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A point $P$ on a line is at a distance of 4 units from the origin $(0,0)$. If the line makes $60^{\circ}$ with the negative direction of the $X$-axis, then $P$ is
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The correct answer is:
$(2 \sqrt{3}, 2)$
$O P$ is perpendicular to given line.
$\therefore$ Slope of $O P=\frac{1}{\sqrt{3}}=\frac{y}{x} \Rightarrow x=\sqrt{3} y$
Also $x^2+y^2=16$
$\therefore 3 y^2+y^2=16 \Rightarrow y= \pm 2$
$\therefore$ Points can be $(2 \sqrt{3}, 2)$ or $(-2 \sqrt{3},-2)$.
$\therefore$ Slope of $O P=\frac{1}{\sqrt{3}}=\frac{y}{x} \Rightarrow x=\sqrt{3} y$
Also $x^2+y^2=16$
$\therefore 3 y^2+y^2=16 \Rightarrow y= \pm 2$
$\therefore$ Points can be $(2 \sqrt{3}, 2)$ or $(-2 \sqrt{3},-2)$.
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