Answer
The point in polar coordinates is $\Big(6,\dfrac{11\pi}{6}\Big)$
Work Step by Step
$(3\sqrt{3},-3)$
This point has $x=3\sqrt{3}$ and $y=-3$. The formulas for converting rectangular to polar coordinates are $r^{2}=x^{2}+y^{2}$ and $\tan\theta=\dfrac{y}{x}$
Substitute the known values into the formulas to obtain $r$ and $\theta$:
$r^{2}=x^{2}+y^{2}$
$r^{2}=(3\sqrt{3})^{2}+(-3)^{2}$
$r^{2}=(9)(3)+9$
$r^{2}=27+9$
$r^{2}=36$
$r=\sqrt{36}$
$r=6$
$\tan\theta=\dfrac{y}{x}$
$\tan\theta=\dfrac{-3}{3\sqrt{3}}$
$\tan\theta=-\dfrac{1}{\sqrt{3}}$
$\tan\theta=-\dfrac{\sqrt{3}}{3}$
$\theta=\tan^{-1}\Big(-\dfrac{\sqrt{3}}{3}\Big)$
$\theta=-\dfrac{\pi}{6}$
To obtain $\theta$ between $0$ and $2\pi$:
$\theta=2\pi-\dfrac{\pi}{6}=\dfrac{11\pi}{6}$
The point in polar coordinates is $\Big(6,\dfrac{11\pi}{6}\Big)$
