## Algebra 1

$y-4 = 3(x+2)$ $y-4 + 4 = 3x+6+4$ $y = 3x + 10$ $2x+6y = 10$ $2x+6y-2x = 10-2x$ $6y = - 2x + 10$ $6y/6 = (- 2x + 10)/6$ $y= -1/3*x + 5/3$ The products of the slopes is -1, so the lines are perpendicular.