1. **State the problem:** We are given the line $y = -3x + 4$ and a point $(-8, -6)$. We need to find: - The equation of the line parallel to $y = -3x + 4$ passing through $(-8, -6)$. - The equation of the line perpendicular to $y = -3x + 4$ passing through $(-8, -6)$. 2. **Recall formulas and rules:** - Parallel lines have the **same slope**. - Perpendicular lines have slopes that are **negative reciprocals** of each other. 3. **Find the slope of the given line:** The given line is $y = -3x + 4$, so its slope is $m = -3$. 4. **Equation of the parallel line:** Since parallel lines share the same slope, the slope of the parallel line is also $m = -3$. Use point-slope form: $$y - y_1 = m(x - x_1)$$ Substitute $m = -3$ and point $(-8, -6)$: $$y - (-6) = -3(x - (-8))$$ Simplify: $$y + 6 = -3(x + 8)$$ $$y + 6 = -3x - 24$$ Subtract 6 from both sides: $$y = -3x - 24 - 6$$ $$y = -3x - 30$$ 5. **Equation of the perpendicular line:** The slope of the perpendicular line is the negative reciprocal of $-3$: $$m_{perp} = -\frac{1}{-3} = \frac{1}{3}$$ Use point-slope form again with $m = \frac{1}{3}$ and point $(-8, -6)$: $$y - (-6) = \frac{1}{3}(x - (-8))$$ Simplify: $$y + 6 = \frac{1}{3}(x + 8)$$ Multiply out: $$y + 6 = \frac{1}{3}x + \frac{8}{3}$$ Subtract 6 from both sides: $$y = \frac{1}{3}x + \frac{8}{3} - 6$$ Convert 6 to thirds: $6 = \frac{18}{3}$ $$y = \frac{1}{3}x + \frac{8}{3} - \frac{18}{3}$$ $$y = \frac{1}{3}x - \frac{10}{3}$$ **Final answers:** - Parallel line: $y = -3x - 30$ - Perpendicular line: $y = \frac{1}{3}x - \frac{10}{3}$