Chapter 13.3, Problem 7PSB

To determine

To show: The graph of the following three equations are concurrent. And hence find the coordinates of the point of intersection.

Answer to Problem 7PSB

The point of intersection is $\left(4,-2\right)$ .

Explanation of Solution

Given:

The given equation of lines are:

  $\begin{array}{c}2x+3y=2\left(1\right)\\ y=2x-10\left(2\right)\\ 3x-y=14\left(3\right)\end{array}$

Calculation:

The point of intersection of equation $\left(1\right)\&\left(2\right)$ is calculated as:

  $\begin{array}{c}y+10=2-3y\left[\text{put equation}\left(1\right)=\text{equation}\left(2\right)\right]\\ y=-2\\ x=\frac{2-3\left(-2\right)}{2}\\ x=4\end{array}$

Therefore the point of intersection $\left(1\right)\&\left(2\right)$ is $\left(4,-2\right)$ .

The point of intersection of equation $\left(2\right)\&\left(3\right)$ is calculated as:

  $\begin{array}{c}2x-10=3x-14\left[\text{put equation}\left(2\right)=\text{equation}\left(3\right)\right]\\ x=4\\ y=3\left(4\right)-14\text{}\left[\text{substitute}x=4\text{ in equation }\left(3\right)\right]\\ y=-2\end{array}$

Therefore the point of intersection $\left(2\right)\&\left(3\right)$ is $\left(4,-2\right)$ .

The point of intersection of equation $\left(1\right)\&\left(3\right)$ is calculated as:

  $\begin{array}{c}\left(2x+9x\right)+\left(3y-3y\right)=2+42\left[\text{multiplying equation}\left(3\right)\text{ by 3 and added to equation}\left(1\right)\right]\\ x=4\\ y=3\left(4\right)-14\text{}\left[\text{substitute}x=4\text{ in equation }\left(3\right)\right]\\ y=-2\end{array}$

Therefore the point of intersection $\left(1\right)\&\left(3\right)$ is $\left(4,-2\right)$ .

As the above solution proved that the graph of three lines are concurrent, therefore the point of intersection of the three lines is $\left(4,-2\right)$ .