Answer
(a) See the graph.
(b) $(8\sqrt2,\frac{\pi}{4})$
(c) $(-8\sqrt2,\frac{5\pi}{4})$
Work Step by Step
(a) $x\gt0$ and $y\gt0$.
The point lies in Quadrant I.
$r^2=x^2+y^2$
$r^2=8^2+8^2=128$
$r=±8\sqrt 2$
(b) $r=8\sqrt 2$
$tan~θ=\frac{y}{x}=\frac{8}{8}=1$
$θ=\frac{\pi}{4}$
(c) $r=-8\sqrt 2$
$tan~θ=\frac{y}{x}=\frac{8}{8}=1$
$θ=\frac{5\pi}{4}$
