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.
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To determine the relationships between the lines, we first need to find their slopes. For Line 1: \( 4y = 3x + 5 \) Divide by 4 to put it in slope-intercept form: \[ y = \frac{3}{4}x + \frac{5}{4} \] Here, the slope (m1) is \( \frac{3}{4} \). For Line 2: \( 6x + 8y = 6 \) Rearranging gives: \[ 8y = -6x + 6 \] Dividing by 8: \[ y = -\frac{3}{4}x + \frac{3}{4} \] Here, the slope (m2) is \( -\frac{3}{4} \). For Line 3: \( y = \frac{4}{3}x - 3 \) The slope (m3) is \( \frac{4}{3} \). Now let's analyze the pairs: 1. **Line 1 and Line 2**: - m1 = \( \frac{3}{4} \) and m2 = \( -\frac{3}{4} \). - Since the product of the slopes \( m1 \cdot m2 = \frac{3}{4} \cdot \left(-\frac{3}{4}\right) = -\frac{9}{16} \), they are neither parallel nor perpendicular. 2. **Line 1 and Line 3**: - m1 = \( \frac{3}{4} \) and m3 = \( \frac{4}{3} \). - The product of the slopes \( m1 \cdot m3 = \frac{3}{4} \cdot \frac{4}{3} = 1 \), indicating that the lines are perpendicular. 3. **Line 2 and Line 3**: - m2 = \( -\frac{3}{4} \) and m3 = \( \frac{4}{3} \). - The product of the slopes \( m2 \cdot m3 = -\frac{3}{4} \cdot \frac{4}{3} = -1 \), indicating that these lines are also perpendicular. To summarize: - Line 1 and Line 2: Neither parallel nor perpendicular - Line 1 and Line 3: Perpendicular - Line 2 and Line 3: Perpendicular