Two lines pass through the point $(2,3)$ and intersect each other at an angle of $60^{\circ}$. If the slope of one line is $2$,find the equations of the other line.

(N/A) Let the slope of the first line be $m_{1} = 2$ and the slope of the other line be $m_{2}$.
The angle $\theta$ between the two lines is given by $\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$.
Substituting $\theta = 60^{\circ}$ and $m_{1} = 2$:
$\tan 60^{\circ} = \left| \frac{2 - m_{2}}{1 + 2m_{2}} \right|$ $\Rightarrow \sqrt{3} = \left| \frac{2 - m_{2}}{1 + 2m_{2}} \right|$.
This gives two cases:
Case $I$: $\frac{2 - m_{2}}{1 + 2m_{2}} = \sqrt{3}$ $\Rightarrow 2 - m_{2} = \sqrt{3} + 2\sqrt{3}m_{2}$ $\Rightarrow m_{2}(1 + 2\sqrt{3}) = 2 - \sqrt{3}$ $\Rightarrow m_{2} = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}$.
The equation of the line passing through $(2,3)$ with this slope is $(y - 3) = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}(x - 2)$.
Simplifying,$(1 + 2\sqrt{3})y - 3(1 + 2\sqrt{3}) = (2 - \sqrt{3})x - 2(2 - \sqrt{3})$ $\Rightarrow (2 - \sqrt{3})x - (1 + 2\sqrt{3})y + (3 + 6\sqrt{3} - 4 + 2\sqrt{3}) = 0$ $\Rightarrow (2 - \sqrt{3})x - (1 + 2\sqrt{3})y + (8\sqrt{3} - 1) = 0$.
Case $II$: $\frac{2 - m_{2}}{1 + 2m_{2}} = -\sqrt{3}$ $\Rightarrow 2 - m_{2} = -\sqrt{3} - 2\sqrt{3}m_{2}$ $\Rightarrow m_{2}(2\sqrt{3} - 1) = -2 - \sqrt{3}$ $\Rightarrow m_{2} = -\frac{2 + \sqrt{3}}{2\sqrt{3} - 1}$.
The equation of the line passing through $(2,3)$ with this slope is $(y - 3) = -\frac{2 + \sqrt{3}}{2\sqrt{3} - 1}(x - 2)$.
Simplifying,$(2\sqrt{3} - 1)y - 3(2\sqrt{3} - 1) = -(2 + \sqrt{3})x + 2(2 + \sqrt{3})$ $\Rightarrow (2 + \sqrt{3})x + (2\sqrt{3} - 1)y - (4 + 2\sqrt{3} + 6\sqrt{3} - 3) = 0$ $\Rightarrow (2 + \sqrt{3})x + (2\sqrt{3} - 1)y - (1 + 8\sqrt{3}) = 0$.