Chapter 8.5, Problem 67E

To determine

To find: the area of the triangle whose vertices are $\left(3,-1\right)$ , $\left(7,-1\right)$ and $\left(7,5\right)$ .

Answer to Problem 67E

  $12\text{ square units}$

Explanation of Solution

Concept used:

The area of a triangle with vertices $\left(x_1,y_1\right)$ , $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ is:

  $Area=\pm \frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Where the sign $\pm$ is chosen so that the area is positive.

Let $\left(x_1,y_1\right)=\left(3,-1\right)$ , $\left(x_2,y_2\right)=\left(7,-1\right)$ and $\left(x_3,y_3\right)=\left(7,5\right)$ .

Then,

  $\begin{array}{c}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\left|\begin{array}{ccc}3& -1& 1\\ 7& -1& 1\\ 7& 5& 1\end{array}\right|\\ =3\left|\begin{array}{cc}-1& 1\\ 5& 1\end{array}\right|-\left(-1\right)\left|\begin{array}{cc}7& 1\\ 7& 1\end{array}\right|+1\left|\begin{array}{cc}7& -1\\ 7& 5\end{array}\right|\\ =3\left(-1-5\right)+1\left(7-7\right)+1\left(35+7\right)\end{array}$

  $\begin{array}{c}=3\left(-6\right)+1\left(0\right)+1\left(42\right)\\ =-18+0+42\\ =24\end{array}$

Since, the determinant is positive, so in formula for finding the area positive sign is taken.

  $\begin{array}{c}Area=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\\ =\frac{1}{2}\left(24\right)\\ =12\text{ square unit}\end{array}$

Check:

Plot the points $\left(3,-1\right)$ , $\left(7,-1\right)$ and $\left(7,5\right)$ . in the coordinate plane.

It can be observed that the length of base of this triangle is 4 units and its height is 6 units.

So,

  $\begin{array}{c}Area=\frac{1}{2}\times base\times height\\ =\frac{1}{2}\times 4\times 6\\ =12\text{ square unit}\end{array}$

Hence verified.