Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 619: 42

$$r= 2\sqrt 2\sec(\theta-(3\pi/4)).$$

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 $$r=d\sec(\theta-\alpha)=2\sqrt 2\sec(\theta-(3\pi/4)).$$