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

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$x-\sqrt{3} y+14=0$

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(C) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$. Here,the perpendicular distance from the origin is $p = 7$ and the angle $\alpha = 120^{\circ}$. Substituting these values into the equation: $x \cos 120^{\circ} + y \sin 120^{\circ} = 7$ Since $\cos 120^{\circ} = -\frac{1}{2}$ and $\sin 120^{\circ} = \frac{\sqrt{3}}{2}$,we get: $x(-\frac{1}{2}) + y(\frac{\sqrt{3}}{2}) = 7$ Multiplying both sides by $2$: $-x + \sqrt{3}y = 14$ Rearranging the terms to the standard form: $x - \sqrt{3}y + 14 = 0$

The equation of a line,whose perpendicular distance from the origin is $7$ units and the angle,which the perpendicular to the line from the origin makes,is $120^{\circ}$ with the positive $X$-axis,is

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