## Intermediate Algebra (12th Edition)

$\text{slope-intercept form: } y=-\dfrac{2}{5}x+\dfrac{13}{5} \\\\ \text{standard form: } 2x+5y=13$
In the slope-intercept form, the given equation, $2x+5y=10 ,$ is equivalent to \begin{array}{l}\require{cancel} 5y=-2x+10 \\\\ y=-\dfrac{2}{5}x+\dfrac{10}{5} \\\\ y=-\dfrac{2}{5}x+2 .\end{array} Hence, the slope of this line is $-\dfrac{2}{5} .$ Since the line passing through the given point, $( 4,1 ),$ is parallel to the previous line, then the slopes of these lines are equal. Using $y-y_1=m(x-x_1)$ or the point-slope form, the equation of the line is \begin{array}{l}\require{cancel} y-1=-\dfrac{2}{5}(x-4) .\end{array} In $y=mx+b$ form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-1=-\dfrac{2}{5}x+\dfrac{8}{5} \\\\ y=-\dfrac{2}{5}x+\dfrac{8}{5}+1 \\\\ y=-\dfrac{2}{5}x+\dfrac{8}{5}+\dfrac{5}{5} \\\\ y=-\dfrac{2}{5}x+\dfrac{13}{5} .\end{array} In $Ax+By=C$ form, the equation above is equivalent to \begin{array}{l}\require{cancel} 5(y)=5\left( -\dfrac{2}{5}x+\dfrac{13}{5} \right) \\\\ 5y=-2x+13 \\\\ 2x+5y=13 .\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{2}{5}x+\dfrac{13}{5} \\\\ \text{standard form: } 2x+5y=13 .\end{array}