$\begin{array}{1 1}(A)\;p=\sqrt 2 \: units ; \alpha = 315^{\circ} \\(B)\; p= 2 \: units ; \alpha = 45^{\circ} \\(C)\; p=2\sqrt 2 \: units ; \alpha = 315^{\circ} \\(D)\;p=2\sqrt 2 \: units ; \alpha = 45^{\circ} \end{array} $

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- Equation of a line in the normal form is $ x \cos \alpha + y \sin \alpha=p$ where p is the perpendicular distance from the origin and $ \alpha$ is the angle between the perpendicular and the positive $x$ - axis.

Given equation is $ x-y=4$

This can be written as

$1.x+(-1).y=4$

$ \sqrt{1^2+(-1)^2}=\sqrt 2$

dividing on both sides by $\sqrt 2$ we get,

$ \large\frac{1}{\sqrt 2}$$x+ \bigg( - \large\frac{1}{\sqrt 2} \bigg)$$y=\large\frac{4}{\sqrt 2}$

$ \Rightarrow x \cos \bigg( 2\pi - \large\frac{\pi}{4} \bigg)$$ + y \sin \bigg( 2\pi - \large\frac{\pi}{4} \bigg)$$= 2 \sqrt 2$

i.e., $x \cos 315^{\circ}+y \sin 315^{\circ}= 2\sqrt 2$----------(1)

This is of the form

$ x \cos \alpha + y \sin \alpha=p$----------(2)

Comparing the two equations we get,

$ p = 2\sqrt 2 \: and \: \alpha = 315^{\circ}$

Hence the perpendicular distance is $2\sqrt 2$ units and angle = $315^{\circ}$

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