# How do you convert (-sqrt3,-1) into polar form?

Feb 12, 2017

The polar coordinates are $\left(2 , \frac{7}{6} \pi\right)$

#### Explanation:

To convert from cartesian coordinates $\left(x , y\right)$ into polar coordinates, $\left(r , \theta\right)$ we use the following equations

$r = \sqrt{{x}^{2} + {y}^{2}}$

$\tan \theta = \frac{y}{x}$

$r = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{2}}$

$= \sqrt{4} = 2$

$\tan \theta = - \frac{1}{-} \sqrt{3} = \frac{1}{\sqrt{3}}$

We are in the 3rd quadrant

$\theta = \pi + \frac{\pi}{6} = \frac{7}{6} \pi$

The polar coordinates are $\left(2 , \frac{7}{6} \pi\right)$