1. **State the problem:** Find the slope of a line parallel and perpendicular to the line given by the equation $$3y + 8x = 9$$. 2. **Rewrite the equation in slope-intercept form:** The slope-intercept form is $$y = mx + b$$ where $$m$$ is the slope. Start with the given equation: $$3y + 8x = 9$$ Subtract $$8x$$ from both sides: $$3y = -8x + 9$$ Divide both sides by 3: $$y = \frac{\cancel{3}y}{\cancel{3}} = \frac{-8x}{3} + \frac{9}{3}$$ Simplify: $$y = -\frac{8}{3}x + 3$$ 3. **Identify the slope:** From the slope-intercept form, the slope $$m$$ is $$-\frac{8}{3}$$. 4. **Slope of a line parallel:** Lines parallel to this line have the **same slope**. So, the slope of a line parallel is: $$m_{parallel} = -\frac{8}{3}$$ 5. **Slope of a line perpendicular:** The slope of a line perpendicular is the **negative reciprocal** of the original slope. Calculate the negative reciprocal: $$m_{perpendicular} = -\frac{1}{m} = -\frac{1}{-\frac{8}{3}} = \frac{3}{8}$$ **Final answers:** - Slope of a line parallel: $$-\frac{8}{3}$$ - Slope of a line perpendicular: $$\frac{3}{8}$$