Answer
(a) See the graph.
(b) $(2\sqrt 3,\frac{5\pi}{6})$
(c) $(-2\sqrt 3,-\frac{\pi}{6})$
Work Step by Step
(a) $x\lt0$ and $y\gt0$.
The point lies in Quadrant II.
$r^2=x^2+y^2$
$r^2=(-3)^2+(\sqrt 3)^2=12$
$r=±2\sqrt 3$
(b) $r=2\sqrt 3$
$tan~θ=\frac{y}{x}=\frac{\sqrt 3}{-3}=-\frac{\sqrt 3}{3}$
$θ=\frac{5\pi}{6}$
(c) $r=-2\sqrt 3$
$tan~θ=\frac{y}{x}=\frac{\sqrt 3}{-3}=-\frac{\sqrt 3}{3}$
$θ=\frac{5\pi}{6}-\pi=-\frac{\pi}{6}$
