Chapter 4, Problem 17RP

To determine

To Prove: EFHJ is a parallelogram, using slops.

Explanation of Solution

Given information: coordinates of E, F, H and J.

Formula Used: slope of line joining $(x_1,y_1)$ and $\therefore EF\parallel HJ.$ is $\frac{y_2-y_1}{x_2-x_1}$ . If two lines are parallel, then their slops are equals.

Proof:

  $\begin{array}{l}\text{Slope of}\text{EF}=\frac{11-3}{8+2}=\frac{4}{5}.\\ \text{Slope of}\text{HF}=\frac{2+6}{7+3}=\frac{4}{5}.\\ \therefore EF\parallel HJ.\\ \text{Slope of}\text{EJ}=\frac{-6-3}{-3+2}=9.\\ \text{Slope of}\text{FH}=\frac{2-11}{7-8}=9.\\ \therefore EJ\parallel FH.\end{array}$

  $\because$ Opposite sides of EFHG are parallel, hence it is a parallelogram.