Question

graph each pair of lines. use slopes to determine whether the lines are parallel, perpendicular, or neither.

1. $overline{ij}$ and $overline{kl}$ for $i(1,0), j(5,3), k(6, - 1),$ and $l(0,2)$
2. $overline{pq}$ and $overline{rs}$ for $p(5,1), q(-1,-1), r(2,1),$ and $s(3,-2)$

Explanation:

Step1: Recall slope - formula

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Calculate slope of $\overline{IJ}$ for $I(1,0)$ and $J(5,3)$

$m_{IJ}=\frac{3 - 0}{5 - 1}=\frac{3}{4}$

Step3: Calculate slope of $\overline{KL}$ for $K(6,-1)$ and $L(0,2)$

$m_{KL}=\frac{2-( - 1)}{0 - 6}=\frac{3}{-6}=-\frac{1}{2}$

Step4: Determine the relationship

Since $m_{IJ}
eq m_{KL}$ and $m_{IJ}\times m_{KL}=\frac{3}{4}\times(-\frac{1}{2})=-\frac{3}{8}
eq - 1$, the lines $\overline{IJ}$ and $\overline{KL}$ are neither parallel nor perpendicular.

Step5: Calculate slope of $\overline{PQ}$ for $P(5,1)$ and $Q(-1,-1)$

$m_{PQ}=\frac{-1 - 1}{-1 - 5}=\frac{-2}{-6}=\frac{1}{3}$

Step6: Calculate slope of $\overline{RS}$ for $R(2,1)$ and $S(3,-2)$

$m_{RS}=\frac{-2 - 1}{3 - 2}=\frac{-3}{1}=-3$

Step7: Determine the relationship

Since $m_{PQ}\times m_{RS}=\frac{1}{3}\times(-3)=-1$, the lines $\overline{PQ}$ and $\overline{RS}$ are perpendicular.

Answer:

1. Neither
2. Perpendicular