Question

Show that the following points form a right angled triangle.

(2, -3), (-6, -7) and (-8, -3)

Answer 1

To form Right Angled $△$

$A{B}^2+B{C}^2=A{C}^2$

or Sum of squares of any two side is Third side.

also $m_{AB}\times m_{BC}=-1$ ($1^{st}$ condition)

$A(2,-3)B(-6,-7)C(-8,-3)$

$AB=\begin{array}{c}\sqrt{(-6-2{)}^2+(-7+3{)}^2}\\ =\sqrt{80}\end{array}|\begin{array}{c}{m}_{AB}=\frac{-7+3}{-8}\\ =\frac{1}{2}\end{array}$

$BC=\begin{array}{c}\sqrt{(-8-(-6{)}^2+4^2}=\sqrt{20}\\ \end{array}|\begin{array}{c}{m}_{BC}=\frac{-3+7}{-2}\\ =-2\end{array}$

$AC=\sqrt{(-8-2{)}^2+0}=10$

also

$A{B}^2+B{C}^2=A{C}^2$ & $m_{AB}\times m_{BC}=-1$

Hence It is Right angled $△$