Question

In Problems55–66, the rectangular coordinates of a point are given. Find polar coordinates for each point.

$(-2,-2\sqrt{3})$

Short answer

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

Step-by-step solution

Step 1. Given Information   

We are given rectangular coordinates of a point,

$(-2,-2\sqrt{3})$

We have to find polar coordinates of given point.

Step 2. Finding the polar coordinates   

The distance of the point $(-2,-2\sqrt{3})$from origin is given by,

role="math" localid="1646791490159" $$\begin{gathered} r=\sqrt{x^2+y^2} \\ r=\sqrt{{(-2)}^2+{\left(-2\sqrt{3}\right)}^2} \\ =\sqrt{4+12} \\ =\sqrt{16} \\ =4 \end{gathered}$$

Step 3. Finding the polar coordinates   

We know that,

$$\begin{gathered} \tan \theta =\frac{y}{x}\text{, so} \\ \theta ={\tan }^{-1}\frac{y}{x},-\frac{\pi }{2}<\theta <\frac{\pi }{2} \end{gathered}$$

Putting values, we get

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

Since the point is on the third quadrant, so we have to add $\pi$to get the third quadrant angle,

$$\begin{gathered} \theta =\pi +\frac{\pi }{3} \\ \theta =\frac{3\pi +\pi }{3} \\ \theta =\frac{4\pi }{3} \end{gathered}$$

So, the polar coordinate is $\left(4,\frac{4\pi }{3}\right)$.