Answer
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) $y+4 = 1/7*(x-0)$
Work Step by Step
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)
$m*-7 = -1$
$m*-7 * -1/7 = -1 *- 1/7$
$m = 1/7$
$y+4 = 1/7*(x-0)$
