## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=-5x+19 \\\\\text{b) Stadard Form: } 5x+y=19$
$\bf{\text{Solution Outline:}}$ To find the equation of the line with the given characeristics: \begin{array}{l}\require{cancel} \text{through } (4,-1) \\ m=-5 ,\end{array} use the Point-Slope Form of linear equations. Express the answer in Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions, \begin{array}{l}\require{cancel} y_1=-1 ,\\x_1=4 ,\\m=-5 ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-(-1)=-5(x-4) \\\\ y+1=-5(x-4) .\end{array} In the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y+1=-5(x-4) \\\\ y+1=-5(x)-5(-4) \\\\ y+1=-5x+20 \\\\ y=-5x+20-1 \\\\ y=-5x+19 .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=-5x+19 \\\\ 5x+y=19 .\end{array} Hence, the equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-5x+19 \\\\\text{b) Stadard Form: } 5x+y=19 .\end{array}