# What is the polar form of ( -1,-2 )?

Mar 26, 2016

polar coordinate $\left(\sqrt{5} , {\tan}^{-} 1 2\right)$

#### Explanation:

$x = r \cos \theta \mathmr{and} y = r \sin \theta$
here $x = - 1 \mathmr{and} y = - 2$
the point is in third quadrant
So$\tan \theta = \frac{y}{x} = \frac{- 2}{-} 1 = 2$
as the point is in 3rd coordinate
$\therefore \theta = \pi + {\tan}^{-} 1 2$
and $r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{5}$

polar coordinate $\left(\sqrt{5} , \pi + {\tan}^{-} 1 2\right)$

Mar 26, 2016

$\left(\sqrt{5} , \pi + {\tan}^{- 1} 2\right)$

#### Explanation:

The equations $x = - 1 = r \cos \theta \mathmr{and} y = - 2 = r \sin \theta$ give
r = $\sqrt{{x}^{2} + {y}^{2}} = \sqrt{5}$.

To find $\theta$, observe that both $\sin \theta$ and $\cos \theta$ are negative. The angle is in the third quadrant. The first quadrant angle $\theta = {\tan}^{- 1} 2$. The angle for the opposite direction is the correct value for $\theta$.i

$\tan \left(\pi + \theta\right) = \tan \theta$
So, $\theta = \pi + {\tan}^{- 1} 2$. .