# Question #a0503

Mar 19, 2017

($\sqrt{3} , 1$)

#### Explanation:

To make this change, we will use two equations:

$x = r \cos \left(\theta\right)$ and $y = r \sin \left(\theta\right)$

Our polar coordinate point is written in the form

$\left(r , \theta\right)$

So for the x coordinate we will plug in r and $\theta$ into the x equation

$x = \left(- 2\right) \left(\cos \left(\frac{7 \pi}{6}\right)\right) = \left(- 2\right) \left(- \frac{\sqrt{3}}{2}\right) = \sqrt{3}$

We will do the same for the y coordinate by plugging in the point values to the y equation

$y = \left(- 2\right) \left(\sin \left(\frac{7 \pi}{6}\right)\right) = \left(- 2\right) \left(- \frac{1}{2}\right) = 1$

And we take those values and write them as a point

$\left(\sqrt{3} , 1\right)$