## Intermediate Algebra (12th Edition)

$y=\dfrac{2}{3}x+1$
The given graph of a line passes through the points $( 0,1 )$ and $( 3,3 ).$ Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the two-point form of linear equations, then the equation of the line is \begin{array}{l}\require{cancel} y-1=\dfrac{1-3}{0-3}(x-0) \\\\ y-1=\dfrac{-2}{-3}(x) \\\\ y-1=\dfrac{2}{3}x \\\\ y=\dfrac{2}{3}x+1 .\end{array}