Question

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

Answer 1

Given vertices of the triangle are $A(4,4),B(3,5)$ and $C(-1,-1)$
We know that slope of line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is given by
$m=\frac{y_2-y_1}{x_2-x_1},x_2\ne x_1$
Slope of $AB$ i.e.$m_1=\frac{5-4}{3-4}=-1$
Slope of $BC$ i.e.$m_2=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$
Slope of $CA$ i.e. $m_3=\frac{4+1}{4+1}=\frac{5}{5}=1$
Clearly, $m_1{m}_3=-1$
$\Rightarrow$ line segments $AB$ and $CA$ are perpendicular to each other i.e; the given triangle is right angled at $A(4,4)$.
Thus the points $(4,4),(3,5)$ and $(1,1)$ are the vertices of a right angled triangle.