# How do you convert (2, -3) into polar form?

Aug 23, 2016

$\left(2 , - 3\right) \to \left(\sqrt{13} , - 0.983\right)$

#### Explanation:

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

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

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

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

here x = 2 and y = -3

$\Rightarrow r = \sqrt{{2}^{2} + {\left(- 3\right)}^{2}} = \sqrt{13}$

now (2 ,-3) is a point in the 4th quadrant so we must ensure that $\theta$ is in the 4th quadrant.

$\theta = {\tan}^{-} 1 \left(- \frac{3}{2}\right) = - 0.983 \text{ in 4th quadrant }$

$\Rightarrow \left(2 , - 3\right) \to \left(\sqrt{13} , - 0.983\right) = \left(\sqrt{13} , - {56.31}^{\circ}\right)$