## Intermediate Algebra (12th Edition)

$\text{a) Slope-Intercept Form: } y=-9x+13 \\\\\text{b) Standard Form: } 9x+y=13$
$\bf{\text{Solution Outline:}}$ To find the equation of the line that passes through the given points $( 2,-5 )$ and $( 1,4 ),$ 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=2 ,\\x_2=1 ,\\y_1=-5 ,\\y_2=4 ,\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-(-5)=\dfrac{-5-4}{2-1}(x-2) \\\\ y+5=\dfrac{-9}{1}(x-2) \\\\ y+5=-9(x-2) .\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+5=-9(x-2) \\\\ y+5=-9(x)-9(-2) \\\\ y+5=-9x+18 \\\\ y=-9x+18-5 \\\\ y=-9x+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=-9x+13 \\\\ 9x+y=13 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-9x+13 \\\\\text{b) Standard Form: } 9x+y=13 .\end{array}