Chapter E, Problem 13.8EX

To determine

To find the vertex of the fourth side of the parallelogram, whose vertices for other sides are $(j,5),(0,0)$ and $(7,0)$ .

Answer to Problem 13.8EX

Option (c)

Explanation of Solution

Given information: Vertices of the parallelogram are $(j,5),(0,0)$ and $(7,0)$ .

Concept used:

In parallelogram, opposite sides are equal and parallel.

That is, the slope of the opposite lines are equal and to find the distance between the points, distance formula is defined as:

The distance between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is

  $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

The slope between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$

Let ABCD be the vertices of the parallelogram with vertices as $A(j,5),B(0,0),C(7,0)$ and $(a,b)$ . The slope of the sides of the parallelogram is

  $\begin{array}{l}AB=\frac{5}{j};BC=0;\\ CD=\frac{b}{a-7};DA=\frac{b-5}{a-j}\end{array}$

For a parallelogram, the slope of AB = CD and the slope of BC = AD

  $\begin{array}{l}\frac{b-5}{a-j}=0\\ \Rightarrow b=5\end{array}$

and

  $\begin{array}{l}\frac{5}{j}=\frac{b}{a-7}\\ \frac{5}{j}=\frac{5}{a-7}\\ a-7=j\\ \Rightarrow a=7+j\end{array}$

Hence, the coordinates of the fourth vertices is $\left(7+j,5\right)$