Suppose P(x,y)lying on $\sqrt{3} x-y+2=0$ or $\sqrt{3} x+y-2=0$ is at a distance of 5 units from their point…
Suppose P(x,y)lying on $\sqrt{3} x-y+2=0$ or $\sqrt{3} x+y-2=0$
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
$2$
$\frac{5 \sqrt{3}}{2}$
$2+\frac{5 \sqrt{3}}{2}$
$2-\frac{5 \sqrt{3}}{2}$
Solution
$\mathrm{L}_1: \sqrt{3} x-y+2=0$
$\begin{aligned} & \mathrm{L}_2: \sqrt{3} x+y-2=0 \\ & \mathrm{~L}_1 \text { and } \mathrm{L}_2 \text { intersect on } A(0,2) \\ & P \text { can be on any of } \mathrm{L}_1 \text { or } \mathrm{L}_2 \\ & \text{ such that } \mathrm{AP}=5\end{aligned}$
For $\mathrm{L}_1: y=\sqrt{3} x+2$
i.e. $m=\sqrt{3}$
$\begin{aligned} & \angle \mathrm{ABO}=60^{\circ} \\ & \angle \mathrm{BAO}=30^{\circ} \\ & \angle \mathrm{PAQ}=30^{\circ}\end{aligned}$
$\begin{aligned} & \text { In } \Delta \text { PAQ, } \\ & \cos 30^{\circ}=A Q / A P\end{aligned}$
$\begin{aligned} & \mathrm{AQ}=\mathrm{AP} \cos 30^{\circ}=\frac{5 \sqrt{3}}{2} \\ & \mathrm{OQ}=\mathrm{OA}+\mathrm{AQ}\end{aligned}$
$=2+\frac{5 \sqrt{3}}{2}$