# What is the polar form of ( -23,-3 )?

Dec 5, 2015

The polar form of $\left(- 23 , - 3\right)$ is

$\left(\sqrt{538} , {\tan}^{- 1} \left(\frac{3}{23}\right) + \pi\right) \approx \left(23.195 , 3.271\right)$

#### Explanation:

This question has a list of equations used when converting between rectangular and polar coordinates.

In this case, we will be using
$\left\{\begin{matrix}{r}^{2} = {x}^{2} + {y}^{2} \\ \tan \left(\theta\right) = \frac{y}{x}\end{matrix}\right.$

$\implies \left\{\begin{matrix}r = \sqrt{{x}^{2} + {y}^{2}} \\ \theta = {\tan}^{- 1} \left(\frac{y}{x}\right)\end{matrix}\right.$

$r = \sqrt{{\left(- 23\right)}^{2} + {\left(- 3\right)}^{2}}$
theta = tan^(-1)((-3)/(-23)))^(color(red)("*"))

" "^color(red)("*")(While calculating $\theta$, the $- 3$ and $- 23$ cancel negatives, causing the resulting angle puts us in quadrant $I$ when we want quadrant $I I I$. To fix this, all we need to do is add or subtract $\pi$ from the angle to put us in the correct quadrant.)

$\implies \left\{\begin{matrix}r = \sqrt{538} \\ \theta = {\tan}^{- 1} \left(\frac{3}{23}\right) + \pi\end{matrix}\right.$

Thus we get the polar form of $\left(- 23 , - 3\right)$ to be

$\left(\sqrt{538} , {\tan}^{- 1} \left(\frac{3}{23}\right) + \pi\right) \approx \left(23.195 , 3.271\right)$