P is a point on either of the two lines y - s | vedclass

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(B) The equation of the lines is $y = \sqrt{3}|x| + 2$. This represents two lines: $y = \sqrt{3}x + 2$ for $x \ge 0$ and $y = -\sqrt{3}x + 2$ for $x <

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$\left(0, \frac{4 + 5\sqrt{3}}{2}\right)$

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(B) The equation of the lines is $y = \sqrt{3}|x| + 2$. This represents two lines: $y = \sqrt{3}x + 2$ for $x \ge 0$ and $y = -\sqrt{3}x + 2$ for $x The point of intersection $A$ is $(0, 2)$. The angle between the lines and the $y$-axis (which is the angle bisector) can be found from the slope. For $y = \sqrt{3}x + 2$,the slope is $\sqrt{3}$,so the angle with the $x$-axis is $60^\circ$. The angle with the $y$-axis is $90^\circ - 60^\circ = 30^\circ$. Let $P$ be a point on the line at a distance $AP = 5$. The foot of the perpendicular $M$ from $P$ onto the $y$-axis lies on the $y$-axis. The distance $AM = AP \cos(30^\circ) = 5 	imes \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$. Since $A$ is at $(0, 2)$ and the lines extend upwards,the $y$-coordinate of $M$ is $2 + \frac{5\sqrt{3}}{2} = \frac{4 + 5\sqrt{3}}{2}$. Thus,the coordinates of $M$ are $\left(0, \frac{4 + 5\sqrt{3}}{2}ight)$.

$P$ is a point on either of the two lines $y - \sqrt{3}|x| = 2$ at a distance of $5 \ units$ from their point of intersection. The coordinates of the foot of the perpendicular from $P$ on the bisector of the angle between them are

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