A is a point on either of two lines y + sqrt{ | vedclass

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(D) The given equation is $y + \sqrt{3} |x| = 2$. This represents two lines: $L_1: y + \sqrt{3}x = 2$ (for $x \ge 0$) and $L_2: y - \sqrt{3}x = 2$ (fo

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$(0, 4)$

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(D) The given equation is $y + \sqrt{3} |x| = 2$. This represents two lines: $L_1: y + \sqrt{3}x = 2$ (for $x \ge 0$) and $L_2: y - \sqrt{3}x = 2$ (for $x Both lines intersect at the point $P(0, 2)$.The angle bisectors of these lines are the $y$-axis $(x=0)$ and the line $y=2$.The point $A$ lies on the lines at a distance of $d = \frac{4}{\sqrt{3}}$ from $P(0, 2)$.For $L_1$,the slope is $m = -\sqrt{3}$,so the angle with the $x$-axis is $120^\circ$. The coordinates of $A$ are $(0 + d \cos 120^\circ, 2 + d \sin 120^\circ) = \left( \frac{4}{\sqrt{3}} \cdot \left( -\frac{1}{2} \right), 2 + \frac{4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \right) = \left( -\frac{2}{\sqrt{3}}, 4 \right)$.However,checking the geometry,the foot of the perpendicular from $A$ to the angle bisector $x=0$ is $(0, 4)$.

$A$ is a point on either of two lines $y + \sqrt{3} |x| = 2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection. The coordinates of the foot of the perpendicular from $A$ on the bisector of the angle between them are

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