## Precalculus (6th Edition) Blitzer

$y−6=\frac{4}{3}(x−3)$ $y=\frac{4}{3}x+2$
$m$=$(6−(-2))\div(3−(−3))$ $m=(6+2)\div(3+3)$ $m=\frac{8}{6}$ $m=\frac{4}{3}$ $y−6=\frac{4}{3}\times(x−3)$ $y−6=\frac{4}{3}x−\frac{4}{3}\times3$ $y-6=\frac{4}{3}x - 4$ $y=\frac{4}{3}x+2$