Answer
$a) |AB| + |BC| = |AC|$
$b) m_{1} = m_{2} = m{3}$
Work Step by Step
$a) |AB| = \sqrt (3+1)^2 + (11-3)^2 = \sqrt 80 = \sqrt (16 \times 5) = 4\sqrt 5$
$ |BC| = \sqrt (5-3)^2 + (15-11)^2 = \sqrt 20 = \sqrt (4 \times 5) = 2 \sqrt 5$
$|AC| = \sqrt (5+1)^2 + (15-3)^2 = \sqrt 180 = \sqrt (36 \times 5) = 6 \sqrt 5$
$|AB| + |BC| = |AC|$
$4 \sqrt 5 + 2 \sqrt 5 = 6 \sqrt 5$
$6 \sqrt 5 = 6 \sqrt 5$
$b) slope AB = m_{1} = \frac{11-3}{3+1} = 2$
$ slope BC = m_{2} = \frac{15-11}{5-3} = 2$
$ slope AC = m_{3} = \frac{15-3}{5+1} = 2$
$m_{1} = m_{2} = m_{3}$