A straight line $L$ through the point $(3, -2)$ is inclined at an angle $60^\circ$ to the line $\sqrt{3}x + y = 1$. If $L$ also intersects the $x$-axis, then the equation of $L$ is

$\text{Given: } L \text{ passes through } (3,\, -2)$ $L \text{ makes } 60^\circ \text{ with } \sqrt{3}x + y = 1$ $L \text{ intersects the } x\text{-axis}$ $\text{To find: the equation of } L$ $\text{(from the four given options)}$ $1.\ \text{slope of the given line}$ $2.\ \text{angle formula for } L$ $3.\ x\text{-axis condition, then the equation}$ $\sqrt{3}x + y = 1 \ \Rightarrow \ y = -\sqrt{3}x + 1$ $m_1 = -\sqrt{3}$ $\tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right|$ $\sqrt{3} = \left|\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2}\right|$ $m_2 + \sqrt{3} = \sqrt{3} - 3m_2$ $4m_2 = 0 \ \Rightarrow \ m_2 = 0$ $m_2 + \sqrt{3} = -\sqrt{3} + 3m_2$ $-2m_2 = -2\sqrt{3} \ \Rightarrow \ m_2 = \sqrt{3}$ $m_2 = 0 \quad \text{or} \quad m_2 = \sqrt{3}$ $\text{inclination of given line} = 120^\circ$ $\theta_L = 60^\circ \text{ or } 180^\circ \ \Rightarrow \ m_2 = \sqrt{3} \text{ or } 0$ $m_2 = 0 \Rightarrow y = -2 \ \text{(parallel to } x\text{-axis)} \Rightarrow \text{discard}$ $y - (-2) = \sqrt{3}\,(x - 3)$ $y + 2 = \sqrt{3}x - 3\sqrt{3}$ $y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$ $\boxed{\text{Answer: (2)}\ \ y - \sqrt{3}x + 2 + 3\sqrt{3} = 0}$