The slope of a line is double the slope of another line. If the tangent of the angle between them is $\frac{1}{3},$ find the slopes of the lines.

(N/A) Let $m_{1}$ and $m_{2}$ be the slopes of the two lines such that $m_{1} = 2m_{2}$.
We know that if $\theta$ is the angle between two lines with slopes $m_{1}$ and $m_{2},$ then $\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$.
Given $\tan \theta = \frac{1}{3},$ we have $\frac{1}{3} = \left| \frac{2m_{2} - m_{2}}{1 + (2m_{2})m_{2}} \right| = \left| \frac{m_{2}}{1 + 2m_{2}^{2}} \right|$.
This implies $\frac{m_{2}}{1 + 2m_{2}^{2}} = \frac{1}{3}$ or $\frac{m_{2}}{1 + 2m_{2}^{2}} = -\frac{1}{3}$.
Case $I$: $\frac{m_{2}}{1 + 2m_{2}^{2}} = \frac{1}{3}$ $\Rightarrow 2m_{2}^{2} - 3m_{2} + 1 = 0$ $\Rightarrow (2m_{2} - 1)(m_{2} - 1) = 0$.
So,$m_{2} = \frac{1}{2}$ (giving $m_{1} = 1$) or $m_{2} = 1$ (giving $m_{1} = 2$).
Case $II$: $\frac{m_{2}}{1 + 2m_{2}^{2}} = -\frac{1}{3}$ $\Rightarrow 2m_{2}^{2} + 3m_{2} + 1 = 0$ $\Rightarrow (2m_{2} + 1)(m_{2} + 1) = 0$.
So,$m_{2} = -\frac{1}{2}$ (giving $m_{1} = -1$) or $m_{2} = -1$ (giving $m_{1} = -2$).
Thus,the possible pairs of slopes are $(1, \frac{1}{2}), (2, 1), (-1, -\frac{1}{2}), (-2, -1)$.