Consider the line \( y=\frac{3}{7} x+6 \). Find the equation of the line that is parallel to this line and passes through the point \( (-9,-5) \). Find the equation of the line that is perpendicular to this line and passes through the point \( (-9,-5) \). Note that the ALEKS graphing calculator may be helpful in checking your answer. Equation of parallel line:

The slope of the line \( y=\frac{3}{7} x+6 \) is \( \frac{3}{7} \). A parallel line will have the same slope. To find the equation of the parallel line that passes through the point \( (-9,-5) \), we can use the point-slope form: \[ y - y_1 = m(x - x_1) \] Substituting \( m = \frac{3}{7} \), \( x_1 = -9 \), and \( y_1 = -5 \): \[ y - (-5) = \frac{3}{7}(x - (-9)) \] Simplifying, \[ y + 5 = \frac{3}{7}(x + 9) \] This gives: \[ y = \frac{3}{7} x + \frac{27}{7} - 5 \] Now converting -5 to have a common denominator: \[ -5 = -\frac{35}{7} \] Thus, \[ y = \frac{3}{7} x + \frac{27}{7} - \frac{35}{7} = \frac{3}{7} x - \frac{8}{7} \] So, the equation of the parallel line is: \[ y = \frac{3}{7} x - \frac{8}{7} \] Now, for the equation of the line that is perpendicular to \( y=\frac{3}{7} x+6 \), we first determine the negative reciprocal of the slope \( \frac{3}{7} \), which is \( -\frac{7}{3} \). Using the point-slope form again for the point \( (-9,-5) \): \[ y - (-5) = -\frac{7}{3}(x - (-9)) \] This simplifies to: \[ y + 5 = -\frac{7}{3}(x + 9) \] Multiplying through by 3 to eliminate the fraction gives: \[ 3(y + 5) = -7(x + 9) \] Distributing: \[ 3y + 15 = -7x - 63 \] Now isolating \( y \): \[ 3y = -7x - 63 - 15 \] \[ 3y = -7x - 78 \] \[ y = -\frac{7}{3} x - 26 \] Thus, the equation of the perpendicular line is: \[ y = -\frac{7}{3} x - 26 \]