Question

Show that the points A(5, 2), B(2, − 2) and C(− 2, t) are the vertices of a right triangle with $\angle B={90}^{\circ }$, then find the value of t.

Answer 1

$$\begin{gathered} \because \angle B={90}^{\circ } \\ \therefore A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow {\left(5+2\right)}^2+{\left(2-t\right)}^2={\left(5-2\right)}^2+{\left(2+2\right)}^2+{\left(2+2\right)}^2+{\left(-2-t\right)}^2 \\ \Rightarrow {\left(7\right)}^2+{\left(t-2\right)}^2={\left(3\right)}^2+{\left(4\right)}^2+{\left(4\right)}^2+{\left(t+2\right)}^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow 49+t^2-4t+4=9+16+16+t^2+4t+4 \\ \Rightarrow 8-4t=4t \\ \Rightarrow 8t=8 \\ \Rightarrow t=1 \end{gathered}$$
Hence, t = 1.