Show that the points A(5, 2), B(2, − 2) and C(− 2, t) are the vertices of a right triangle with

$\because \angle B=90^{\circ}$

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

$\Rightarrow(5+2)^{2}+(2-t)^{2}=(5-2)^{2}+(2+2)^{2}+(2+2)^{2}+(-2-t)^{2}$

$\Rightarrow(7)^{2}+(t-2)^{2}=(3)^{2}+(4)^{2}+(4)^{2}+(t+2)^{2}$

$\Rightarrow 49+t^{2}-4 t+4=9+16+16+t^{2}+4 t+4$

$\Rightarrow 8-4 t=4 t$

$\Rightarrow 8 t=8$

$\Rightarrow t=1$

Hence, t = 1.