1. **Problem:** Determine if the pairs of lines are parallel, perpendicular, or neither. 2. **Recall:** - Lines are parallel if their gradients (slopes) are equal. - Lines are perpendicular if the product of their gradients is -1. - Gradient of a line perpendicular to one with gradient $m$ is $-\frac{1}{m}$. --- **Exercise 5:** **1.** Given lines: $$y=3x-4$$ $$y=3x+7$$ Gradients: $m_1=3$, $m_2=3$ Since $m_1=m_2$, lines are **parallel**. **2.** Given lines: $$y=4x+7$$ $$y=\frac{1}{4}x-5$$ Gradients: $m_1=4$, $m_2=\frac{1}{4}$ Product: $4 \times \frac{1}{4} = 1 \neq -1$ Lines are **neither** parallel nor perpendicular. **3.** Given lines: $$y = -\frac{1}{3}x + 2$$ $$x + 3y + 2 = 0$$ Rewrite second line in slope-intercept form: $$3y = -x - 2$$ $$y = -\frac{1}{3}x - \frac{2}{3}$$ Gradients: $m_1 = -\frac{1}{3}$, $m_2 = -\frac{1}{3}$ Since $m_1 = m_2$, lines are **parallel**. **4.** Given lines: $$2x + 3y + 3 = 0$$ $$3x - 2y + 6 = 0$$ Rewrite both in slope-intercept form: $$3y = -2x - 3$$ $$y = -\frac{2}{3}x - 1$$ $$-2y = -3x - 6$$ $$y = \frac{3}{2}x + 3$$ Gradients: $m_1 = -\frac{2}{3}$, $m_2 = \frac{3}{2}$ Product: $$-\frac{2}{3} \times \frac{3}{2} = -1$$ Lines are **perpendicular**. --- **Final answers for Exercise 5:** 1. Parallel 2. Neither 3. Parallel 4. Perpendicular