# What is the polar form of (-2,10)?

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{4 + 100} = \sqrt{104} = 2 \sqrt{26}$
$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right) = {\tan}^{-} 1 \left(\frac{10}{-} 2\right) = \left({\tan}^{-} 1 - 5\right) = - 1.373 \ldots \to \pi - 1.373 \ldots \mathmr{and} - 1.373 \ldots + \pi \to \theta \approx 1.768 \to \left(2 \sqrt{26} , 1.768\right)$
Use the formulas ${r}^{2} = {x}^{2} + {y}^{2} \mathmr{and} \tan \theta = \frac{y}{x}$to change the point from rectangular to polar. Note that the point (-2,10) is in quadrant 2 hence to find $\theta$ we have to add $\pi$ or 180 to the negative answer we got or ignore the negative sign and subtract from $\pi$