Answer
(a) See the graph.
(b) $(12,\frac{5\pi}{4})$
(c) $(-12,\frac{\pi}{4})$
Work Step by Step
(a) $x\lt0$ and $y\lt0$.
The point lies in Quadrant III.
$r^2=x^2+y^2$
$r^2=(-6\sqrt 2)^2+(-6\sqrt 2)^2=144$
$r=±12$
(b) $r=12$
$tan~θ=\frac{y}{x}=\frac{-6\sqrt 2}{-6\sqrt 2}=1$
$θ=\frac{5\pi}{4}$
(c) $r=-12$
$tan~θ=\frac{y}{x}=\frac{-6\sqrt 2}{-6\sqrt 2}=1$
$θ=\frac{5\pi}{4}-\pi=\frac{\pi}{4}$
