Chapter 13.7, Problem 18WE

To determine

To find : an equation of line described

Answer to Problem 18WE

The equation is $y=\frac{4}{3}x-\frac{5}{3}$

Explanation of Solution

Given information :

Line through $\left(-1,-3\right)$ and $\left(2,1\right)$

Formula used :

  $m=\frac{y-y_1}{x-x_1}$

Calculation :

To find the slope

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

Then use the point and slope

Using $\left(-1,-3\right)$ , the equation is

  $\begin{array}{l}y-\left(-3\right)=\frac{4}{3}\left(x-\left(-1\right)\right)\\ y+3=\frac{4}{3}x+\frac{4}{3}\\ y=\frac{4}{3}x+\left(\frac{4}{3}-3\right)\\ y=\frac{4}{3}x+\left(\frac{4-9}{3}\right)\\ y=\frac{4}{3}x+\left(\frac{-5}{3}\right)\\ y=\frac{4}{3}x-\frac{5}{3}\end{array}$

Using $\left(2,1\right)$ , the equation is

  $\begin{array}{l}y-1=\frac{4}{3}\left(x-2\right)\\ y=\frac{4}{3}x-\frac{8}{3}+1\\ y=\frac{4}{3}x+\left(\frac{-8+3}{3}\right)\\ y=\frac{4}{3}x+\left(\frac{-5}{3}\right)\\ y=\frac{4}{3}x-\frac{5}{3}\end{array}$

Hence, the equation is $y=\frac{4}{3}x-\frac{5}{3}$ .