## Precalculus (6th Edition) Blitzer

The rectangular equation is $y=x+2\sqrt{2}$, with slope $1$ and y intercept $2\sqrt{2}$.
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 $2\sqrt{2}$