Answer
(a) See the graph.
(b) $(2\sqrt 2,\frac{2\pi}{3})$
(c) $(-2\sqrt 2,\frac{5\pi}{3})$
Work Step by Step
(a) $x\lt0$ and $y\gt0$.
The point lies in Quadrant II.
$r^2=x^2+y^2$
$r^2=(-\sqrt 2)^2+(\sqrt 6)^2=8$
$r=±2\sqrt 2$
(b) $r=2\sqrt 2$
$tan~θ=\frac{y}{x}=\frac{\sqrt 6}{-\sqrt 2}=-\sqrt 3$
$θ=\frac{2\pi}{3}$
(c) $r=-2\sqrt 2$
$tan~θ=\frac{y}{x}=\frac{\sqrt 6}{-\sqrt 2}=-\sqrt 3$
$θ=\frac{2\pi}{3}+\pi=\frac{5\pi}{3}$
