Consider the line \( y=2 x-8 \) Find the equation of the line that is perpendicular to this line and passes through the point \( (-5,6) \). Find the equation of the line that is parallel to this line and passes through the point \( (-5,6) \).

To find a line that is perpendicular to \( y = 2x - 8 \), we need to determine the slope. The slope of the given line is \( 2 \). The slope of a line that is perpendicular to it will be the negative reciprocal, so we have \( -\frac{1}{2} \). Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point \( (-5, 6) \): \[ y - 6 = -\frac{1}{2}(x + 5) \] Distributing gives: \[ y - 6 = -\frac{1}{2}x - \frac{5}{2} \] Adding \( 6 \) will yield: \[ y = -\frac{1}{2}x + \frac{7}{2} \] Now, for the parallel line, we use the same slope \( 2 \) and the point \( (-5, 6) \): \[ y - 6 = 2(x + 5) \] Distributing gives: \[ y - 6 = 2x + 10 \] Adding \( 6 \) gives: \[ y = 2x + 16 \] Thus, the equations of the lines are: 1. Perpendicular: \( y = -\frac{1}{2}x + \frac{7}{2} \) 2. Parallel: \( y = 2x + 16 \)