Answer
$x+y=-\sqrt{2}$
Work Step by Step
Conversion formulas:
$\left\{\begin{array}{ll}
(x,y)=(r\cos\theta,r\sin\theta) & \\
r^{2}=x^{2}+y^{2}, & \tan\theta=\frac{y}{x}
\end{array}\right.$
Apply additive identity for cosine
$\displaystyle \cos(\theta+\frac{3\pi}{4})=\cos\theta\cos\frac{3\pi}{4}-\sin\theta\sin\frac{3\pi}{4}$
$\displaystyle \cos(\theta+\frac{3\pi}{4})=-\frac{1}{\sqrt{2}}\cdot\cos\theta-\frac{1}{\sqrt{2}}\sin\theta\qquad/\times r$
$r\displaystyle \cos(\theta+\frac{3\pi}{4})=-\frac{1}{\sqrt{2}}\cdot(r\cos\theta)-\frac{1}{\sqrt{2}}\cdot(r\sin\theta)$
Apply the conversion formula
$r\displaystyle \cos(\theta-\frac{\pi}{4})=-\frac{1}{\sqrt{2}}\cdot x-\frac{1}{\sqrt{2}}\cdot y$
So the line, in Cartesian coordinates, is
$-\displaystyle \frac{1}{\sqrt{2}}\cdot x-\frac{1}{\sqrt{2}}\cdot y=1\qquad/\times(-\sqrt{2})$
$x+y=-\sqrt{2}$