## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=-\dfrac{4}{3}x+\dfrac{29}{3} \\\\\text{b) Standard Form: } 4x+3y=29$
$\bf{\text{Solution Outline:}}$ To find the equation of the line with the following properties: \begin{array}{l}\require{cancel} \text{Slope: } -\dfrac{4}{3} \\\text{Through: } (2,7) ,\end{array} use the Point-Slope Form of linear equations. Give the equation in the 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=7 ,\\x_1=2 ,\\m=-\dfrac{4}{3} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-7=-\dfrac{4}{3}(x-2) .\end{array} Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-7=-\dfrac{4}{3}(x-2) \\\\ y-7=-\dfrac{4}{3}(x)-\dfrac{4}{3}(-2) \\\\ y-7=-\dfrac{4}{3}x+\dfrac{8}{3} \\\\ y=-\dfrac{4}{3}x+\dfrac{8}{3}+7 \\\\ y=-\dfrac{4}{3}x+\dfrac{8}{3}+\dfrac{21}{3} \\\\ y=-\dfrac{4}{3}x+\dfrac{29}{3} .\end{array} Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=-\dfrac{4}{3}x+\dfrac{29}{3} \\\\ 3(y)=3\left( -\dfrac{4}{3}x+\dfrac{29}{3} \right) \\\\ 3y=-4x+29 \\\\ 4x+3y=29 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-\dfrac{4}{3}x+\dfrac{29}{3} \\\\\text{b) Standard Form: } 4x+3y=29 .\end{array}