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: \( y= \)

**Step 1. Find the slope of the given line** The given line is: \[ y+7x=2 \] Rewriting in slope-intercept form: \[ y = -7x + 2 \] Thus, the slope of the given line is \(-7\). --- **Step 2. Equation of the line parallel to the given line** A line parallel to the given line must have the same slope, \(-7\). Since it passes through \((3,8)\), we use the point-slope form: \[ y - 8 = -7(x - 3) \] Expanding and simplifying: \[ y - 8 = -7x + 21 \quad \Rightarrow \quad y = -7x + 29 \] --- **Step 3. Equation of the line perpendicular to the given line** A line perpendicular to the given line will have a slope equal to the negative reciprocal of \(-7\), which is: \[ \frac{1}{7} \] Using the point-slope form for the point \((3,8)\): \[ y - 8 = \frac{1}{7}(x - 3) \] Expanding and simplifying: \[ y - 8 = \frac{1}{7}x - \frac{3}{7} \] Adding 8 (which is \(\frac{56}{7}\)) to both sides: \[ y = \frac{1}{7}x - \frac{3}{7} + \frac{56}{7} = \frac{1}{7}x + \frac{53}{7} \] --- **Final Answers:** Parallel: \[ y = -7x+29 \] Perpendicular: \[ y = \frac{1}{7}x+\frac{53}{7} \]