The equation of a line, whose perpendicular distance from the origin is 5 units and the angle, which the…
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 positive $\mathrm{X}$-axis, is
$-x \sqrt{3}+y+10=0$
$x \sqrt{3}+y-10=0$
$x \sqrt{3}+y+10=0$
$x \sqrt{3}-y+10=0$
Solution
Here, $P=5$ and $\alpha=210^{\circ}$
Writing the equation of straight line in normal form
$\begin{aligned} & \Rightarrow x \cos \left(210^{\circ}\right)+y \sin 30^{\circ}=5 \\ & \Rightarrow-x \cos 30^{\circ}-y \sin 30^{\circ}=5 \\ & \Rightarrow-x \cdot \frac{\sqrt{3}}{2}-y \cdot \frac{1}{2}=5 \\ & \Rightarrow-\sqrt{3} x-y=10 \\ & \Rightarrow \sqrt{3} x+y+10=0\end{aligned}$