## Algebra 1

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