# How do you find the polar coordinates given (-2, -2sqrt3)?

Nov 17, 2016

#### Explanation:

Polar coordinates are an order pair of $\left(r , \theta\right)$

The conversion from $\left(x , y\right)$ to r is:

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

The conversion from $\left(x , y\right)$ to $\theta$ is more complicated:

If $x > 0 \mathmr{and} y \ge 0$, then $\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right) \text{ [1]}$

If $x = 0 \mathmr{and} y > 0$, then $\theta = \frac{\pi}{2} \text{ [2]}$

If $x = 0 \mathmr{and} y < 0$, then $\theta = \frac{3 \pi}{2} \text{ [3]}$

If $x < 0$, then $\theta = \pi + {\tan}^{-} 1 \left(\frac{y}{x}\right) \text{ [4]}$

If $x > 0 \mathmr{and} y < 0$, then $\theta = 2 \pi + {\tan}^{-} 1 \left(\frac{y}{x}\right) \text{ [5]}$

For the given point $\left(- 2 , - 2 \sqrt{3}\right)$

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

$r = \sqrt{4 + 12}$

$r = \sqrt{16}$

$r = 4$

Because the x coordinate is less than zero, we use equation [4]:

$\theta = \pi + {\tan}^{-} 1 \left(\frac{- 2 \sqrt{3}}{- 2}\right)$

$\theta = \pi + \frac{\pi}{3}$

$\theta = \frac{4 \pi}{3}$

The polar point is $\left(4 , \frac{4 \pi}{3}\right)$