Question

If the triangle formed by $A(8,-10),B(7,-3)$ and $C(0,p)$ is right angled at $B$,find the value of $p$.

Answer 1

Let $A(x_1,y_1)=(8,-10),B(x_2,y_2)=(7,-3)$ and $C(x_3,y_3)=(0,p)$ be the vertices of a right angled triangle.

$\therefore$ Distance between two points $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{{(7-8)}^2+{(-3+10)}^2}=\sqrt{1+49}=\sqrt{50}$units

$BC=\sqrt{{(0-7)}^2+{(p+3)}^2}=\sqrt{49+p^2+6p+9}=\sqrt{p^2+6p+58}$units

$CA=\sqrt{{(0-8)}^2+{(p+10)}^2}=\sqrt{64+p^2+20p+100}=\sqrt{p^2+20p+164}$units

The triangle is right angled at $B$

$\therefore {AB}^2+{BC}^2={AC}^2$

$\Rightarrow {\left(\sqrt{50}\right)}^2+{\left(\sqrt{p^2+6p+58}\right)}^2={\left(\sqrt{p^2+20p+164}\right)}^2$

$\Rightarrow p^2+6p+108=p^2+20p+164$

$\Rightarrow 108+6p=20p+164$

$\Rightarrow 108-164=20p-6p$

$\Rightarrow -56=14p$

$\therefore p=\frac{-56}{14}=-4$