Consider the line \( 4 x+7 y=-6 \). Find the equation of the line that is parallel to this line and passes through the point \( (-8,-1) \). Find the equation of the line that is perpendicular to this line and passes through the point \( (-8,-1) \).

To find the equation of the line that is parallel to \( 4x + 7y = -6 \) and passes through the point \( (-8, -1) \), we start by determining the slope of the original line. We rearrange the equation into slope-intercept form \( y = mx + b \): \[ 7y = -4x - 6 \implies y = -\frac{4}{7}x - \frac{6}{7} \] The slope (m) is \(-\frac{4}{7}\). Since parallel lines have the same slope, the equation for the parallel line can be written using point-slope form: \[ y - (-1) = -\frac{4}{7}(x - (-8)) \implies y + 1 = -\frac{4}{7}(x + 8) \] Distributing gives: \[ y + 1 = -\frac{4}{7}x - \frac{32}{7} \] Subtracting 1 from both sides: \[ y = -\frac{4}{7}x - \frac{32}{7} - \frac{7}{7} = -\frac{4}{7}x - \frac{39}{7} \] Thus, the equation for the parallel line is: \[ y = -\frac{4}{7}x - \frac{39}{7} \] For the perpendicular line, the slope will be the negative reciprocal of \(-\frac{4}{7}\), which is \(\frac{7}{4}\). Using the point \((-8, -1)\) in point-slope form: \[ y - (-1) = \frac{7}{4}(x - (-8)) \implies y + 1 = \frac{7}{4}(x + 8) \] Distributing gives: \[ y + 1 = \frac{7}{4}x + \frac{56}{4} \implies y + 1 = \frac{7}{4}x + 14 \] Subtracting 1 from both sides: \[ y = \frac{7}{4}x + 14 - 1 = \frac{7}{4}x + 13 \] Thus, the equation for the perpendicular line is: \[ y = \frac{7}{4}x + 13 \] In summary, the equations are: - Parallel line: \( y = -\frac{4}{7}x - \frac{39}{7} \) - Perpendicular line: \( y = \frac{7}{4}x + 13 \)