1. **Problem statement:** We are given two lines where the slope of one line is double the slope of the other. The tangent of the angle between these two lines is $\frac{1}{3}$. We need to find the slopes of both lines. 2. **Formula used:** The tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ 3. **Given:** Let the slope of the first line be $m$. Then the slope of the second line is $2m$. 4. Substitute into the formula: $$\frac{1}{3} = \left| \frac{m - 2m}{1 + m \cdot 2m} \right| = \left| \frac{-m}{1 + 2m^2} \right| = \frac{|m|}{|1 + 2m^2|}$$ 5. Since $1 + 2m^2 > 0$ for all real $m$, we can write: $$\frac{1}{3} = \frac{|m|}{1 + 2m^2}$$ 6. Multiply both sides by $1 + 2m^2$: $$1 + 2m^2 = 3|m|$$ 7. Let $x = |m|$, then: $$1 + 2x^2 = 3x$$ 8. Rearrange to form a quadratic equation: $$2x^2 - 3x + 1 = 0$$ 9. Solve the quadratic using the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=2$, $b=-3$, $c=1$: $$x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}$$ 10. So the two possible values for $x$ are: $$x_1 = \frac{3 + 1}{4} = 1$$ $$x_2 = \frac{3 - 1}{4} = \frac{1}{2}$$ 11. Recall $x = |m|$, so $m = \pm 1$ or $m = \pm \frac{1}{2}$. 12. Since the slope of the second line is $2m$, the pairs of slopes are: - If $m = 1$, then slopes are $1$ and $2$. - If $m = -1$, then slopes are $-1$ and $-2$. - If $m = \frac{1}{2}$, then slopes are $\frac{1}{2}$ and $1$. - If $m = -\frac{1}{2}$, then slopes are $-\frac{1}{2}$ and $-1$. 13. These pairs satisfy the condition that the tangent of the angle between the lines is $\frac{1}{3}$. **Final answer:** The slopes of the lines are either $(1, 2)$, $(-1, -2)$, $(\frac{1}{2}, 1)$, or $(-\frac{1}{2}, -1)$.