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

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

$\sqrt{3} x + y = 1$

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

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

answer is B.

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Detailed Solution

Equation of the line is $x \cos a + y \sin a = p$ where

$p = 3, α = 30^{\circ}$ So the required equation is

$(\frac{\sqrt{3}}{2}) x + (\frac{1}{2}) y = 3$ or $\sqrt{3} x + y = 6$

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