## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=14 \\\\\text{b) Standard Form: } y=14$
$\bf{\text{Solution Outline:}}$ To find the equation of the line with the given characeristics: \begin{array}{l}\require{cancel} \text{through } (-3,14) \\ \text{horizontal} ,\end{array} use the Point-Slope Form of linear equations. Express the answer in Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ Horizontal lines have a slope of $0$. 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=14 ,\\x_1=-3 ,\\m=0 ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-14=0(x-(-3)) \\\\ y-14=0 .\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-14=0 \\\\ y-14=0x \\\\ y=0x+14 \\\\ y=14 .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=14 .\end{array} Hence, the equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=14 \\\\\text{b) Standard Form: } y=14 .\end{array}