Question

for questions 4 - 7, determine if $overleftrightarrow{jk}$ and $overleftrightarrow{lm}$ are parallel, perpendicular, or neither.

1. $j(1,9),k(7,4),l(8,13),m(-2,1)$
2. $j(13, - 5),k(2,6),l(-1,-5),m(-4,-2)$
3. $j(-10,-7),k(-4,1),l(-3,2),m(-6,-2)$
4. $j(11,-2),k(3,-2),l(1,-7),m(1,-2)$

Explanation:

Step1: Recall slope - formula

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

Step2: Calculate slope of $\overrightarrow{JK}$ for question 4

For $J(1,9)$ and $K(7,4)$, $m_{JK}=\frac{4 - 9}{7 - 1}=\frac{-5}{6}$.

Step3: Calculate slope of $\overrightarrow{LM}$ for question 4

For $L(8,13)$ and $M(-2,1)$, $m_{LM}=\frac{1 - 13}{-2 - 8}=\frac{-12}{-10}=\frac{6}{5}$.

Step4: Check relationship for question 4

Since $m_{JK}\times m_{LM}=\frac{-5}{6}\times\frac{6}{5}=- 1$, $\overrightarrow{JK}$ and $\overrightarrow{LM}$ are perpendicular.

Step5: Calculate slope of $\overrightarrow{JK}$ for question 5

For $J(13,-5)$ and $K(2,6)$, $m_{JK}=\frac{6+5}{2 - 13}=\frac{11}{-11}=-1$.

Step6: Calculate slope of $\overrightarrow{LM}$ for question 5

For $L(-1,-5)$ and $M(-4,-2)$, $m_{LM}=\frac{-2 + 5}{-4+1}=\frac{3}{-3}=-1$.

Step7: Check relationship for question 5

Since $m_{JK}=m_{LM}=-1$, $\overrightarrow{JK}$ and $\overrightarrow{LM}$ are parallel.

Step8: Calculate slope of $\overrightarrow{JK}$ for question 6

For $J(-10,-7)$ and $K(-4,1)$, $m_{JK}=\frac{1 + 7}{-4 + 10}=\frac{8}{6}=\frac{4}{3}$.

Step9: Calculate slope of $\overrightarrow{LM}$ for question 6

For $L(-3,2)$ and $M(-6,-2)$, $m_{LM}=\frac{-2 - 2}{-6+3}=\frac{-4}{-3}=\frac{4}{3}$.

Step10: Check relationship for question 6

Since $m_{JK}=m_{LM}=\frac{4}{3}$, $\overrightarrow{JK}$ and $\overrightarrow{LM}$ are parallel.

Step11: Calculate slope of $\overrightarrow{JK}$ for question 7

For $J(11,-2)$ and $K(3,-2)$, $m_{JK}=\frac{-2+2}{3 - 11}=0$.

Step12: Calculate slope of $\overrightarrow{LM}$ for question 7

For $L(1,-7)$ and $M(1,-2)$, the denominator of $m_{LM}=\frac{-2 + 7}{1 - 1}$ is $0$, so the slope of $\overrightarrow{LM}$ is undefined.

Step13: Check relationship for question 7

Since the slope of $\overrightarrow{JK}$ is $0$ (horizontal line) and the slope of $\overrightarrow{LM}$ is undefined (vertical line), $\overrightarrow{JK}$ and $\overrightarrow{LM}$ are perpendicular.

Answer:

1. Perpendicular
2. Parallel
3. Parallel
4. Perpendicular