Show that the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).

$A B=\sqrt{(3-(-5))^{2}+(0-6)^{2}}=\sqrt{(8)^{2}+(-6)^{2}}=\sqrt{64+36}=\sqrt{100}=10$ units

$B C=\sqrt{(9-3)^{2}+(8-0)^{2}}=\sqrt{(6)^{2}+(8)^{2}}=\sqrt{36+64}=\sqrt{100}=10$ units

$A C=\sqrt{(9-(-5))^{2}+(8-6)^{2}}=\sqrt{(14)^{2}+(2)^{2}}=\sqrt{196+4}=\sqrt{200}=10 \sqrt{2}$ units

Therefore, $A B=B C=10$ units

Also, $(A B)^{2}+(B C)^{2}=(10)^{2}+(10)^{2}=200$

and $(A C)^{2}=(10 \sqrt{2})^{2}=200$

Thus, $(A B)^{2}+(B C)^{2}=(A C)^{2}$

This show that $\Delta A B C$ is right- angled at $B$.

Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle

Also, a rea of a triangle $=\frac{1}{2} \times$ base $\times$ height

If $A B$ is the height and $B C$ is the base,

Area $=\frac{1}{2} \times 10 \times 10$

$=50$ square units