## Intermediate Algebra (12th Edition)

$\text{slope-intercept form: } y=-\dfrac{5}{2}x+2 \\\\ \text{standard form: } 5x+2y=4$
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 $\left( 4,-8 \right)$ and $\left( -4,12 \right)$ is \begin{array}{l}\require{cancel} y-(-8)=\dfrac{-8-12}{4-(-4)}\left( x-4 \right) \\\\ y+8=\dfrac{-8-12}{4+4}\left( x-4 \right) \\\\ y+8=\dfrac{-20}{8}\left( x-4 \right) \\\\ y+8=-\dfrac{5}{2}\left( x-4 \right) .\end{array} In the form $y=mx+b$, the equation above is equivalent to \begin{array}{l}\require{cancel} y+8=-\dfrac{5}{2}\left( x-4 \right) \\\\ y+8=-\dfrac{5}{2}x+10 \\\\ y=-\dfrac{5}{2}x+10-8 \\\\ y=-\dfrac{5}{2}x+2 .\end{array} In the form $Ax+By=C$, the equation above is equivalent to \begin{array}{l}\require{cancel} y=-\dfrac{5}{2}x+2 \\\\ 2(y)=2\left( -\dfrac{5}{2}x+2 \right) \\\\ 2y=-5x+4 \\\\ 5x+2y=4 .\end{array} Hence, the different forms of the equation of the line with the given conditions are \begin{array}{l}\require{cancel} \text{slope-intercept form: } y=-\dfrac{5}{2}x+2 \\\\ \text{standard form: } 5x+2y=4 .\end{array}