Chapter 11.8, Problem 10PSB

To determine

To find: The area of a triangle with vertices $\left(15,5\right),\left(-2,12\right)\text{and}\left(-2,3\right).$

Answer to Problem 10PSB

Area of the triangle with vertices $\left(15,5\right),\left(-2,12\right)\text{and}\left(-2,3\right)$ is $\mathrm{76.5.}$

Explanation of Solution

Given Information:

Vertices of the triangle are:

  $\left(x_1,y_1\right)=\left(15,5\right),\left(x_2,y_2\right)=\left(-2,12\right)\text{and}\left(x_3,y_3\right)=\left(-2,3\right)$

Formula used:

Area of a triangle $\left(A\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

Here, $\left(x_1,y_1\right),\left(x_2,y_2\right)\text{and}\left(x_3,y_3\right)$ are the vertices of the triangle

Calculation:

Vertices of the triangle are given as

  $\left(x_1,y_1\right)=\left(15,5\right),\left(x_2,y_2\right)=\left(-2,12\right)\text{and}\left(x_3,y_3\right)=\left(-2,3\right)$

We know that, Area of a triangle $\left(A\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

  $\left(x_1,y_1\right),\left(x_2,y_2\right)\text{and}\left(x_3,y_3\right)$ are the vertices of the triangle

  $\begin{array}{l}\therefore A=\frac{1}{2}\left|15\left(12-3\right)+\left(-2\right)\left(3-5\right)+\left(-2\right)\left(5-12\right)\right|\\ A=\frac{1}{2}\left|15\left(9\right)+\left(-2\right)\left(-2\right)+\left(-2\right)\left(-7\right)\right|\\ A=\frac{1}{2}\left|135+4+14\right|\\ A=\frac{1}{2}\left|153\right|\\ A=76.5\end{array}$

Hence, area of the triangle is $\mathrm{76.5.}$