## Intermediate Algebra (12th Edition)

$\text{slope-intercept form: } y=\dfrac{1}{4}x-\dfrac{15}{4} \\\\ \text{standard form: } x-4y=15$
Using $y-y_1=m(x-x_1)$ or the point-slope form of linear equations, the equation of the line passing through $( 7,-2 )$ and with a slope of $m= \dfrac{1}{4}$ is \begin{array}{l}\require{cancel} y-(-2)=\dfrac{1}{4}(x-7) \\\\ y+2=\dfrac{1}{4}(x-7) .\end{array} In the form $y=mx+b$, the equation above is equivalent to \begin{array}{l}\require{cancel} y+2=\dfrac{1}{4}(x-7) \\\\ y+2=\dfrac{1}{4}x-\dfrac{7}{4} \\\\ y=\dfrac{1}{4}x-\dfrac{7}{4}-2 \\\\ y=\dfrac{1}{4}x-\dfrac{7}{4}-\dfrac{8}{4} \\\\ y=\dfrac{1}{4}x-\dfrac{15}{4} .\end{array} In the form $Ax+By=C$, the equation above is equivalent to \begin{array}{l}\require{cancel} y=\dfrac{1}{4}x-\dfrac{15}{4} \\\\ 4(y)=\left(\dfrac{1}{4}x-\dfrac{15}{4}\right)4 \\\\ 4y=x-15 \\\\ -x+4y=-15 \\\\ x-4y=15 .\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{1}{4}x-\dfrac{15}{4} \\\\ \text{standard form: } x-4y=15 .\end{array}