If the length of the perpendicular to a line from the origin is \(2 \sqrt{2}\) units, which makes an angle of \(135^{\circ}\) with the X -axis, then the equation of line is

Given that the perpendicular to the line makes an angle of \(135^{\circ}\) with the X -axis.
\(\therefore \quad\) Slope of perpendicular \(=\tan \left(135^{\circ}\right)=1\)
\(\therefore \quad\) Slope of the required line \(=1\), and it passes through the second quadrant.
\(\therefore \quad\) Equation of the line is of the form \(y=x+c\)
i.e., \(x-y+c=0\)
Distance of this line from the origin is \(2 \sqrt{2}\) units.
\(\begin{array}{ll}
\therefore & 2 \sqrt{2}=\left|\frac{0-0+c}{\sqrt{(1)^2+(-1)^2}}\right| \\
\therefore & 2 \sqrt{2}=\left|\frac{c}{\sqrt{2}}\right|
\end{array}\)
\(\therefore \quad \mathrm{c}=4\) \(\ldots[\mathrm{c}=-4\) is not possible as line passes through second quadrant]