9. Are the following lines parallel, perpendicular or neither? a) \( y=3 x+2 \) and \( y=-\frac{1}{3} x+4 \) b) \( y=\frac{1}{2} x-1 \) and \( 2 y-x=2 \) c) \( 3 y-x=9 \) and \( y=3 x-7 \) d) \( 2 y+3 x=8 \) and \( 3 y=2 x+1 \)

To determine the relationship between the lines (parallel, perpendicular, or neither), we need to analyze their slopes. a) The slopes of the lines are \(3\) and \(-\frac{1}{3}\). Since the product of these slopes is \(-1\) (which means they are negative reciprocals), the lines are perpendicular. b) The slope of the first line is \(\frac{1}{2}\). To find the slope of the second line, we rewrite it in slope-intercept form: \[ 2y = x + 2 \Rightarrow y = \frac{1}{2}x + 1 \] Both lines have the same slope \(\frac{1}{2}\), which means they are parallel. c) For the line \(3y - x = 9\), we can rewrite it as: \[ y = \frac{1}{3}x + 3 \] So, the slopes of the lines are \(\frac{1}{3}\) and \(3\). The product of these slopes is \(1\), meaning they are neither parallel nor perpendicular. d) For the first line \(2y + 3x = 8\), we rewrite it as: \[ y = -\frac{3}{2}x + 4 \] And the second line \(3y = 2x + 1\) can be rewritten as: \[ y = \frac{2}{3}x + \frac{1}{3} \] Here, the slopes are \(-\frac{3}{2}\) and \(\frac{2}{3}\). Since the product of the slopes is \(-1\), these lines are also perpendicular. In summary: - a) Perpendicular - b) Parallel - c) Neither - d) Perpendicular