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