The equation of line, where length of the perpendicular segment from origin to the line is 4 and the…

The equation of line, where length of the perpendicular segment from origin to the line is 4 and the inclination of this perpendicular segment with the positive direction of $\mathrm{X}$-axis is $30^{\circ}$, is
  1. $x+\sqrt{3} y=8$
  2. $x-\sqrt{3} y=8$
  3. $\sqrt{3} x-y=8$
  4. $\sqrt{3} x+y=8$

Solution

We have $\ell(\mathrm{OP})=4$ and $\mathrm{m} \angle \mathrm{POX}=30^{\circ}$ $\therefore \mathrm{P} \equiv\left(4 \cos 30^{\circ}, 4 \sin 30^{\circ}\right) \equiv(2 \sqrt{3}, 2)$ From figure, we conclude that angle made be line with +ve $\mathrm{X}$ axis is $120^{\circ}$. $\therefore$ Slope of line $=\tan \left(120^{\circ}\right)=-\sqrt{3}$ Hence required equation of line $\mathrm{L}$ is $(y-2)=(-\sqrt{3})(x-2 \sqrt{3}) \Rightarrow \sqrt{3} x+y=8$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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