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(B) The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the origin and

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$x-y+4=0$

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(B) The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the origin and $\alpha$ is the angle the perpendicular makes with the positive $X$-axis. Given $p = 2 \sqrt{2}$ and $\alpha = 135^{\circ}$. The equation of the line is $x \cos(135^{\circ}) + y \sin(135^{\circ}) = 2 \sqrt{2}$. Since $\cos(135^{\circ}) = -\frac{1}{\sqrt{2}}$ and $\sin(135^{\circ}) = \frac{1}{\sqrt{2}}$,we have: $x \left(-\frac{1}{\sqrt{2}}\right) + y \left(\frac{1}{\sqrt{2}}\right) = 2 \sqrt{2}$. Multiplying both sides by $\sqrt{2}$,we get: $-x + y = 4$ or $x - y + 4 = 0$.

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 the line is

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