## Algebra 1

a) line AB: $y -1 = 1/3*(x+3)$ line BC: $y-3 = -3*(x-3)$ line CD: $y+2 = 1/3*(x+2)$ line AD: $y+2 = -3*(x+2)$ b) My conjecture is that the slopes of perpendicular lines have a product (multiplication product) of -1. c) $y+4 = 1/7*(x-0)$
a) line AB: $m = (1-3)/(-3-3)$ $m = -2/-6$ $m= 1/3$ $y -1 = 1/3*(x+3)$ line BC: $m = (0-3)/(4-3)$ $m = -3/1$ $m= -3$ $y -3 = -3*(x-3)$ line CD:$m = (0--2)/(4--2)$ $m = 2/6$ $m= 1/3$ $y+2 = 1/3*(x+2)$ line AD $m = (1--2)/(-3--2)$ $m = 3/-1$ $m= -3$ b) lines AB and BC: $1/3 * -3 = -1$ lines BC and CD: $-3 * 1/3 = -1$ lines CD and AD: $1/3 * -3 = -1$ lines AB and AD: $1/3 * -3 = -1$ c) $m*-7 = -1$ $m*-7 * -1/7 = -1 *- 1/7$ $m = 1/7$ $y+4 = 1/7*(x-0)$