC is a isosceles, right angled triangle.

Hence Proved.