Chapter 13.5, Problem 20WE

a.

To determine

To find:The coordinates of $D$ and $E$ .

Answer to Problem 20WE

Coordinates of point $D$ is $\left(7,5\right)$ and point $E$ is $\left(2,4\right)$ .

Explanation of Solution

Given information:Points $A(1,1),B(13,9),\text{ and }C(3,7)$ . D is the midpoint of $\overline{AB}$ , and E the midpoint of $\overline{AC}$ .

Formula Used:The midpoint of segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is the point $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Calculation:

Since, D is the midpoint of $\overline{AB}$ .

So, coordinates of point D is $\left(\frac{1+13}{2},\frac{1+9}{2}\right)\equiv \left(7,5\right)$

Since, E the midpoint of $\overline{AC}$ .

So, coordinates of point E is $\left(\frac{1+3}{2},\frac{1+7}{2}\right)\equiv \left(2,4\right)$

Hence, coordinates of point $D$ is $\left(7,5\right)$ and point $E$ is $\left(2,4\right)$ .

b.

To determine

To find:Use slopes to show that $\overline{DE}\left|\right|\overline{BC}$ .

Explanation of Solution

Given information:Points $A(1,1),B(13,9),\text{ and }C(3,7)$ . D is the midpoint of $\overline{AB}$ , and E the midpoint of $\overline{AC}$ .

Formula used: Theslope of a line passing through point $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

From part a, $D$ is $\left(7,5\right)$ and $E$ is $\left(2,4\right)$ .

  $\begin{array}{l}\text{Slope of }\overline{DE}=\frac{4-5}{2-7}\\ \Rightarrow \text{Slope of }\overline{DE}=\frac{-1}{-5}\\ \Rightarrow \text{Slope of }\overline{DE}=\frac{1}{5}\\ \text{Slope of }\overline{BC}=\frac{7-9}{3-13}\\ \Rightarrow \text{Slope of }\overline{BC}=\frac{-2}{-10}\\ \Rightarrow \text{Slope of }\overline{BC}=\frac{1}{5}\end{array}$

Since, slopes of $\overline{DE}$ and slope of $\overline{BC}$ are equal.

Hence, $\overline{DE}\left|\right|\overline{BC}$

c.

To determine

To find:Use the distance formula to show that $DE=\frac{1}{2}BC$

Explanation of Solution

Given information:Points $A(1,1),B(13,9),\text{ and }C(3,7)$ . D is the midpoint of $\overline{AB}$ , and E the midpoint of $\overline{AC}$ .

Formula used: The distance between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by:

  $AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

From part a, $D$ is $\left(7,5\right)$ and $E$ is $\left(2,4\right)$ .

  $\begin{array}{l}DE=\sqrt{{\left(2-7\right)}^2+{\left(5-4\right)}^2}\\ \Rightarrow DE=\sqrt{{\left(-5\right)}^2+{\left(1\right)}^2}\\ \Rightarrow DE=\sqrt{25+1}\\ \Rightarrow DE=\sqrt{26}\mathrm{....}\text{(1)}\\ BC=\sqrt{{\left(3-13\right)}^2+{\left(7-9\right)}^2}\\ \Rightarrow BC=\sqrt{{\left(-10\right)}^2+{\left(-2\right)}^2}\\ \Rightarrow BC=\sqrt{100+4}\\ \Rightarrow BC=\sqrt{104}\\ \Rightarrow BC=2\sqrt{26}\mathrm{....}\text{(2)}\\ \text{Obtain from (1) and (2),}\\ BC=2DE\\ \Rightarrow DE=\frac{1}{2}BC\end{array}$

Hence proved.