L equiv x cos alpha + y sin alpha - p = 0 rep | vedclass

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(A) The slope of line $L$ is $-\cot \alpha$. The slope of line $x + y + 1 = 0$ is $-1$. Since the lines are perpendicular,the product of their slopes

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$5$

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(A) The slope of line $L$ is $-\cot \alpha$. The slope of line $x + y + 1 = 0$ is $-1$. Since the lines are perpendicular,the product of their slopes is $-1$: $(-\cot \alpha)(-1) = -1$ $\Rightarrow \cot \alpha = -1$ $\Rightarrow 	an \alpha = -1$. Since $\alpha$ lies in the fourth quadrant,$\alpha = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$. The perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to $x \cos \alpha + y \sin \alpha - p = 0$ is given by: $\left| \sqrt{2} \cos \alpha + \sqrt{2} \sin \alpha - p ight| = 5$. Substituting $\alpha = \frac{7\pi}{4}$: $\cos \frac{7\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{7\pi}{4} = -\frac{1}{\sqrt{2}}$. $\left| \sqrt{2} \left( \frac{1}{\sqrt{2}} ight) + \sqrt{2} \left( -\frac{1}{\sqrt{2}} ight) - p ight| = 5$. $|1 - 1 - p| = 5 \Rightarrow |-p| = 5$. Since $p$ is positive,$p = 5$.

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

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