Find the equation of the line that is perpendicular to this line and passes through the point \( (-5,6) \). Find the equation of the line that is parallel to this line and passes through the point \( (-5,6) \).

Assume the given line has the equation \[ y=3x+2. \] **Step 1. Find the slope of the given line.** Since the line is written in slope‐intercept form, the slope is \[ m=3. \] **Step 2. Find the slope of the line perpendicular to the given line.** For two perpendicular lines the slopes satisfy \[ m\cdot m_\perp=-1. \] Thus, the slope of the perpendicular line is \[ m_\perp=-\frac{1}{3}. \] **Step 3. Write the equation of the perpendicular line through \((-5,6)\).** Using the point–slope form \[ y-y_1=m(x-x_1), \] with \((x_1,y_1)=(-5,6)\) and \(m=-\frac{1}{3}\), we get \[ y-6=-\frac{1}{3}(x+5). \] To express this in slope–intercept form, solve for \(y\): \[ y= -\frac{1}{3}(x+5)+6. \] Expanding, \[ y= -\frac{1}{3}x-\frac{5}{3}+6. \] Write 6 as \(\frac{18}{3}\): \[ y=-\frac{1}{3}x-\frac{5}{3}+\frac{18}{3}=-\frac{1}{3}x+\frac{13}{3}. \] So, the equation of the perpendicular line is \[ y=-\frac{1}{3}x+\frac{13}{3}. \] **Step 4. Find the slope of the line parallel to the given line.** Lines that are parallel have the same slope, so the slope of the parallel line is \[ m=3. \] **Step 5. Write the equation of the parallel line through \((-5,6)\).** Again, applying the point–slope form: \[ y-6=3(x+5). \] Solve for \(y\): \[ y=3(x+5)+6. \] Expanding, \[ y=3x+15+6=3x+21. \] Thus, the equation of the parallel line is \[ y=3x+21. \] **Final Answers:** - Perpendicular line: \(\displaystyle y=-\frac{1}{3}x+\frac{13}{3}\). - Parallel line: \(\displaystyle y=3x+21\).