1. **Stating the problem:** We need to describe the slopes of parallel and perpendicular lines and determine the relationship between line segments AB and CD for given points. 2. **Slope rules:** - Parallel lines have the **same slope**. - Perpendicular lines have slopes that are **negative reciprocals** of each other, i.e., if slope of one line is $m$, the other is $-\frac{1}{m}$. 3. **Calculate slopes for problem 18:** Points: $A(-4,3)$, $B(2,-12)$, $C(10,5)$, $D(0,1)$ Slope of $AB$: $$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-12 - 3}{2 - (-4)} = \frac{-15}{6} = -\frac{15}{6}$$ Simplify: $$m_{AB} = -\frac{\cancel{15}}{\cancel{6}} = -\frac{5}{2}$$ Slope of $CD$: $$m_{CD} = \frac{1 - 5}{0 - 10} = \frac{-4}{-10} = \frac{4}{10}$$ Simplify: $$m_{CD} = \frac{\cancel{4}}{\cancel{10}} = \frac{2}{5}$$ Check if $AB$ and $CD$ are perpendicular: $$m_{AB} \times m_{CD} = -\frac{5}{2} \times \frac{2}{5} = -1$$ Since the product is $-1$, lines $AB$ and $CD$ are **perpendicular**. 4. **Calculate slopes for problem 19:** Points: $A(2,3)$, $B(8,-15)$, $C(-2,2)$, $D(-5,11)$ Slope of $AB$: $$m_{AB} = \frac{-15 - 3}{8 - 2} = \frac{-18}{6} = -3$$ Slope of $CD$: $$m_{CD} = \frac{11 - 2}{-5 - (-2)} = \frac{9}{-3} = -3$$ Since $m_{AB} = m_{CD} = -3$, lines $AB$ and $CD$ are **parallel**. **Final answers:** - Problem 18: $AB$ and $CD$ are perpendicular. - Problem 19: $AB$ and $CD$ are parallel.