Chapter 4.3, Problem 52E

To determine

To show: that the line segment connecting the midpoints of sides which is parallel to another side.

Answer to Problem 52E

The line segment connecting the midpoints of side $\overline{TR}\text{ and }\overline{TI}$ is parallel to $\overline{RI}$ .

Explanation of Solution

Given:

  

Concept used:

Midpoint of the line segment is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Slope of straight line is: $m=\frac{y_2-y_1}{x_2-x_1}$ .

If the slope are equal it means the lines are parallel to each other.

Calculation:

Consider the figure shown below:

  

Consider the points $R\left(-7,2\right),T\left(-2,6\right)\text{ and }I\left(-2,-3\right)$ .

The midpoint of the line segment $\overline{TR}$ is:

  $\begin{array}{l}M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).\\ \Rightarrow \left(\frac{-2-7}{2},\frac{6+2}{2}\right).\\ \Rightarrow \left(-4.5,4\right).\end{array}$

Now the midpoint of the line segment $\overline{TI}$ is:

  $\begin{array}{l}N=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).\\ \Rightarrow \left(\frac{-2-2}{2},\frac{6-3}{2}\right).\\ \Rightarrow \left(-2.1,5\right).\end{array}$

Now the slope of the line segment $\overline{RI}$ is:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}.\\ \Rightarrow \frac{2-\left(-3\right)}{-7-\left(-2\right)}.\\ \Rightarrow \frac{5}{-5}.\\ \Rightarrow -1.\end{array}$

Now the slope of the line segment joining the point $\overline{MN}$ is:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}.\\ \Rightarrow \frac{1.5-5}{-2-\left(-4.5\right)}.\\ \Rightarrow \frac{-3.5}{3.5}.\\ \Rightarrow -1.\end{array}$

Now the two line are parallel if their slopes are same or equal.

The slopes of the line segment $\overline{RI}\text{ and }\overline{MN}$ are equal.

Hence, the line segment connecting the midpoints of side $\overline{TR}\text{ and }\overline{TI}$ is parallel to $\overline{RI}$ .