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Equation of straight line which is 10 units from origin and normal ray from origin to the line makes an angle of $135^{°}$ with positive direction of x-axis in anti clock direction is

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$x + y = 10 \sqrt{2}$

$x + y + 10 \sqrt{2} = 0$

$x - y - 10 \sqrt{2} = 0$

$x - y + 10 \sqrt{2} = 0$

answer is B.

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Detailed Solution

Let $p = 10 a n d α = 135^{°}$

Equation of the line in normal form is $x \cos α + y \sin α = p$

$\Rightarrow x \cos 135 + y \sin 135 = 10 \Rightarrow x (\frac{- 1}{\sqrt{2}}) + y (\frac{1}{\sqrt{2}}) = 10 \Rightarrow - x + y = 10 \sqrt{2} \Rightarrow x - y + 10 \sqrt{2} = 0$

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