Chapter 13.3, Problem 12PSB

To determine

To find: The point of intersection of lines 2 and 3.

Answer to Problem 12PSB

The point of intersection is $\left(-8,-23\right)$ .

Explanation of Solution

Given:

The equation of given is:

  $y=2x+1$ .

The condition of line 2 is given as:

The line 2 is parallel to the given line and contains $\left(5,3\right)$ .

The condition of line 3 is given as:

The line 3 has the same $y$ -intercept as the given line and contains $\left(5,16\right)$ .

Calculation:

The equation of line 2 is given as:

  $\begin{array}{l}y-3=2\left(x-5\right)\left[\because \text{slope of given line is same as the slope of line 2}\right]\\ y=2x-7\left(1\right)\end{array}$

The equation of line 3 is given as:

  $\begin{array}{c}y-1=\frac{16-1}{5}\left(x-0\right)\left[\begin{array}{l}\because \text{as line 3 and given line have same }y\text{-intercept,}\\ \text{so line 3 passes through}\left(0,1\right)\&\left(5,16\right)\end{array}\right]\\ y=3x+1\left(2\right)\end{array}$

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

  $\begin{array}{c}2x-7=3x+1\left[\text{put equation}\left(1\right)=\text{equation}\left(2\right)\right]\\ x=-8\\ y=3\left(-8\right)+1\left[\text{substitute }x=-8\text{ in equation}\left(2\right)\right]\\ y=-23\end{array}$

Therefore the point of intersection is $\left(-8,-23\right)$ .