Consider the line \( y=7 x-2 \). (a) Find the equation of the line that is parallel to this line and passes through the point \( (-3,5) \). (b) Find the equation of the line that is perpendicular to this line and passes through the point \( (-3,5) \).

To find the parallel line, recall that parallel lines have the same slope. The slope from the equation \( y = 7x - 2 \) is \( 7 \). Using point-slope form \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) = (-3, 5) \), we have: \[ y - 5 = 7(x + 3) \] Distributing gives: \[ y - 5 = 7x + 21 \] Adding \( 5 \) to both sides yields: \[ y = 7x + 26 \] Now for the perpendicular line, the slope of a line perpendicular to \( y = 7x - 2 \) is the negative reciprocal of \( 7 \), which is \( -\frac{1}{7} \). Using the same point-slope form again: \[ y - 5 = -\frac{1}{7}(x + 3) \] Distributing gives: \[ y - 5 = -\frac{1}{7}x - \frac{3}{7} \] Adding \( 5 \) (or \( \frac{35}{7} \)) to both sides yields: \[ y = -\frac{1}{7}x + \frac{32}{7} \] ### Summary (a) The equation of the line parallel to \( y = 7x - 2 \) passing through \( (-3, 5) \) is \( y = 7x + 26 \). (b) The equation of the line perpendicular to \( y = 7x - 2 \) passing through \( (-3, 5) \) is \( y = -\frac{1}{7}x + \frac{32}{7} \).