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

To determine the relationship between the pairs of lines, let's convert all lines to the slope-intercept form \( y = mx + b \) and compare their slopes. **Line 1:** Already given as \( y = -\frac{2}{3} x - 6 \). The slope (m1) is \( -\frac{2}{3} \). **Line 2:** Let's rewrite it: \[ 6x - 4y = -6 \implies -4y = -6 - 6x \implies y = \frac{3}{2} x + \frac{3}{2} \] The slope (m2) is \( \frac{3}{2} \). **Line 3:** Rewrite it as well: \[ -2y = 3x + 7 \implies y = -\frac{3}{2} x - \frac{7}{2} \] The slope (m3) is \( -\frac{3}{2} \). Now, let's analyze the pairs of lines: 1. **Line 1 and Line 2:** - \( m1 = -\frac{2}{3} \) and \( m2 = \frac{3}{2} \) - Since the product of their slopes \( (-\frac{2}{3}) \times (\frac{3}{2}) = -1 \), they are **Perpendicular.** 2. **Line 1 and Line 3:** - \( m1 = -\frac{2}{3} \) and \( m3 = -\frac{3}{2} \) - Their slopes are both negative and different, so they are **Neither** parallel nor perpendicular. 3. **Line 2 and Line 3:** - \( m2 = \frac{3}{2} \) and \( m3 = -\frac{3}{2} \) - The product of their slopes \( (\frac{3}{2}) \times (-\frac{3}{2}) = -\frac{9}{4} \), so they are **Neither** parallel nor perpendicular. In conclusion: - **Line 1 and Line 2:** \( \bigcirc \) Parallel \( \bigcirc \) Perpendicular ✅ \( \bigcirc \) Neither - **Line 1 and Line 3:** \( \bigcirc \) Parallel \( \bigcirc \) Perpendicular \( \bigcirc \) Neither ✅ - **Line 2 and Line 3:** \( \bigcirc \) Parallel \( \bigcirc \) Perpendicular \( \bigcirc \) Neither ✅