Answer
$(\sqrt{2}, \displaystyle \frac{3\pi}{4})$
Work Step by Step
To change from rectangular to polar coordinates, use
$r^{2}=x^{2}+y^{2}$ and $\displaystyle \tan\theta=\frac{y}{x}$.
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$r^{2}=(-1)^{2}+1^{2}=2$
$r=\pm\sqrt{2}$
$\displaystyle \tan^{-1}(\frac{1}{-1})=\tan^{-1}(-1)=-\frac{\pi}{4}$, 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: $(\sqrt{2}, \displaystyle \frac{3\pi}{4})$
