The equation of a line, whose perpendicular distance from the origin is 7 units and the angle, which the…
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 positive $\mathrm{X}$-axis, is
$x+\sqrt{3} y-14=0$
$x+\sqrt{3} y+14=0$
$x-\sqrt{3} y+14=0$
$x-\sqrt{3} y-14=0$
Solution
Normal form of the equation of line is
$\begin{array}{ll} & x \cos \alpha+y \sin \alpha=\mathrm{p} \\ & \text { Here, } \alpha=120^{\circ} \text { and } \mathrm{p}=7 \\ \therefore \quad & x \cos 120^{\circ}+y \sin 120^{\circ}=7 \\ \therefore \quad & x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=7 \\ \therefore \quad & \frac{-x+\sqrt{3} y}{2}=7 \\ \therefore \quad & -x+\sqrt{3} y=14 \\ \therefore \quad & -x+\sqrt{3} y-14=0 \\ \therefore \quad & x-\sqrt{3} y+14=0\end{array}$