Prove that the triangle formed by the points representing the complex numbers $\left(10 + 8i\right)$, $\left(-2 + 4i\right)$ and $\left(-11 + 31i\right)$ on the Argand plane is right angled.
Let $A$, $B$ and $C$ represent the complex numbers $\left(10 + 8i\right)$, $\left(-2 + 4i\right)$ and $\left(-11 + 31i\right)$ respectively on the Argand diagram.
Now,
$\begin{aligned}
AB & = \left|\left(10 + 8i\right) - \left(-2 + 4i\right)\right| \\\\
& = \left|12 + 4i\right| \\\\
& = \sqrt{\left(12\right)^2 + \left(4\right)^2} = \sqrt{160} \;\;\; \cdots \; (1)
\end{aligned}$
$\begin{aligned}
BC & = \left|\left(-2 + 4i\right) - \left(-11 + 31i\right)\right| \\\\
& = \left|9 - 27i\right| \\\\
& = \sqrt{\left(9\right)^2 + \left(-27\right)^2} = \sqrt{810} \;\;\; \cdots \; (2)
\end{aligned}$
$\begin{aligned}
CA & = \left|\left(-11 + 31i\right) - \left(10 + 8i\right)\right| \\\\
& = \left|-21 + 23i\right| \\\\
& = \sqrt{\left(-21\right)^2 + \left(23\right)^2} = \sqrt{970} \;\;\; \cdots \; (3)
\end{aligned}$
$\therefore$ $\;$ We have from equations $(1)$, $(2)$ and $(3)$
$CA^2 = AB^2 + BC^2$
$\implies$ $\angle ABC = 90^{\circ}$
i.e. $\;$ $\triangle ABC$ is a right angled triangle.