The equation of the straight line in symmetric form having the slope −1 and passing through the point (1,1) is

Symmetric form of an equation is

\frac{x - x_{1}}{\cos θ} = \frac{y - y_{1}}{\sin θ}

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

$\tan θ = \tan (90 + 45 °) \Rightarrow θ = 90 ° + 45 ° \Rightarrow θ = 135 °$

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

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

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

$x - 1 = (y - 1)$

$x - 1 = - y + 1$

$x + y = 2$

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