Question

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

Answer 1

R.E.F image

We can safely assume that AB and BC are the isosceles sides.

By distance formula, $AB=\sqrt{(-1-3{)}^2+(3-0{)}^2}=\sqrt{16+9}=5$

$BC=\sqrt{(3-6{)}^2+(0-4{)}^2}=\sqrt{9+16}=5$

Since AB=BC, the $\Delta ABC$ is isosceles.

For proving $\angle ABC$ is right, we use the relation $m_1\times m_2=-1$

where $m_1$ is slope of line AB and $m_2$ of BC :

$m_1=\frac{3-0}{-1-4}=\frac{3}{-4}$ $m_2=\frac{4-0}{6-3}=\frac{4}{3}$

$m_1\times m_2=\frac{-3}{4}\times \frac{4}{3}=-1$

Hence proved, $\Delta ABC$ is right angled isosceles.