Chapter 13.2, Problem 32WE

To determine

To find the point of intersection of line with $x-axis$ and $y-axis$

Answer to Problem 32WE

The point of intersection of line with $x-axis$ is $\left(-\frac{1}{2},0\right)$ and $y-axis$ is $\left(0,\frac{1}{3}\right)$

Explanation of Solution

Given:

A line passes through points $\left(-2,-1\right)$ and $\left(4,3\right)$

Calculation:

Since, equation of line when two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are given is

  $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Points given are $\left(-2,-1\right)$ and $\left(4,3\right)$

  $\begin{array}{l}y-\left(-1\right)=\frac{3-\left(-1\right)}{4-\left(-2\right)}\left(x-\left(-2\right)\right)\\ y+1=\frac{2}{3}\left(x+2\right)\\ 3y+3=2x+4\\ 3y-2x=1\end{array}$

Now, equation of $x-\text{axis}$ is $y=0$ , solving it with $3y-2x=1$

  $\begin{array}{l}3y-2x=1\\ 3\left(0\right)-2x=1\\ -2x=1\\ x=-\frac{1}{2}\end{array}$

Equation of $y-\text{axis}$ is $x=0$ , solving it with $3y-2x=1$

  $\begin{array}{l}3y-2x=1\\ 3y-2\left(0\right)=1\\ 3y=1\\ y=\frac{1}{3}\end{array}$

Therefore, the point of intersection of line with $x-axis$ is $\left(-\frac{1}{2},0\right)$ and $y-axis$ is $\left(0,\frac{1}{3}\right)$ .