The angle made by a line L with the positive | vedclass

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(D) The slope of the line $L$ is $m = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.The equation of the line in slope-intercept form is $y = \frac{1}{\sqr

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$3$

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(D) The slope of the line $L$ is $m = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.The equation of the line in slope-intercept form is $y = \frac{1}{\sqrt{3}}x + c$,which simplifies to $x - \sqrt{3}y + \sqrt{3}c = 0$.Given that the distance from the origin $(0, 0)$ to the line is $5$,we have $\frac{|\sqrt{3}c|}{\sqrt{1^2 + (-\sqrt{3})^2}} = 5$.$\frac{|\sqrt{3}c|}{2} = 5 \implies |\sqrt{3}c| = 10 \implies \sqrt{3}c = \pm 10$.Since the $Y$-intercept $c$ is negative,we take $\sqrt{3}c = -10$.The equation of the line $L$ is $x - \sqrt{3}y - 10 = 0$.The perpendicular distance from the point $(1, -\sqrt{3})$ to the line $L$ is given by $d = \frac{|1(1) - \sqrt{3}(-\sqrt{3}) - 10|}{\sqrt{1^2 + (-\sqrt{3})^2}}$.$d = \frac{|1 + 3 - 10|}{\sqrt{1 + 3}} = \frac{|-6|}{2} = 3$.

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