## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$y=\dfrac{11}{5}x+3$
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form, then the equation of the line passing through the given points, $( -1,-2.5 )\text{ and }( 4,8.5 )$, is \begin{array}{l} y-(-2.5)=\dfrac{-2.5-8.5}{-1-4}(x-(-1)) \\\\ y+2.5=\dfrac{-11}{-5}(x+1) \\\\ y+2.5=\dfrac{11}{5}(x+1) \\\\ y+2.5=\dfrac{11}{5}x+\dfrac{11}{5} \\\\ y=\dfrac{11}{5}x+5.5-2.5 \\\\ y=\dfrac{11}{5}x+3 .\end{array}