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$\mathrm{P}_1, \mathrm{P}_2$ are points in either of the two lines $y-\sqrt{3}|\mathrm{x}|=2$ at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from $\mathrm{P}_1, \mathrm{P}_2$ on the bisector of the angle between the given lines.
Solution:
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Verified Answer
Lines are $\mathrm{y}=2+\sqrt{3} \mathrm{x}, \mathrm{y}=2-\sqrt{3} \mathrm{x}$ which intersect at $\mathrm{A}(0,2)$.
Now, $\mathrm{OQ}=\mathrm{OA}+\mathrm{AQ}$
$$
=2+5 \cos 30^{\circ} \quad\left[\begin{array}{ll}
\because & \cos 30^{\circ}=\frac{\mathrm{AQ}}{5}
\end{array}\right]
$$
$$
\mathrm{OQ}=2+\frac{5 \sqrt{3}}{2}
$$
$\therefore$ Coordinates of required point are
$$
\left(0,2+\frac{5 \sqrt{3}}{2}\right)
$$
Now, $\mathrm{OQ}=\mathrm{OA}+\mathrm{AQ}$
$$
=2+5 \cos 30^{\circ} \quad\left[\begin{array}{ll}
\because & \cos 30^{\circ}=\frac{\mathrm{AQ}}{5}
\end{array}\right]
$$
$$
\mathrm{OQ}=2+\frac{5 \sqrt{3}}{2}
$$
$\therefore$ Coordinates of required point are
$$
\left(0,2+\frac{5 \sqrt{3}}{2}\right)
$$
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