Answer
The rectangular equation is $y=x+2\sqrt{2}$, with slope $1$ and y intercept $2\sqrt{2}$.
Work Step by Step
Consider the given equation,
$r\sin \left( \theta -\frac{\pi }{4} \right)=2$
Using, $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$ we will simplify given equation as,
$\begin{align}
& r\left[ \sin \theta \cos \frac{\pi }{4}-\cos \theta \sin \frac{\pi }{4} \right]=2 \\
& r\left[ \frac{\sin \theta }{\sqrt{2}}-\frac{\cos \theta }{\sqrt{2}} \right]=2 \\
& r\sin \theta -r\cos \theta =2\sqrt{2}
\end{align}$
Using $x=r\cos \theta $ and $y=r\sin \theta $ we will convert it to the rectangular equation. Therefore,
$\begin{align}
& y-x=2\sqrt{2} \\
& y=x+2\sqrt{2}
\end{align}$
Compare this equation with the equation of the line, $y=mx+c$, where m is the slope of line and c is the y-intercept. Therefore, the slope of the line is $1$ and to get the the y intercept, put $x=0$, in the equation.
$\begin{align}
& y=0+2\sqrt{2} \\
& =2\sqrt{2}
\end{align}$
Thus, the y intercept is $2\sqrt{2}$.
Hence, the rectangular equation of $r\sin \left( \theta -\frac{\pi }{4} \right)=2$ is $y=x+2\sqrt{2}$ with slope $1$ and y intercept $2\sqrt{2}$
