# What is the polar form of ( -13,14 )?

Oct 20, 2017

$\left(r , \theta\right) \approx \left(19.1 , 2.3 \text{ [radians]}\right)$

#### Explanation:

$\left(x , y\right) = \left(- 13 , 14\right)$ is a point in Quadrant II
So $\theta$ must be $\in \left[0 , \pi\right]$

The radius (distance from the origin) for the point $\left(x , y\right)$
is $\sqrt{{x}^{2} + {y}^{2}}$
In this case
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{{\left(- 13\right)}^{2} + {14}^{2}} \approx 19.10497 \ldots$ (using a calculator)

The tangent for this angle $\left(\theta\right)$ is $\frac{y}{x}$
In this case
$\textcolor{w h i t e}{\text{XXX}} \tan \left(\theta\right) = \frac{14}{- 13}$
which implies
$\textcolor{w h i t e}{\text{XXX}} \theta = \arctan \left(- \frac{14}{13}\right) \approx 2.3191744 \ldots$ (again using a calculator
Note:
you might need to adjust for the quadrant if your $\arctan$ function gives a value in the range $\left[- \frac{\pi}{2} , + \frac{\pi}{2}\right]$