# How do you convert  (-6,-3) into polar coordinates?

Feb 13, 2016

Polar coordinates of (−6,−3) are $\left(3 \sqrt{5} , {206.565}^{o}\right)$

#### Explanation:

If $\left(x , y\right)$ are converted into polar coordinates $\left(r , \theta\right)$,

while $\left(x , y\right)$ in terms of $r$ and $\theta$ are $x = r \cos \theta$ and $y = r \sin \theta$, $\left(r , \theta\right)$ in terms of $x$ and $y$ are $r = \sqrt{{x}^{2} + {y}^{2}}$ and $\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

Hence, converting (−6,−3) into polar coordinates

r=sqrt((-6)^2+(-3^2) = $\sqrt{45}$ = $3 \sqrt{5}$

$\theta = {\tan}^{-} 1 \left(\frac{- 3}{-} 6\right)$ = ${\tan}^{-} 1 \left(\frac{1}{2}\right)$. Further, as (−6,−3) is in third quadrant $\theta = \left(180 + 26.565\right) = {206.565}^{o}$

Hence polar coordinates of (−6,−3) are $\left(3 \sqrt{5} , {206.565}^{o}\right)$