## Algebra 1

$y-2 = 4/3*(x-3)$ $y = 4/3*x -2$
given: (0, -2) and (3,2) $y-y_{1} = m*(x-x_{1})$ $m = (2--2)/(3-0)$ $m = 4/3$ $y-y_{1} = m*(x-x_{1})$ $y-2 = 4/3*(x-3)$ $y-2 = 4/3*(x-3)$ $y-2 = 4/3*x - 4$ $y-2 + 2 = 4/3*x - 4 + 2$ $y = 4/3*x -2$