Chapter 13.3, Problem 10PSB

To determine

To find: The equation of the line that is parallel to the graph of the given two lines and contains the point of intersection of the graph of another two given lines.

Answer to Problem 10PSB

The equation of the required line is $3y=-2x+10$ .

Explanation of Solution

Given:

The given line that is parallel to the graph of the required line is as:

  $2x+3y=5\left(A\right)$

The given lines of that point of intersection are contained by the required line are:

  $\begin{array}{l}y=4x+8\left(1\right)\\ y=x+5\left(2\right)\end{array}$

Calculation:

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

  $\begin{array}{c}4x+8=x+5\left[\text{put equation}\left(2\right)=\text{equation}\left(1\right)\right]\\ 3x=-3\\ x=-1\\ y=-1+5\left[\text{put }x=-1\text{ in equation }\left(2\right)\right]\end{array}$

Further simplified as:

  $y=4$

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

The equation of required line that is parallel to the graph of the given two lines and contains the point of intersection of the graph of another two given lines is as:

  $\begin{array}{c}y-4=\frac{-2}{3}\left(x-\left(-1\right)\right)\left[\because \text{slope}=\frac{-2}{3}\text{ , as the line parallel to the line }A\right]\\ 3y-12=-2x-2\\ 3y=-2x+10\end{array}$

Therefore the equation of the required lineis $3y=-2x+10$ .