Question

Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle.

Answer 1

We have, A (0, 4), B (1, 2) and C (3, 3)
Now,
$m_1=\mathrm{Slope}\mathrm{of}AB=\frac{2-4}{1-0}=-2$

$m_2=\mathrm{Slope}\mathrm{of}BC=\frac{3-2}{3-1}=\frac{1}{2}$

$m_3=\mathrm{Slope}\mathrm{of}CA=\frac{4-3}{0-3}=-\frac{1}{3}$

$\therefore m_1{m}_2=-2\times \frac{1}{2}=-1$

Therefore, AB is perpendicular to BC, i.e. $\angle ABC={90}^{\circ }$.

Thus, the given points are the vertices of a right angled triangle.