A straight line makes an angle $\frac{π}{3}$ with positive direction of x-axis measured in anti clock direction and makes positive intercept on y-axis and the line is at a distance of 5 units from origin then its equation is

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$y = \sqrt{3} x + 8$

$y = \sqrt{3} x - 8$

$y = \sqrt{3} x + 10$

$y = \sqrt{3} x - 10$

answer is B.

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Detailed Solution

Let angle made by the line is $θ = \frac{π}{3}$

$∴$ Normal makes an angle $α = 90 + θ$

$\begin{array}{l} = \frac{π}{2} + \frac{π}{3} \\ = \frac{5 π}{6} \end{array}$

Let p=5

Equation of line in Normal form is $x \cos α + y \sin α = p$

$\begin{array}{l} x \cos \frac{5 π}{6} + y \sin \frac{5 π}{6} = p \\ \Rightarrow \frac{- \sqrt{3} x}{2} + \frac{y}{2} = 5 \\ \Rightarrow y = \sqrt{3} x + 10 \end{array}$

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