A point P on a line is at a distance of 4 uni | vedclass

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(B) The line makes an angle of $60^{\circ}$ with the negative direction of the $X$-axis. Therefore,the angle it makes with the positive direction of t

Vedclass · Accepted Answer

$(2\sqrt{3}, 2)$

Vedclass · Answer

(B) The line makes an angle of $60^{\circ}$ with the negative direction of the $X$-axis. Therefore,the angle it makes with the positive direction of the $X$-axis is $180^{\circ} - 60^{\circ} = 120^{\circ}$.The slope of the line is $m = 	an(120^{\circ}) = -\sqrt{3}$.The line segment $OP$ is perpendicular to this line,so the slope of $OP$ is $m' = -\frac{1}{m} = \frac{1}{\sqrt{3}}$.Let $P = (x, y)$. Since $OP$ makes an angle $	heta$ with the $X$-axis where $	an(	heta) = \frac{1}{\sqrt{3}}$,we have $	heta = 30^{\circ}$ or $210^{\circ}$.Using polar coordinates,$x = r \cos(	heta)$ and $y = r \sin(	heta)$ with $r = 4$.For $	heta = 30^{\circ}$,$P = (4 \cos(30^{\circ}), 4 \sin(30^{\circ})) = (4 \cdot \frac{\sqrt{3}}{2}, 4 \cdot \frac{1}{2}) = (2\sqrt{3}, 2)$.For $	heta = 210^{\circ}$,$P = (4 \cos(210^{\circ}), 4 \sin(210^{\circ})) = (-2\sqrt{3}, -2)$.Comparing with the options,the correct point is $(2\sqrt{3}, 2)$.

$A$ point $P$ on a line is at a distance of $4$ units from the origin $(0,0)$. If the line makes an angle of $60^{\circ}$ with the negative direction of the $X$-axis,then the coordinates of $P$ are

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