## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=-\dfrac{5}{2}x+13 \\\\\text{b) Standard Form: } 5x+2y=26$
$\bf{\text{Solution Outline:}}$ Find the slope of the given equation, $2x-5y=7 .$ Then use the negative reciprocal of this slope and the given point , $( 4,3 )$ to find the equation of the needed line. Give the equation in the Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the given equation is equivalent to \begin{array}{l}\require{cancel} 2x-5y=7 \\\\ -5y=-2x+7 \\\\ \dfrac{-5y}{-5}=\dfrac{-2x}{{-5}}+\dfrac{7}{-5} \\\\ y=\dfrac{2}{5}x-\dfrac{7}{5} .\end{array} Hence, the slope of the given line is $m=\dfrac{2}{5} .$ Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=-\dfrac{5}{2} \\\text{Through: } (4,3) .\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=3 ,\\x_1=4 ,\\m=-\dfrac{5}{2} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-3=-\dfrac{5}{2}(x-4) .\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-3=-\dfrac{5}{2}(x)-\dfrac{5}{2}(-4) \\\\ y-3=-\dfrac{5}{2}x+10 \\\\ y=-\dfrac{5}{2}x+10+3 \\\\ y=-\dfrac{5}{2}x+13 .\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{5}{2}x+13 \\\\ 2(y)=2\left( -\dfrac{5}{2}x+13 \right) \\\\ 2y=-5x+26 \\\\ 5x+2y=26 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-\dfrac{5}{2}x+13 \\\\\text{b) Standard Form: } 5x+2y=26 .\end{array}