# How do you convert  (-1,1) into polar form?

Sep 22, 2016

$\left(\sqrt{2} , \frac{3 \pi}{4}\right)$

#### Explanation:

To convert from $\textcolor{b l u e}{\text{cartesian to polar form}}$

That is $\left(x , y\right) \to \left(r , \theta\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here x = -1 and y = 1

$\Rightarrow r = \sqrt{{\left(- 1\right)}^{2} + {1}^{2}} = \sqrt{2}$

Now (-1 ,1) is in the 2nd quadrant so we must ensure that $\theta$ is in the 2nd quadrant.

$\theta = {\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4} \leftarrow \text{ related acute angle}$

$\Rightarrow \theta = \left(\pi - \frac{\pi}{4}\right) = \frac{3 \pi}{4}$

Thus $\left(- 1 , 1\right) \to \left(\sqrt{2} , \frac{3 \pi}{4}\right)$