$Normalformofthelinex-y+\sqrt{2}=0is$

$\begin{array}{l}GivenEquationx-y+\sqrt{2}=0\\ \Rightarrow x-y=-\sqrt{2}\\ dividingwith\sqrt{a^2+b^2}=\sqrt{1+1}=\sqrt{2}\\ \therefore \frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=\frac{-\sqrt{2}}{\sqrt{2}}\\ \Rightarrow (-\frac{1}{\sqrt{2}})x+\left(\frac{1}{\sqrt{2}}\right)y=1\\ comparingwithx\cos \alpha +y\sin \alpha =p\\ \cos \alpha =\frac{-1}{\sqrt{2}},\sin \alpha =\frac{1}{\sqrt{2}},p=1\\ \cos \alpha is(-)and\sin \alpha is(+)\\ \therefore \alpha \in Q_2\Rightarrow \alpha =\pi -\frac{\pi }{4}=\frac{3\pi }{4}\\ \therefore x\cos \frac{3\pi }{4}+y\sin \frac{3\pi }{4}=1\end{array}$