Chapter 4.6, Problem 7PSA

To determine

To show: The given figure FHJK is a parallelogram.

Explanation of Solution

Given information:

A figure is given as below.

  

Concept used: Slope of a line passing through two points $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is given as $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$ .

Slope of two parallel lines are equal.

Calculation:

  

In above figure,

  $\begin{array}{l}\text{Slope of line }\overleftrightarrow{FH}=\frac{y_2-y_1}{x_2-x_1}=\frac{10-8}{9-2}=\frac{2}{7}\\ \text{Slope of line }\overleftrightarrow{JK}=\frac{y_2-y_1}{x_2-x_1}=\frac{3-5}{6-13}=\frac{2}{7}\\ \text{Therefore, }\overleftrightarrow{FH}\parallel \overleftrightarrow{JK}.\\ \text{Similarly,}\\ \text{Slope of line }\overleftrightarrow{FK}=\frac{y_2-y_1}{x_2-x_1}=\frac{8-3}{2-6}=-\frac{5}{4}\\ \text{Slope of line }\overleftrightarrow{HJ}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-10}{13-9}=-\frac{5}{4}\\ \text{Therefore, }\overleftrightarrow{FK}\parallel \overleftrightarrow{HJ}.\end{array}$

Thus opposite sides of $FHJK$ are parallel. This figure is a parallelogram.

Hence, value of $k$ is $32$ .