The equations of three lines are given below. Line 1: \( 3 y=5 x+4 \) Line 2: \( y=\frac{5}{3} x-7 \) Line 3: \( 6 x-10 y=-4 \) For each pair of lines, determine whether they are parallel, perpendicular, or neither.

To determine the relationships between the lines, we need to convert all the equations into slope-intercept form (\(y = mx + b\)), where \(m\) represents the slope. 1. **Line 1:** \[ 3y = 5x + 4 \quad \Rightarrow \quad y = \frac{5}{3}x + \frac{4}{3} \] - Slope (m) = \(\frac{5}{3}\) 2. **Line 2:** \[ y = \frac{5}{3}x - 7 \] - Slope (m) = \(\frac{5}{3}\) 3. **Line 3:** \[ 6x - 10y = -4 \quad \Rightarrow \quad -10y = -6x - 4 \quad \Rightarrow \quad y = \frac{3}{5}x + \frac{2}{5} \] - Slope (m) = \(\frac{3}{5}\) Now, let's analyze each pair of lines: **Line 1 and Line 2:** Both have slopes of \(\frac{5}{3}\). Since they have identical slopes, they are **parallel**. **Line 1 and Line 3:** Line 1 has a slope of \(\frac{5}{3}\), and Line 3 has a slope of \(\frac{3}{5}\). The product of the slopes is \(\frac{5}{3} \cdot \frac{3}{5} = 1\), indicating that they are **perpendicular**. **Line 2 and Line 3:** Again, Line 2 has a slope of \(\frac{5}{3}\) and Line 3 has a slope of \(\frac{3}{5}\). As established, \(\frac{5}{3} \cdot \frac{3}{5} = 1\), meaning these lines are also **perpendicular**. In summary: - Line 1 and Line 2: **Parallel** - Line 1 and Line 3: **Perpendicular** - Line 2 and Line 3: **Perpendicular**