Question
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and C(-1, -1) are the vertices of a right angled triangle.

Answer

Let A (4, 4), B (3, 5) and C(-1, -1) be three vertices of a $\Delta {\rm A}{\rm B}C$.
$\therefore$ Slope of AB $ = \frac{{5 - 4}}{{3 - 4}} = \frac{1}{{ - 1}} = - 1$
$\therefore$ Slope of BC $= \frac{{ - 1 - 5}}{{ - 1 - 3}} = \frac{{ - 6}}{{ - 4}} = \frac{3}{2}$
$\therefore$ Slope of AC $ = \frac{{ - 1 - 4}}{{ - 1 - 4}} = \frac{{ - 5}}{{ - 5}} = 1$
Now slope of AB $\times$ slope of AC = -1$\times$1 = -1
This shows thatAB$\bot$AC. Thus $\Delta {\rm A}{\rm B}C$ is right angled at point A.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free