Question

1. if q is the mid - point of $overline{pr}$, find the coordinates of r if p(11, - 2) and q(4, 3).
2. a high school soccer team is going to columbus, ohio to see a professional soccer game. a coordinate grid is superimposed on a highway map of ohio. the high school is at point (3, 4) and the stadium in columbus is at point (7, 1). the map shows a highway rest stop halfway between the high school and the stadium.

what are the coordinates of the rest stop?
what is the approximate distance between the high school and the stadium? (one unit is approximately equal to 6.4 miles.)

Explanation:

Step1: Recall mid - point formula

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Let $P(x_1,y_1)=(11,-2)$, $Q(x_m,y_m)=(4,3)$ and $R(x_2,y_2)$.
For the $x$ - coordinate of the mid - point: $x_m=\frac{x_1 + x_2}{2}$, so $4=\frac{11 + x_2}{2}$.

Step2: Solve for $x_2$

Multiply both sides of the equation $4=\frac{11 + x_2}{2}$ by $2$: $4\times2=11 + x_2$, which gives $8 = 11+x_2$. Then subtract $11$ from both sides: $x_2=8 - 11=-3$.

Step3: Solve for $y_2$ using the $y$ - coordinate of the mid - point formula

For the $y$ - coordinate of the mid - point: $y_m=\frac{y_1 + y_2}{2}$, so $3=\frac{-2 + y_2}{2}$. Multiply both sides by $2$: $3\times2=-2 + y_2$, which gives $6=-2 + y_2$. Add $2$ to both sides: $y_2=6 + 2 = 8$. So the coordinates of $R$ are $(-3,8)$.

Step4: For the rest - stop coordinates (mid - point of $(3,4)$ and $(7,1)$)

Use the mid - point formula $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here $x_1 = 3$, $y_1 = 4$, $x_2 = 7$, $y_2 = 1$.
The $x$ - coordinate of the mid - point is $\frac{3+7}{2}=\frac{10}{2}=5$.
The $y$ - coordinate of the mid - point is $\frac{4 + 1}{2}=\frac{5}{2}=2.5$. So the coordinates of the rest - stop are $(5,2.5)$.

Step5: Calculate the distance between $(3,4)$ and $(7,1)$

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
Here $x_1 = 3$, $y_1 = 4$, $x_2 = 7$, $y_2 = 1$.
$d=\sqrt{(7 - 3)^2+(1 - 4)^2}=\sqrt{4^2+( - 3)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.
Since one unit is approximately $6.4$ miles, the distance in miles is $5\times6.4 = 32$ miles.

Answer:

1. The coordinates of $R$ are $(-3,8)$.
2. The coordinates of the rest - stop are $(5,2.5)$ and the distance between the high school and the stadium is $32$ miles.