# How do you convert (11,-9) into polar coordinates?

May 6, 2018

$\left(\sqrt{202} , {\tan}^{-} 1 \left(- \frac{9}{11}\right) + 2 \pi\right) \mathmr{and} \left(14.2 , {5.60}^{c}\right)$

#### Explanation:

(x,y)->(r,theta);(r,theta)=(sqrt(x^2+y^2),tan^-1(y/x))

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{11}^{2} + {\left(- 9\right)}^{2}} = \sqrt{121 + 81} = \sqrt{202} \approx 14.2$

$\theta = {\tan}^{-} 1 \left(- \frac{9}{11}\right)$

However, $\left(11 , - 9\right)$ is in quadrant 4, and so we must add $2 \pi$ to our answer.

$\theta = {\tan}^{-} 1 \left(- \frac{9}{11}\right) + 2 \pi \approx {5.60}^{c}$

$\left(\sqrt{202} , {\tan}^{-} 1 \left(- \frac{9}{11}\right) + 2 \pi\right) \mathmr{and} \left(14.2 , {5.60}^{c}\right)$