Find the equations to the straight lines which pass through the origin and are inclined at $75^\circ$ to the straight line $x \left(1 - \sqrt{3}\right) + y \left(1 + \sqrt{3}\right) + c = 0$
Let the given point $= P \left(x_1, y_1\right) = \left(0,0\right)$
Equation of given line: $\;$ $x \left(1 - \sqrt{3}\right) + y \left(1 + \sqrt{3}\right) + c = 0$
i.e. $\;$ $y = \left(\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}\right) x - \dfrac{c}{\sqrt{3} + 1}$
$\therefore \;$ Slope of given line $= m = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$
i.e. $\;$ $m = \dfrac{\left(\sqrt{3} - 1\right)^2}{\left(\sqrt{3} + 1\right) \left(\sqrt{3} - 1\right)} = \dfrac{3 + 1 - 2 \sqrt{3}}{3 - 1} = 2 - \sqrt{3}$
Let the angle between the given line and the required lines $= \alpha = 75^\circ$
Then the equations of the required two lines are
$y - y_1 = \left(\dfrac{m - \tan \alpha}{1 + m \tan \alpha}\right) \left(x - x_1\right)$ $\;$ and $\;$ $y - y_1 = \left(\dfrac{m + \tan \alpha}{1 - m \tan \alpha}\right) \left(x - x_1\right)$
$\therefore \;$ The equations of the two lines are
$y - 0 = \dfrac{2 - \sqrt{3} - \tan 75^\circ}{1 + \left(2 - \sqrt{3}\right) \times \tan 75^\circ} \left(x - 0\right)$ $\;$ and $\;$ $y - 0 = \dfrac{2 - \sqrt{3} + \tan 75^\circ}{1 - \left(2 - \sqrt{3}\right) \times \tan 75^\circ} \left(x - 0\right)$
i.e. $\;$ $y = \left[\dfrac{2 - \sqrt{3} - \left(2 + \sqrt{3}\right)}{1 + \left(2 - \sqrt{3}\right) \times \left(2 + \sqrt{3}\right)}\right] x$ $\;$ and $\;$ $y = \left[\dfrac{2 - \sqrt{3} + 2 + \sqrt{3}}{1 - \left(2 - \sqrt{3}\right) \times \left(2 + \sqrt{3}\right)}\right] x$
i.e. $\;$ $y = \left(\dfrac{-2 \sqrt{3}}{1 + 4 - 3}\right) x$ $\;$ and $\;$ $y = \left(\dfrac{4}{1 - 4 + 3}\right) x$
i.e. $\;$ $y = -\sqrt{3} x$ $\;$ and $\;$ $x = \left(\dfrac{0}{4}\right) y$
i.e. $\;$ $\sqrt{3} x + y = 0$ $\;$ and $\;$ $x = 0$