Chapter 8.5, Problem 42E

To determine

To find the area of the parallelogram formed by the given vertices.

Answer to Problem 42E

Area of the parallelogram is 10 square units.

Explanation of Solution

Given:

The vertices of the parallelogram are $\left(0,0\right),\left(-2,0\right),\left(3,5\right),\left(1,5\right)$ .

Concept used:

The area of a parallelogram with vertices $\left(0,0\right),\left(a,b\right),\left(c,d\right),\left(a+c,b+d\right)$ is,

  $\text{Area}=\left|\det \left(A\right)\right|,\text{where}A=\left[\begin{array}{l}ab\\ cd\end{array}\right]$

Calculation:

Given vertices are $\left(0,0\right),\left(-2,0\right),\left(3,5\right),\left(1,5\right)$ .

These can be written as,

  $\left(0,0\right),\left(-2,0\right),\left(3,5\right),\left(-2+3,0+5\right)$

Now comparing with $\left(0,0\right),\left(a,b\right),\left(c,d\right),\left(a+c,b+d\right)$ , we get

  $\begin{array}{l}a=-2,b=0,c=3\&d=5\\ \therefore A=\left[\begin{array}{l}ab\\ cd\end{array}\right]=\left[\begin{array}{l}-20\\ 35\end{array}\right]\\ \Rightarrow \det \left(A\right)=\left|\begin{array}{l}-20\\ 35\end{array}\right|=-2\left(5\right)-3\left(0\right)=-10\end{array}$

Therefore area of the parallelogram is,

  $\text{Area}=\left|\det \left(A\right)\right|=\left|-10\right|=10\text{square}\text{units}\text{.}$