Suppose P(x,y) lies on sqrt{3} x-y+2=0 or sqr | vedclass

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(C) Let $L_1: \sqrt{3} x-y+2=0$ and $L_2: \sqrt{3} x+y-2=0$. Solving these,we find the point of intersection $A(0,2)$. $P$ lies on $L_1$ or $L_2$ such

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$2+\frac{5 \sqrt{3}}{2}$

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(C) Let $L_1: \sqrt{3} x-y+2=0$ and $L_2: \sqrt{3} x+y-2=0$. Solving these,we find the point of intersection $A(0,2)$. $P$ lies on $L_1$ or $L_2$ such that $AP=5$. For $L_1$,the slope is $\sqrt{3}$,so the angle with the $y$-axis is $30^{\circ}$. Let $Q$ be the foot of the perpendicular from $P$ onto the $y$-axis. In $	riangle PAQ$,$AQ = AP \cos 30^{\circ} = 5 	imes \frac{\sqrt{3}}{2} = \frac{5 \sqrt{3}}{2}$. The $y$-coordinate of $P$ is $y_P = y_A + AQ = 2 + \frac{5 \sqrt{3}}{2}$ (or $2 - \frac{5 \sqrt{3}}{2}$ depending on the side). The foot of the perpendicular $Q$ on the $y$-axis is $(0, y_P)$. The distance from $(0,0)$ to $Q$ is $|y_P| = |2 \pm \frac{5 \sqrt{3}}{2}|$. Given the options,the distance is $2 + \frac{5 \sqrt{3}}{2}$.

Suppose $P(x,y)$ lies on $\sqrt{3} x-y+2=0$ or $\sqrt{3} x+y-2=0$ and is at a distance of $5$ units from their point of intersection. Then the distance from $(0,0)$ to the foot of the perpendicular of $P$ onto the $y$-axis is

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