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

Let say $A(4,4)$ , $B(3,5)$ , $C(-1,-1)$ be the three points.

Slope of line AB$=m_1=\frac{(5-4)}{(3-4)}=-1$

Slope of BC$=m_2=\frac{(5+1)}{(3+1)}=\frac{6}{4}=\frac{3}{2}$

Slope of AC$=m_3=\frac{(-1-4)}{(-1-4)}=1$

Since, $m_1\times m_3=(-1)\left(1\right)=-1$

$\therefore m_1$ and $m_3$ are perpendicular.

$\Rightarrow$ Angle between AB and AC$={90}^o$

$\therefore$ ABC is right triangle right angled at 'A' $(4,4)$.