Question

1. p(-3, 14), q(2, -1), r(4, 8), s(-2, -10)
2. p(2, -1), q(-3, -1), r(-11, 9), s(-7, 9)
3. given c(x, 16), d(2, -4), e(-6, 14), and f(-2, 4), find the value of x so that $overline{cd}paralleloverline{ef}$.
4. given j(x, -8) and k(-1, -5) and the graph of line l below, find the value of x so that $overline{jk}parallel l$.
5. given p(12, -2), q(5, -10), r(-4, 10), and s(4, y), find the value of y so that $overline{pq}perpoverline{rs}$.
6. given a(4, 2) and b(-1, y) and the graph of line t below, find the value of y so that $overrightarrow{ab}perp t$.

Explanation:

Step1: Recall slope - formula

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Solve problem 16

For $\overrightarrow{PQ}$ with $P(2,-1)$ and $Q(-3,-1)$, $m(\overrightarrow{PQ})=\frac{-1-(-1)}{-3 - 2}=\frac{-1 + 1}{-5}=0$.
For $\overrightarrow{RS}$ with $R(-11,9)$ and $S(-7,9)$, $m(\overrightarrow{RS})=\frac{9 - 9}{-7+11}=0$. Since $m(\overrightarrow{PQ})=m(\overrightarrow{RS}) = 0$, the lines are parallel.

Step3: Solve problem 17

The slope of $\overrightarrow{EF}$ with $E(-6,14)$ and $F(-2,4)$ is $m(\overrightarrow{EF})=\frac{4 - 14}{-2+6}=\frac{-10}{4}=-\frac{5}{2}$.
The slope of $\overrightarrow{CD}$ with $C(x,16)$ and $D(2,-4)$ is $m(\overrightarrow{CD})=\frac{-4 - 16}{2 - x}=\frac{-20}{2 - x}$.
Since $\overrightarrow{CD}\parallel\overrightarrow{EF}$, then $m(\overrightarrow{CD})=m(\overrightarrow{EF})$. So $\frac{-20}{2 - x}=-\frac{5}{2}$.
Cross - multiply: $-20\times2=-5\times(2 - x)$.
$-40=-10 + 5x$.
$5x=-40 + 10=-30$.
$x=-6$.

Step4: Solve problem 18

First, find the slope of line $l$. Let's assume two points on line $l$: say $(0,0)$ and $(1,2)$ (from the grid), then $m_l = 2$.
The slope of $\overrightarrow{JK}$ with $J(x,-8)$ and $K(-1,-5)$ is $m(\overrightarrow{JK})=\frac{-5+8}{-1 - x}=\frac{3}{-1 - x}$.
Since $\overrightarrow{JK}\parallel l$, then $m(\overrightarrow{JK})=m_l$. So $\frac{3}{-1 - x}=2$.
Cross - multiply: $3=2\times(-1 - x)$.
$3=-2-2x$.
$2x=-2 - 3=-5$.
$x=-\frac{5}{2}$.

Step5: Solve problem 19

The slope of $\overrightarrow{PQ}$ with $P(12,-2)$ and $Q(5,-10)$ is $m(\overrightarrow{PQ})=\frac{-10 + 2}{5 - 12}=\frac{-8}{-7}=\frac{8}{7}$.
The slope of $\overrightarrow{RS}$ with $R(-4,10)$ and $S(4,y)$ is $m(\overrightarrow{RS})=\frac{y - 10}{4 + 4}=\frac{y - 10}{8}$.
Since $\overrightarrow{PQ}\perp\overrightarrow{RS}$, then $m(\overrightarrow{PQ})\times m(\overrightarrow{RS})=-1$.
$\frac{8}{7}\times\frac{y - 10}{8}=-1$.
$\frac{y - 10}{7}=-1$.
$y-10=-7$.
$y = 3$.

Step6: Solve problem 20

First, find the slope of line $t$. Let's assume two points on line $t$: say $(0,0)$ and $(1,1)$ (from the grid), then $m_t = 1$.
The slope of $\overrightarrow{AB}$ with $A(4,2)$ and $B(-1,y)$ is $m(\overrightarrow{AB})=\frac{y - 2}{-1 - 4}=\frac{y - 2}{-5}$.
Since $\overrightarrow{AB}\perp t$, then $m(\overrightarrow{AB})\times m_t=-1$.
$\frac{y - 2}{-5}\times1=-1$.
$y-2 = 5$.
$y=7$.

Answer:

1. $m(\overrightarrow{PQ}) = 0$, $m(\overrightarrow{RS}) = 0$, parallel
2. $x=-6$
3. $x=-\frac{5}{2}$
4. $y = 3$
5. $y=7$