# How do you convert (6, -6) into polar coordinates?

Jan 31, 2016

The point that has coordinates $\left(6 , - 6\right)$ in rectangular coordinates has the polar coordinates $\left(\sqrt{72} , - \frac{\pi}{4}\right)$ or $\left(8.5 , - 0.79\right)$ or (to give a positive value to $\theta$) as $\left(\sqrt{72} , \frac{7 \pi}{4}\right)$.

#### Explanation:

Polar coordinates are in the form $\left(r , \theta\right)$ where $r$ is the distance from the origin $\left(0 , 0\right)$ to the point and $\theta$ is the angle in radians from the positive x-axis.

To find the radius, use:

$r = \sqrt{{6}^{2} + {\left(- 6\right)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 8.5$

(some may prefer to leave it in the form $\sqrt{72}$)

To find the value of $\theta$, know that $6$ is the opposite and $- 6$ is the adjacent side of a right-angled triangle, so:

$\tan \theta = \frac{6}{-} 6 = - 1$

Therefore $\theta = {\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4}$ $r a d$.

This means the polar coordinates can be expressed as $\left(\sqrt{72} , - \frac{\pi}{4}\right)$ or $\left(8.5 , - 0.79\right)$ or (to give a positive value to $\theta$) as $\left(\sqrt{72} , \frac{7 \pi}{4}\right)$.