The equation of a straight line,where the len | vedclass

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Question

(C) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle the normal makes with the positive $x$-axis.Given the line makes an angle of $120^{\circ}$ with the $x$-axis,its slope is $m = 	an(120^{\circ}) = -\sqrt{3}$.The normal to this line makes an angle $\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$ or $120^{\circ} + 90^{\circ} = 210^{\circ}$ with the $x$-axis.Using $p = 4$,the equations are:$x \cos(30^{\circ}) + y \sin(30^{\circ}) = 4$ $\Rightarrow x(\frac{\sqrt{3}}{2}) + y(\frac{1}{2}) = 4$ $\Rightarrow x\sqrt{3} + y = 8$.$x \cos(210^{\circ}) + y \sin(210^{\circ}) = 4$ $\Rightarrow x(-\frac{\sqrt{3}}{2}) + y(-\frac{1}{2}) = 4$ $\Rightarrow x\sqrt{3} + y = -8$.Comparing with the given options,$x\sqrt{3} + y = 8$ is option $C$.

The equation of a straight line,where the length of the perpendicular from the origin is $4$ units and the line makes an angle of $120^{\circ}$ with the $x$-axis,is:

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