## Calculus (3rd Edition)

$$r= 2\sqrt 2\sec(\theta-(3\pi/4)).$$
Since the point has the coordinates $x=-2$, $y=2$, then the polar coordinates are $d=\sqrt{x^2+y^2}=\sqrt 8=2\sqrt2$, and $\alpha=\tan^{-1}\frac{2}{-2}=-\pi/4+\pi=3\pi/4$. Now, the polar equation of the line $\mathcal{L}$ is given by $$r=d\sec(\theta-\alpha)=2\sqrt 2\sec(\theta-(3\pi/4)).$$