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Question

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

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$x \sqrt{3}+y+10=0$

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(C) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle made by the perpendicular with the positive $X$-axis.Given $p = 5$ and $\alpha = 210^{\circ}$.Substituting these values into the equation:$x \cos(210^{\circ}) + y \sin(210^{\circ}) = 5$Since $\cos(210^{\circ}) = \cos(180^{\circ} + 30^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2}$ and $\sin(210^{\circ}) = \sin(180^{\circ} + 30^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}$,$x(-\frac{\sqrt{3}}{2}) + y(-\frac{1}{2}) = 5$Multiplying both sides by $-2$:$\sqrt{3}x + y = -10$$\sqrt{3}x + y + 10 = 0$

The equation of a line,whose perpendicular distance from the origin is $5$ units and the angle,which the perpendicular to the line from the origin makes,is $210^{\circ}$ with the positive $X$-axis,is

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