Chapter 13.3, Problem 13PSC

To determine

To find: The point of intersection of given equation if two lines.

Answer to Problem 13PSC

The point of intersection is $\left(\frac{ce-bf}{ea-bd},\frac{af-ed}{ea-bd}\right)$ .

Explanation of Solution

Given:

The given equations are:

  $\begin{array}{c}ax+by=c\left(1\right)\\ dx+ey=f\left(2\right)\end{array}$ .

Calculation:

The equation of $\left(1\right)\&\left(2\right)$ can be written as:

  $\begin{array}{c}y=\frac{-ax+c}{b}\left(3\right)\\ y=-\frac{dx+f}{e}\left(4\right)\end{array}$

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

  $\begin{array}{c}\frac{-ax+c}{b}=-\frac{dx+f}{e}\left[\text{put equation}\left(3\right)=\text{equation}\left(4\right)\right]\\ \left(-bd+ea\right)x=ce-bf\\ x=\frac{ce-bf}{ea-bd}\\ y=\frac{-a\left(\frac{ce-bf}{ea-bd}\right)+c}{b}\left[\text{substitute }x=\frac{ce-bf}{ea-bd}\text{ in equation}\left(3\right)\right]\end{array}$

Further simplified as:

  $\begin{array}{c}y=\frac{-a\left(\frac{ce-bf}{ea-bd}\right)+c\left(\frac{ea-bd}{ea-bd}\right)}{b}\\ y=\frac{af-ed}{ea-bd}\end{array}$

Therefore the point of intersection is $\left(\frac{ce-bf}{ea-bd},\frac{af-ed}{ea-bd}\right)$ .