Question

Find the equation of the line whose perpendiculat distance from origin is 3 units and the angle which the normal makes with the positive direction of x-axis is 30°.

• A. $\sqrt{3}x+y=6$
• B. $x+\sqrt{3}y=6$
• C. $\sqrt{3}x+y=3$
• D. None of These

Answer 1

The correct option is A $\sqrt{3}x+y=6$

The perpendicular distance from origin = p = 3 units

DC = 30°

The required equation is

$x\cos \alpha +y\sin \alpha =p$

$\Rightarrow x\cos {30}^{\circ }+y\sin {30}^{\circ }=3$

$\Rightarrow x\left(\frac{\sqrt{3}}{2}\right)+y\left(\frac{1}{2}\right)=3$

$\Rightarrow \frac{\sqrt{3}x+y}{2}=3$

$\therefore \sqrt{3}x+y=6$

So, option a is correct.