1. **Problem statement:** Find the relative position of the lines $\varepsilon_1$ and $\varepsilon_2$ in case (α): $\varepsilon_1: 2x - 4y = 4$ $\varepsilon_2: 2x + 4y - 2 = 0$ 2. **Rewrite equations in slope-intercept form $y = mx + b$ to compare slopes:** For $\varepsilon_1$: $$2x - 4y = 4 \implies -4y = -2x + 4 \implies y = \frac{-2x + 4}{-4} = \frac{1}{2}x - 1$$ For $\varepsilon_2$: $$2x + 4y - 2 = 0 \implies 4y = -2x + 2 \implies y = \frac{-2x + 2}{4} = -\frac{1}{2}x + \frac{1}{2}$$ 3. **Compare slopes:** Slope of $\varepsilon_1$ is $m_1 = \frac{1}{2}$ Slope of $\varepsilon_2$ is $m_2 = -\frac{1}{2}$ 4. **Determine relative position:** - If slopes are equal and intercepts differ, lines are parallel. - If slopes are negative reciprocals, lines are perpendicular. - Otherwise, lines intersect at some angle. Since $m_1 = \frac{1}{2}$ and $m_2 = -\frac{1}{2}$, and $m_1 \cdot m_2 = \frac{1}{2} \times -\frac{1}{2} = -\frac{1}{4} \neq -1$, the lines are neither parallel nor perpendicular. 5. **Find intersection point to confirm they intersect:** Solve system: $$\begin{cases} y = \frac{1}{2}x - 1 \\ y = -\frac{1}{2}x + \frac{1}{2} \end{cases}$$ Set equal: $$\frac{1}{2}x - 1 = -\frac{1}{2}x + \frac{1}{2}$$ Add $\frac{1}{2}x$ to both sides: $$\frac{1}{2}x + \frac{1}{2}x - 1 = \frac{1}{2}$$ $$x - 1 = \frac{1}{2}$$ Add 1 to both sides: $$x = \frac{3}{2}$$ Find $y$: $$y = \frac{1}{2} \times \frac{3}{2} - 1 = \frac{3}{4} - 1 = -\frac{1}{4}$$ 6. **Conclusion:** The lines intersect at point $\left( \frac{3}{2}, -\frac{1}{4} \right)$ and are neither parallel nor perpendicular. **Final answer:** The lines $\varepsilon_1$ and $\varepsilon_2$ intersect at $\left( \frac{3}{2}, -\frac{1}{4} \right)$ and are neither parallel nor perpendicular.