#### Answer

a) Yes
b) $(-1, 5)$

#### Work Step by Step

a)
$AB = (5-1)/(1-1)$
$AB = 4/0 =$ vertical line
$BC = (5-5)/(5-1)$
$BC = 0/4 = 0$
$CD = (5-1)/(5-7)$
$CD = 4/-2 = -2$
$AD = (1-1)/(7-1)$
$AD = 0/6 = 0$
AD and BC are parallel since they have the same slope. A vertical line has an undefined slope (and does not equal a slope of -2).
b)
$CD = (5-1)/(5-7)$
$CD = 4/-2 = -2$
$AD = (1-1)/(7-1)$
$AD = 0/6 = 0$
$BC = (5-y)/(5-x)$
$AB = (1-y)/(1-x) = -2$
$AB = 1-y = -2*(1-x)$
$AB = 1-y = -2+2x$
$AB = 3 = 2x+y$
$BC = (5-y)/(5-x)$
$BC = (5-y)/(5-x)=0$
$BC = 5-y = 0*(5-x)$
$BC = 5-y=0$
$BC = 5=y$
$3 = 2x+y$
$3 = 2x+5$
$-2 = 2x$
$-1 = x$
$BC = (5-y)/(5-x)$
$BC = (5-5)/(5--1)$
$BC = 0/6 = 0$ (parallel to AD)
$AB = (1-y)/(1-x)$
$AB = (1-5)/(1--1)$
$AB = -4/2 = -2$ (parallel to CD)