Question

1. p(-4, 17), q(1, -3), r(-9, 3), s(-5, 4)

	m(\overline{pq})
	m(\overline{rs})
	types of lines

Explanation:

Step1: Calculate slope of PQ

The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. For points $P(-4, 17)$ and $Q(1, -3)$, substitute into the formula:
$m(\overline{PQ}) = \frac{-3 - 17}{1 - (-4)} = \frac{-20}{5} = -4$

Step2: Calculate slope of RS

For points $R(-9, 3)$ and $S(-5, 4)$, use the slope formula:
$m(\overline{RS}) = \frac{4 - 3}{-5 - (-9)} = \frac{1}{4}$

Step3: Determine line type

Since the slopes $-4$ and $\frac{1}{4}$ are not equal (so not parallel) and their product is $-4\times\frac{1}{4} = -1$ (so perpendicular? Wait, no, product of slopes for perpendicular lines is -1. Wait, -4 and 1/4: -4(1/4)= -1, so they are perpendicular? Wait, no, wait the calculation: Wait, PQ slope: (-3 -17)/(1 - (-4)) = (-20)/5 = -4. RS slope: (4 - 3)/(-5 - (-9)) = 1/4. Then -4(1/4) = -1, so they are perpendicular. Wait, but let's check again. Wait, maybe I made a mistake. Wait, PQ: from (-4,17) to (1,-3): change in y is -3 -17 = -20, change in x is 1 - (-4)=5, so slope -20/5=-4. RS: from (-9,3) to (-5,4): change in y is 4-3=1, change in x is -5 - (-9)=4, so slope 1/4. Then -4*(1/4)= -1, so they are perpendicular. So the types of lines: perpendicular.

Answer:

$m(\overline{PQ}) = -4$, $m(\overline{RS}) = \frac{1}{4}$, Types of Lines: Perpendicular