Question

If A (5, 2), B (2,-2) and C (-2, t) are the vertices of a right angled triangle with the right angle at B, then find the value of t.

Answer 1

Using distance formula, we have
$AB=\sqrt{(2-5{)}^2+(-2-2{)}^2}=\sqrt{9+16}=\mathrm{5.....}\left(1\right)$
$BC=\sqrt{(-2-2{)}^2+(t+2{)}^2}=\sqrt{t^2+4t+20}.....\left(2\right)$
$AC=\sqrt{(-2-5{)}^2+(t-2{)}^2}=\sqrt{t^2-4t+53}.....\left(3\right)$
Now, it is given that $\Delta$ ABC is right angled at B.
Using the Pythagorean theorem, we have
$A{B}^2+B{C}^2=A{C}^2$
$\therefore 25+t^2+4t+20=t^2-4t+53$ [From (1) (2) and (3)]
$\Rightarrow 45+4t=-4t+53$
$\Rightarrow 8t=8$
$\Rightarrow t=1$
Hence, the value of t is 1.