The equations of three lines are given below. \[ \begin{array}{l}\text { Line } 1: y=\frac{4}{3} x+6 \\ \text { Line } 2: 3 y=4 x+5 \\ \text { Line } 3: 6 x-8 y=8 \\ \text { Line } 1 \text { and Line 2: Parallel. Perpendicular Neither } 1 \text { and Line 3: Parallel Perpendic. in Neither } \\ \text { Line } 2 \text { and Line 3: Parallel Perpendicular Neither }\end{array} \]

To determine the relationships between the lines, we should first express them in the slope-intercept form \(y = mx + b\), where \(m\) is the slope. - **Line 1:** Already in the form \(y = \frac{4}{3}x + 6\), so the slope \(m_1 = \frac{4}{3}\). - **Line 2:** Rearranging \(3y = 4x + 5\) gives \(y = \frac{4}{3}x + \frac{5}{3}\), so the slope \(m_2 = \frac{4}{3}\). - **Line 3:** Rearranging \(6x - 8y = 8\) gives \(8y = 6x - 8\) or \(y = \frac{3}{4}x - 1\), so the slope \(m_3 = \frac{3}{4}\). **Line 1 and Line 2:** Since their slopes \(m_1\) and \(m_2\) are equal, they are *parallel*. **Line 1 and Line 3:** \(m_1 = \frac{4}{3}\) and \(m_3 = \frac{3}{4}\). The product of the slopes \(m_1 \times m_3 = \frac{4}{3} \times \frac{3}{4} = 1\), hence they are *perpendicular*. **Line 2 and Line 3:** Again, using the slopes \(m_2 = \frac{4}{3}\) and \(m_3 = \frac{3}{4}\), the product \(m_2 \times m_3 = \frac{4}{3} \times \frac{3}{4} = 1\), making them *perpendicular* as well. In summary: - Line 1 and Line 2: Parallel - Line 1 and Line 3: Perpendicular - Line 2 and Line 3: Perpendicular