# What is the polar form of (-2,3)?

Mar 19, 2016

$\left(\sqrt{13} , \pi - {\tan}^{- 1} \left(\frac{3}{2}\right)\right)$

#### Explanation:

To write in polar form, you need to know

1. the distance from the point to the origin
2. the angle the line passing through it and the origin makes with the positive $x$ axis.

To solve 1. we use Pythagoras Theorem

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

$= \sqrt{13}$

To solve 2. we first find the quadrant that the point lies in.

$y$ is positive while $x$ is negative $\implies$ quadrant II

Then we find the basic angle by taking inverse tangent of $| \frac{y}{x} |$.

$\alpha = {\tan}^{- 1} \left(| \frac{3}{- 2} |\right)$

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

The angle that we are looking for would be

$\theta = \pi - \alpha$

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

$\approx 2.16$

Therefore, the polar coordinate is $\left(\sqrt{13} , \pi - {\tan}^{- 1} \left(\frac{3}{2}\right)\right)$.

Note that the answer above is not unique. You can add any integer multiples of $2 \pi$ to $\theta$ to get other representations of the same point.