1. **Problem statement:** Determine if the pairs of lines are parallel, perpendicular, or neither. 2. **Recall:** For lines in the form $Ax+By+C=0$, the slope is $m=-\frac{A}{B}$. 3. **Rules:** - Lines are parallel if their slopes are equal. - Lines are perpendicular if the product of their slopes is $-1$. --- **a)** Lines: $2x+y-3=0$ and $4x+2y+5=0$ - Slope of first line: $m_1 = -\frac{2}{1} = -2$ - Slope of second line: $m_2 = -\frac{4}{2} = -2$ Since $m_1 = m_2 = -2$, the lines are **parallel**. --- **b)** Lines: $3y=2x+5$ and $3x+2y-8=0$ Rewrite first line: $3y=2x+5 \Rightarrow y=\frac{2}{3}x + \frac{5}{3}$ - Slope of first line: $m_1 = \frac{2}{3}$ - Slope of second line: $m_2 = -\frac{3}{2}$ (from $3x+2y-8=0 \Rightarrow 2y=-3x+8 \Rightarrow y=-\frac{3}{2}x + 4$) Check product: $m_1 \times m_2 = \frac{2}{3} \times -\frac{3}{2} = -1$ Since product is $-1$, the lines are **perpendicular**. --- **c)** Lines: $4y+2x-1=0$ and $x-2y-7=0$ Rewrite first line: $4y = -2x + 1 \Rightarrow y = -\frac{2}{4}x + \frac{1}{4} = -\frac{1}{2}x + \frac{1}{4}$ - Slope of first line: $m_1 = -\frac{1}{2}$ - Slope of second line: $m_2 = \frac{1}{2}$ (from $x - 2y -7=0 \Rightarrow -2y = -x + 7 \Rightarrow y = \frac{1}{2}x - \frac{7}{2}$) Check product: $m_1 \times m_2 = -\frac{1}{2} \times \frac{1}{2} = -\frac{1}{4} \neq -1$ Slopes are not equal and product is not $-1$, so lines are **neither parallel nor perpendicular**. --- **Final answers:** - a) Parallel - b) Perpendicular - c) Neither