The equations of three lines are given below. Line \( 1: 8 x+6 y=-2 \) Line \( 2: 4 y=-3 x+5 \) Line \( 3: y=-\frac{3}{4} x-4 \) For each pair of lines, determine whether they are parallel, perpendicular, or neither.

To determine the relationship between the lines (whether they are parallel, perpendicular, or neither), we need to find the slopes of each line. Lines are parallel if they have the same slope and perpendicular if the product of their slopes is -1. Let's rewrite each line in slope-intercept form \( y = mx + b \), where \( m \) is the slope. ### Line 1: \( 8x + 6y = -2 \) 1. Rearranging to solve for \( y \): \[ 6y = -8x - 2 \] \[ y = -\frac{8}{6}x - \frac{2}{6} \] \[ y = -\frac{4}{3}x - \frac{1}{3} \] - **Slope of Line 1**: \( m_1 = -\frac{4}{3} \) ### Line 2: \( 4y = -3x + 5 \) 1. Rearranging to solve for \( y \): \[ y = -\frac{3}{4}x + \frac{5}{4} \] - **Slope of Line 2**: \( m_2 = -\frac{3}{4} \) ### Line 3: \( y = -\frac{3}{4}x - 4 \) - The equation is already in slope-intercept form. - **Slope of Line 3**: \( m_3 = -\frac{3}{4} \) ### Comparing the slopes: 1. **Line 1 and Line 2**: - \( m_1 = -\frac{4}{3} \) - \( m_2 = -\frac{3}{4} \) - Product: \( m_1 \cdot m_2 = -\frac{4}{3} \cdot -\frac{3}{4} = 1 \) (not perpendicular) - Since \( m_1 \neq m_2 \), they are **neither parallel nor perpendicular**. 2. **Line 1 and Line 3**: - \( m_1 = -\frac{4}{3} \) - \( m_3 = -\frac{3}{4} \) - Product: \( m_1 \cdot m_3 = -\frac{4}{3} \cdot -\frac{3}{4} = 1 \) (not perpendicular) - Since \( m_1 \neq m_3 \), they are **neither parallel nor perpendicular**. 3. **Line 2 and Line 3**: - \( m_2 = -\frac{3}{4} \) - \( m_3 = -\frac{3}{4} \) - Since \( m_2 = m_3 \), they are **parallel**. ### Summary of Relationships: - Line 1 and Line 2: **Neither** - Line 1 and Line 3: **Neither** - Line 2 and Line 3: **Parallel**