Answer
$AB\parallel CD$, $BC\parallel DA$, $AB\perp BC$
Work Step by Step
Determine the slope of the line passing through $A(1,1)$ and $B(11,3)$:
$$m_{AB}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{3-1}{11-1}=\dfrac{1}{5}.$$
Determine the slope of the line passing through $B(11,3)$ and $C(10,8)$:
$$m_{BC}=\dfrac{y_C-y_B}{x_C-x_B}=\dfrac{8-3}{10-11}=-5.$$
Determine the slope of the line passing through $C(10,8)$ and $D(0,6)$:
$$m_{CD}=\dfrac{y_D-y_C}{x_D-x_C}=\dfrac{6-8}{0-10}=\dfrac{1}{5}.$$
Determine the slope of the line passing through $D(0,6)$ and $A(1,1)$:
$$m_{DA}=\dfrac{y_A-y_D}{x_A-x_D}=\dfrac{1-6}{1-0}=-5.$$
Because $m_{AB}=m_{CD}$, it means that $AB$ is parallel to $CD$.
Because $m_{BC}=m_{DA}$, it means that $BC$ is parallel to $DA$.
From $AB\parallel CD$ and $BC\parallel DA$ it follows that $ABCD$ is a parallelogram
Because $m_{AB}m_{BC}=\dfrac{1}{5}\cdot (-5)=-1$, it means that $AB$ is perpendicular to $BC$.
Therefore $ABCD$ is a rectangle.
