# How do you graph the point L(3,-pi/6)?

Jan 20, 2018

This is equivalent to plotting the rectangular point $\left(3 \sqrt{3} , - 3\right)$.

#### Explanation:

We want to graph the polar point $\left(3 , - \frac{\pi}{6}\right)$.

First consider the angle, $- \frac{\pi}{6}$. Rotate the angle, which has you at the origin facing $- \frac{\pi}{6}$, which is coterminal with $\frac{11 \pi}{6}$. You're facing into QIV.

Now move 3 units out from the origin. That's the point.

You can also convert to rectangular coordinates using $x = r \cos \left(\theta\right)$ and $y = r \sin \left(\theta\right)$ to convert:

$x = 6 \cos \left(- \frac{\pi}{6}\right) = 6 \left(\frac{\sqrt{3}}{2}\right) = 3 \sqrt{3}$
$x = 6 \sin \left(- \frac{\pi}{6}\right) = 6 \left(- \frac{1}{2}\right) = - 3$,

So the rectangular point is $\left(3 \sqrt{3} , - 3\right)$