Answer
$(\sqrt{5}, \tan^{-1}(-2))$
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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$r^{2}=1^{2}+(-2)^{2}=5$
$r=\pm\sqrt{5}$
$\displaystyle \tan^{-1}(\frac{-2}{1})=\tan^{-1}(-2)\approx-1.10715$
is in quadrant IV.
The point P$(1,-2)$ is in quadrant IV,
so we take $\theta=\tan^{-1}( -2)$.
The terminal end of $\theta$ passes through P, so for this angle, r is positive.
Polar coordinates: $(\sqrt{5}, \tan^{-1}(-2))$
