The slope of a line is double of the slope of another line.

Solution:

Let $m_{1}$ and $m$ be the slopes of the two given lines such that $m_{1}=2 m$.

We know that if $\theta$ isthe angle between the lines $I_{1}$ and $I_{2}$ with slopes $m_{1}$ and $m_{2}$, then $\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|$.

It is given that the tangent of the angle between the two lines is $\frac{1}{3}$.

$\therefore \frac{1}{3}=\left|\frac{m-2 m}{1+(2 m) \cdot m}\right|$

$\Rightarrow \frac{1}{3}=\left|\frac{-m}{1+2 m^{2}}\right|$

$\Rightarrow \frac{1}{3}=\frac{-m}{1+2 m^{2}}$ or $\frac{1}{3}=-\left(\frac{-m}{1+2 m^{2}}\right)=\frac{m}{1+2 m^{2}}$

Case I

$\Rightarrow \frac{1}{3}=\frac{-m}{1+2 m^{2}}$

$\Rightarrow 1+2 m^{2}=-3 m$

$\Rightarrow 2 m^{2}+3 m+1=0$

$\Rightarrow 2 m^{2}+2 m+m+1=0$

$\Rightarrow 2 m(m+1)+1(m+1)=0$

$\Rightarrow(m+1)(2 m+1)=0$

$\Rightarrow m=-1$ or $m=-\frac{1}{2}$

If $m=-1$, then the slopes of the lines are $-1$ and $-2$.

If $m=-\frac{1}{2}$, then the slopes of the lines are $-\frac{1}{2}$ and $-1 .$

Case II

$\frac{1}{3}=\frac{m}{1+2 m^{2}}$

$\Rightarrow 2 m^{2}+1=3 m$

$\Rightarrow 2 m^{2}-3 m+1=0$

$\Rightarrow 2 m^{2}-2 m-m+1=0$

$\Rightarrow 2 m(m-1)-1(m-1)=0$

$\Rightarrow(m-1)(2 m-1)=0$

$\Rightarrow m=1$ or $m=\frac{1}{2}$

If $m=1$, then the slopes of the lines are 1 and 2 .

If $m=\frac{1}{2}$, then the slopes of the lines are $\frac{1}{2}$ and 1 .

Hence, the slopes of the lines are $-1$ and $-2$ or $-\frac{1}{2}$ and $-1$ or 1 and 2 or $\frac{1}{2}$ and 1 .