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(B) The line makes an angle of $120^{\circ}$ with the positive direction of the $X$-axis. The slope of the line is $m = 	an(120^{\circ}) = -\sqrt{3}$

Vedclass · Accepted Answer

$\sqrt{3}x + y = 8$

Vedclass · Answer

(B) The line makes an angle of $120^{\circ}$ with the positive direction of the $X$-axis. The slope of the line is $m = 	an(120^{\circ}) = -\sqrt{3}$.Let the normal to the line make an angle $\alpha$ with the positive $X$-axis. The normal is perpendicular to the line,so the angle of the normal is $\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$.The length of the perpendicular from the origin to the line is given as $p = 4$.The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$.Substituting the values,we get $x \cos(30^{\circ}) + y \sin(30^{\circ}) = 4$.$x(\frac{\sqrt{3}}{2}) + y(\frac{1}{2}) = 4$.Multiplying by $2$,we get $\sqrt{3}x + y = 8$.

Suppose a line makes an angle of $120^{\circ}$ with the positive direction of $X$-axis. If the length of the perpendicular from the origin to that line is $4$,then the equation of the line is

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