## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=\dfrac{7}{5}x+\dfrac{16}{5} \\\\\text{b) Standard Form: } 7x-5y=-16$
$\bf{\text{Solution Outline:}}$ To find the equation of the line that passes through the given points $( -3,-1 )$ and $( 2,6 ),$ use the Two-Point Form of linear equations. Give the equation in the Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, where \begin{array}{l}\require{cancel} x_1=-3 ,\\x_2=2 ,\\y_1=-1 ,\\y_2=6 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1) \\\\ y-(-1)=\dfrac{-1-6}{-3-2}(x-(-3)) \\\\ y+1=\dfrac{-7}{-5}(x+3) \\\\ y+1=\dfrac{7}{5}(x+3) .\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+1=\dfrac{7}{5}(x+3) \\\\ y+1=\dfrac{7}{5}(x)+\dfrac{7}{5}(3) \\\\ y+1=\dfrac{7}{5}x+\dfrac{21}{5} \\\\ y=\dfrac{7}{5}x+\dfrac{21}{5}-1 \\\\ y=\dfrac{7}{5}x+\dfrac{21}{5}-\dfrac{5}{5} \\\\ y=\dfrac{7}{5}x+\dfrac{16}{5} .\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{7}{5}x+\dfrac{16}{5} \\\\ 5(y)=5\left( \dfrac{7}{5}x+\dfrac{16}{5} \right) \\\\ 5y=7x+16 \\\\ -7x+5y=16 \\\\ -1(-7x+5y)=-1(16) \\\\ 7x-5y=-16 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=\dfrac{7}{5}x+\dfrac{16}{5} \\\\\text{b) Standard Form: } 7x-5y=-16 .\end{array}