## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=\dfrac{1}{2}x-1 \\\text{b) Standard Form: } x-2y=2$
$\bf{\text{Solution Outline:}}$ Find the slope of the line defined by the given equation, $-x+2y=10 .$ Then use the Point-Slope Form of linear equations with the given point, $(-2,-2)$ to find the equation of the needed line. Finally, Express the answer in both the Slope-Intercept and Standard forms. $\bf{\text{Solution Details:}}$ In the form $y=mx+b,$ the given equation is equivalent to \begin{array}{l}\require{cancel} -x+2y=10 \\\\ 2y=x+10 \\\\ \dfrac{2y}{2}=\dfrac{x}{2}+\dfrac{10}{2} \\\\ y=\dfrac{1}{2}x+5 .\end{array} Using $y=mx+b$ or the Slope-Intercept Form, where $m$ is the slope, then the slope of the equation above is \begin{array}{l}\require{cancel} m=\dfrac{1}{2} .\end{array} Since parallel lines have the same slope, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} \text{Through: } (-2,-2) \\ m=\dfrac{1}{2} .\end{array} 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=-2 ,\\x_1= -2 ,\\m= \dfrac{1}{2} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-(-2)=\dfrac{1}{2}(x-(-2)) \\\\ y+2=\dfrac{1}{2}(x+2) .\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+2=\dfrac{1}{2}(x)+\dfrac{1}{2}(2) \\\\ y+2=\dfrac{1}{2}(x)+1 \\\\ y=\dfrac{1}{2}x+1-2 \\\\ y=\dfrac{1}{2}x-1 .\end{array} In the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} 2(y)=2\left( \dfrac{1}{2}(x)-1 \right) \\\\ 2y=x-2 \\\\ -x+2y=-2 \\\\ -1(-x+2y)=-1(-2) \\\\ x-2y=2 .\end{array} Hence, $\text{a) Slope-Intercept Form: } y=\dfrac{1}{2}x-1 \\\text{b) Standard Form: } x-2y=2 .$