# How do you find the polar coordinates given (1,-1)?

Aug 29, 2016

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

#### Explanation:

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

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

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ and } \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here x = 1 and y = -1

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

Now, (1 ,-1) is in the 4th quadrant so we must ensure that $\theta$ is in the 4th quadrant.

$\theta = {\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4} \text{ in 4th quadrant}$

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