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

$y=2x-1$
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, $( -1,-3 )\text{ and }( -4,-9 )$, is \begin{array}{l} y-(-3)=\dfrac{-3-(-9)}{-1-(-4)}(x-(-1)) \\\\ y+3=\dfrac{-3+9}{-1+4}(x+1) \\\\ y+3=\dfrac{6}{3}(x+1) \\\\ y+3=2(x+1) \\\\ y+3=2x+2 \\\\ y=2x+2-3 \\\\ y=2x-1 .\end{array}