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Consider a line $' L'$ which makes an angle $30 °$ with the positive direction of $X -$ axis and $Y -$ intercept equal to unity find the equation of line $' L'$

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

$y = x - \sqrt{3}$

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

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

answer is B, D.

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Detailed Solution

$\begin{array}{l} θ = 30 ° \\ m = \tan θ = \tan 30 ° = \frac{1}{\sqrt{3}} \end{array}$

$y -$ intercept $= 1$

$y = m x + c$

$\begin{array}{l} y = \frac{1}{\sqrt{3}} x + 1 \to \sqrt{3} y = x + \sqrt{3} \\ \Rightarrow x - \sqrt{3} y + \sqrt{3} = 0 \end{array}$

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