Find the equations of the lines that pass through the point \( (3,8) \) and are parallel to and perpendicular to the line with equation \( y+7 x=2 \). Parallel: \( y= \) Perpendicular: \( y= \)

First, let's rewrite the given line equation \( y + 7x = 2 \) in slope-intercept form \( y = mx + b \). Rearranging gives us: \[ y = -7x + 2. \] The slope \( m \) of this line is \( -7 \). Lines that are parallel to this line will have the same slope. Therefore, the equation of the line that is parallel to it and passes through the point \( (3,8) \) is given by: \[ y - 8 = -7(x - 3). \] Solving this gives: \[ y = -7x + 21 + 8 = -7x + 29. \] So, the equation of the parallel line is: \[ y = -7x + 29. \] For the perpendicular line, we need the negative reciprocal of the slope. The perpendicular slope is \( \frac{1}{7} \). Using the point-slope form again for the line perpendicular to the original line through \( (3,8) \): \[ y - 8 = \frac{1}{7}(x - 3). \] Solving this yields: \[ y = \frac{1}{7}x - \frac{3}{7} + 8 = \frac{1}{7}x + \frac{53}{7}. \] Thus, the equation of the perpendicular line is: \[ y = \frac{1}{7}x + \frac{53}{7}. \] In summary: Parallel: \( y = -7x + 29 \) Perpendicular: \( y = \frac{1}{7}x + \frac{53}{7} \)