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

$y=\dfrac{5}{3}x$
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, $( -6,-10 )\text{ and }( -3,-5 )$, is \begin{array}{l} y-(-10)=\dfrac{-10-(-5)}{-6-(-3)}(x-(-6)) \\\\ y+10=\dfrac{-10+5}{-6+3}(x+6) \\\\ y+10=\dfrac{-5}{-3}(x+6) \\\\ y+10=\dfrac{5}{3}(x+6) \\\\ y+10=\dfrac{5}{3}x+10 \\\\ y=\dfrac{5}{3}x+10-10 \\\\ y=\dfrac{5}{3}x .\end{array}