The equation of the straight line in symmetric form having the slope $-1$  and passing through the point $\left(1,1\right)$  is

Symmetric form of an equation is $\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }$

Slope $m=-1\Rightarrow \tan \theta =-1=-\cot 45°$

$\tan \theta =\tan \left(90+45°\right)\Rightarrow \theta =90°+45°\Rightarrow \theta =135°$

$\cos \theta =\cos 135°=\cos \left(90°+45°\right)=-\sin 45°=-\frac{1}{\sqrt{2}}$

$\sin \theta =\sin 135°=\sin \left(90°+45°\right)=\cos 45°=\frac{1}{\sqrt{2}}$

equation of the line is $\frac{x-1}{-\frac{1}{\sqrt{2}}}=\frac{y-1}{\frac{1}{\sqrt{2}}}$ $\begin{array}{c}\\ \end{array}$

$x-1=\left(y-1\right)$

$x-1=-y+1$

$x+y=2$