Question

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

Answer 1

Let, $A(4,4),B(3,5),C(-1,-1)$.

Lets find slopes of lines $AB,AC,BC$

We have the slope point form, $slope,m=\frac{y_2-y_1}{x_2-x_1}$ for tow points $(x_1,y_1),(x_2,y_2)$

$\therefore SlopeofAB=\frac{5-4}{3-4}=-1$

$SlopeofAC=\frac{-1-4}{-1-4}=1$

$SlopeofBC=\frac{-1-5}{-1-3}=\frac{3}{2}$

We know that, if the product of the slopes of two lines is equal to $-1$, then the lines are perpendicular to each other.

Here, $SlopeofAB\times slopeofAC=-1$

$\therefore AB$ is perpendicular to $AC$ and $\angle A={90}^{\circ }$

Hence the given triangle is right angled