## Intermediate Algebra (12th Edition)

$2x-y=2$
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the two-point form of linear equations, the equation of the line passing through $( 3,4 )$ and $( 5,8 )$ is \begin{array}{l}\require{cancel} y-4=\dfrac{4-8}{3-5}(x-3) \\\\ y-4=\dfrac{-4}{-2}(x-3) \\\\ y-4=2(x-3) \\\\ y-4=2x-6 \\\\ -2x+y=-6+4 \\\\ -2x+y=-2 \\\\ 2x-y=2 .\end{array}