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

To determine the relationships between the lines, we start by expressing each line in slope-intercept form (\(y = mx + b\)), where \(m\) is the slope. **Line 1:** From \(4y = 3x + 5\), we divide by 4: \[ y = \frac{3}{4}x + \frac{5}{4} \] Slope (\(m_1\)) = \(\frac{3}{4}\) **Line 2:** Starting with \(6x + 8y = 6\), solve for \(y\): \[ 8y = -6x + 6 \] \[ y = -\frac{3}{4}x + \frac{3}{4} \] Slope (\(m_2\)) = \(-\frac{3}{4}\) **Line 3:** The equation is already in slope-intercept form: \[ y = \frac{4}{3}x - 3 \] Slope (\(m_3\)) = \(\frac{4}{3}\) **Comparing Slopes:** 1. **Line 1 and Line 2:** \(m_1 = \frac{3}{4}\) and \(m_2 = -\frac{3}{4}\) Since the product of the slopes \(m_1 \cdot m_2 = \frac{3}{4} \cdot -\frac{3}{4} = -\frac{9}{16} \neq -1\), they are **neither parallel nor perpendicular**. 2. **Line 1 and Line 3:** \(m_1 = \frac{3}{4}\) and \(m_3 = \frac{4}{3}\) Here, \(m_1 \cdot m_3 = \frac{3}{4} \cdot \frac{4}{3} = 1\), indicating they are **perpendicular**. 3. **Line 2 and Line 3:** \(m_2 = -\frac{3}{4}\) and \(m_3 = \frac{4}{3}\) The product \(m_2 \cdot m_3 = -\frac{3}{4} \cdot \frac{4}{3} = -1\), thus they are **perpendicular** as well. In summary: - Lines 1 and 2: Neither - Lines 1 and 3: Perpendicular - Lines 2 and 3: Perpendicular