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

To determine whether the lines are parallel, perpendicular, or neither, we need to first convert each equation into the slope-intercept form, \(y = mx + b\), where \(m\) is the slope. **Line 1:** \( 4x + 6y = -4\) Solving for \(y\): \[ 6y = -4x - 4 \quad \Rightarrow \quad y = -\frac{2}{3}x - \frac{2}{3} \] So the slope \(m_1 = -\frac{2}{3}\). **Line 2:** \( 2y = -3x + 5\) Solving for \(y\): \[ y = -\frac{3}{2}x + \frac{5}{2} \] So the slope \(m_2 = -\frac{3}{2}\). **Line 3:** \(y = -\frac{3}{2}x + 8\) The slope here is \(m_3 = -\frac{3}{2}\). Now, let's analyze the slopes: - For Line 1 and Line 2: - \( m_1 = -\frac{2}{3} \) and \( m_2 = -\frac{3}{2} \) - Since \(m_1 \cdot m_2 = -\frac{2}{3} \cdot -\frac{3}{2} = 1\), they are **perpendicular**. - For Line 1 and Line 3: - \( m_1 = -\frac{2}{3} \) and \( m_3 = -\frac{3}{2} \) - Again, since \(m_1 \cdot m_3 = -\frac{2}{3} \cdot -\frac{3}{2} = 1\), they are **perpendicular**. - For Line 2 and Line 3: - \( m_2 = -\frac{3}{2} \) and \( m_3 = -\frac{3}{2} \) - Since \(m_2 = m_3\), they are **parallel**. So the answers are: Line 1 and Line 2: **Perpendicular** Line 1 and Line 3: **Perpendicular** Line 2 and Line 3: **Parallel**