Answer
$x+y=-\sqrt{2}$
Work Step by Step
Apply identity for $\cos(\alpha+\beta)$
$\displaystyle \cos(\theta+\frac{3\pi}{4})=\cos\theta\cos\frac{3\pi}{4}-\sin\theta\sin\frac{3\pi}{4}$
$\displaystyle \cos(\theta+\frac{3\pi}{4})=-\frac{1}{\sqrt{2}}\cdot\cos\theta-\frac{1}{\sqrt{2}}\sin\theta\qquad/\times r$
$r\displaystyle \cos(\theta+\frac{3\pi}{4})=-\frac{1}{\sqrt{2}}\cdot(r\cos\theta)-\frac{1}{\sqrt{2}}\cdot(r\sin\theta)$
Apply the conversion formula $(x,y)=(r\cos\theta,r\sin\theta)$
$r\displaystyle \cos(\theta-\frac{\pi}{4})=-\frac{1}{\sqrt{2}}\cdot x-\frac{1}{\sqrt{2}}\cdot y$
So the line, in Cartesian coordinates, is
$-\displaystyle \frac{1}{\sqrt{2}}\cdot x-\frac{1}{\sqrt{2}}\cdot y=1\qquad/\times(-\sqrt{2})$
$x+y=-\sqrt{2}$
