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

To analyze the relationship between the lines, we first need to put them in slope-intercept form \(y = mx + b\), where \(m\) is the slope. **Line 1:** Starting with the equation \(4y = -3x + 5\), divide by 4. \[y = -\frac{3}{4}x + \frac{5}{4}\] Slope (m1) = \(-\frac{3}{4}\). **Line 2:** From \(8x - 6y = -2\), rearranging gives us \(6y = 8x + 2\), then divide by 6. \[y = \frac{4}{3}x + \frac{1}{3}\] Slope (m2) = \(\frac{4}{3}\). **Line 3:** The equation \(y = -\frac{3}{4}x - 6\) is already in slope-intercept form. Slope (m3) = \(-\frac{3}{4}\). Now, we form pairs and analyze their slopes: 1. **Line 1 and Line 2:** Slopes are \(-\frac{3}{4}\) and \(\frac{4}{3}\). The product of the slopes: \(-\frac{3}{4} \cdot \frac{4}{3} = -1\). Thus, Lines 1 and 2 are **perpendicular**. 2. **Line 1 and Line 3:** Both have a slope of \(-\frac{3}{4}\). Since their slopes are equal, Lines 1 and 3 are **parallel**. 3. **Line 2 and Line 3:** Slopes are \(\frac{4}{3}\) and \(-\frac{3}{4}\). The product of the slopes: \(\frac{4}{3} \cdot -\frac{3}{4} = -1\). Thus, Lines 2 and 3 are also **perpendicular**. In summary: - Lines 1 and 2: Perpendicular - Lines 1 and 3: Parallel - Lines 2 and 3: Perpendicular