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

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a
$x+y=2$
b
$x+2y=3$
c
$x-4y=7$
d
$x-y=10$

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detailed solution

Correct option is A

Symmetric form of an equation is $\frac{x-{x}_{1}}{\mathrm{cos}\theta }=\frac{y-{y}_{1}}{\mathrm{sin}\theta }$

Slope $m=-1⇒\mathrm{tan}\theta =-1=-\mathrm{cot}45°$

$\mathrm{tan}\theta =\mathrm{tan}\left(90+45°\right)⇒\theta =90°+45°⇒\theta =135°$

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

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

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

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

$x-1=-y+1$

$x+y=2$

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