The equations of three lines are given below. Line \( 1: y=-\frac{4}{3} x+7 \) Line \( 2:-4 y=3 x+2 \) Line \( 3: 6 x-8 y=-8 \) 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. ### Line 1: The equation is given as: \[ y = -\frac{4}{3} x + 7 \] The slope \( m_1 \) of Line 1 is: \[ m_1 = -\frac{4}{3} \] ### Line 2: The equation is given as: \[ -4y = 3x + 2 \] We can rearrange this into slope-intercept form \( y = mx + b \): \[ y = -\frac{3}{4}x - \frac{1}{2} \] The slope \( m_2 \) of Line 2 is: \[ m_2 = -\frac{3}{4} \] ### Line 3: The equation is given as: \[ 6x - 8y = -8 \] We can rearrange this into slope-intercept form: \[ -8y = -6x - 8 \quad \Rightarrow \quad y = \frac{3}{4}x + 1 \] The slope \( m_3 \) of Line 3 is: \[ m_3 = \frac{3}{4} \] ### Comparing the Slopes: 1. **Line 1 and Line 2:** - \( m_1 = -\frac{4}{3} \) - \( m_2 = -\frac{3}{4} \) - The product of the slopes: \[ m_1 \cdot m_2 = -\frac{4}{3} \cdot -\frac{3}{4} = 1 \] Since the product of the slopes is 1, Line 1 and Line 2 are **perpendicular**. 2. **Line 1 and Line 3:** - \( m_1 = -\frac{4}{3} \) - \( m_3 = \frac{3}{4} \) - The product of the slopes: \[ m_1 \cdot m_3 = -\frac{4}{3} \cdot \frac{3}{4} = -1 \] Since the product of the slopes is -1, Line 1 and Line 3 are **perpendicular**. 3. **Line 2 and Line 3:** - \( m_2 = -\frac{3}{4} \) - \( m_3 = \frac{3}{4} \) - The product of the slopes: \[ m_2 \cdot m_3 = -\frac{3}{4} \cdot \frac{3}{4} = -\frac{9}{16} \] Since the product of the slopes is not -1 or 0, Line 2 and Line 3 are **neither** parallel nor perpendicular. ### Summary: - Line 1 and Line 2: **Perpendicular** - Line 1 and Line 3: **Perpendicular** - Line 2 and Line 3: **Neither**