Question

use slopes to determine whether the lines are parallel, perpendicular, or neither. 11. $overline{ab}$ and $overline{cd}$ for a(2,-1), b(7,2), c(2,-3), and d(-3,-6) 12. $overline{xy}$ and $overline{zw}$ for x(-2,5), y(6,-2), z(-3,6), and w(4,0) 13. $overline{jk}$ and $overline{jl}$ for j(-4,-2), k(4,-2), and l(-4,6) 14. $overline{lm}$ and $overline{np}$ for l(-2,2), m(2,5), n(0,2), and p(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 $\overline{AB}$

For $A(2,-1)$ and $B(7,2)$, $m_{AB}=\frac{2 - (-1)}{7 - 2}=\frac{3}{5}$.

Step3: Calculate slope of $\overline{CD}$

For $C(2,-3)$ and $D(-3,-6)$, $m_{CD}=\frac{-6 - (-3)}{-3 - 2}=\frac{-3}{-5}=\frac{3}{5}$.

Step4: Determine relationship

Since $m_{AB}=m_{CD}=\frac{3}{5}$, the lines $\overline{AB}$ and $\overline{CD}$ are parallel.

Step1: Recall slope formula

$m = \frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Calculate slope of $\overline{XY}$

For $X(-2,5)$ and $Y(6,-2)$, $m_{XY}=\frac{-2 - 5}{6 - (-2)}=\frac{-7}{8}$.

Step3: Calculate slope of $\overline{ZW}$

For $Z(-3,6)$ and $W(4,0)$, $m_{ZW}=\frac{0 - 6}{4 - (-3)}=\frac{-6}{7}$.

Step4: Determine relationship

Since $m_{XY}
eq m_{ZW}$ and $m_{XY}\times m_{ZW}
eq - 1$, the lines $\overline{XY}$ and $\overline{ZW}$ are neither parallel nor perpendicular.

Step1: Recall slope formula

$m = \frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Calculate slope of $\overline{JK}$

For $J(-4,-2)$ and $K(4,-2)$, $m_{JK}=\frac{-2-(-2)}{4 - (-4)} = 0$.

Step3: Calculate slope of $\overline{JL}$

For $J(-4,-2)$ and $L(-4,6)$, $m_{JL}=\frac{6 - (-2)}{-4-(-4)}$, the denominator is 0, so the slope is undefined.

Step4: Determine relationship

A line with slope 0 is horizontal and a line with undefined slope is vertical, so $\overline{JK}$ and $\overline{JL}$ are perpendicular.

Step1: Recall slope formula

$m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Calculate slope of $\overline{LM}$

For $L(-2,2)$ and $M(2,5)$, $m_{LM}=\frac{5 - 2}{2-(-2)}=\frac{3}{4}$.

Step3: Calculate slope of $\overline{NP}$

For $N(0,2)$ and $P(3,-2)$, $m_{NP}=\frac{-2 - 2}{3 - 0}=-\frac{4}{3}$.

Step4: Determine relationship

Since $m_{LM}\times m_{NP}=\frac{3}{4}\times(-\frac{4}{3})=-1$, the lines $\overline{LM}$ and $\overline{NP}$ are perpendicular.

Answer:

Parallel

For the second - part (12):