## Precalculus: Mathematics for Calculus, 7th Edition

$(\sqrt{2}, \displaystyle \frac{3\pi}{4})$
To change from rectangular to polar coordinates, use $r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$. --------- $r^{2}=(-1)^{2}+1^{2}=2$ $r=\pm\sqrt{2}$ $\displaystyle \tan^{-1}(\frac{1}{-1})=\tan^{-1}(-1)=-\frac{\pi}{4}$, which is in quadrant IV. The point P$(-1,1)$ is in quadrant II, so we take $\theta= -\displaystyle \frac{\pi}{4}+\pi=\frac{3\pi}{4}$. The terminal end of $\theta$ passes through P, so for this angle, r is positive. Polar coordinates: $(\sqrt{2}, \displaystyle \frac{3\pi}{4})$