Answer
If ABCD is a Rectangle, than d(AB) = d(CD), d(DA) = d(BC) and AB is perpendicular to DA (a parallelogram with two adjacent perpendicular sides)
Work Step by Step
1. d(AB) = d(CD)
2. d(DA) = d(BC)
3. AB $\perp$ DA
1.
d(AB) = d(CD)
d(AB) = $\sqrt{(x_b - x_a) + (y_b - y_b)}$
d(AB) = $\sqrt {10^{2}+2^2} $
d(AB) = $\sqrt {104}$
d(CD) = $\sqrt{(x_d - x_c) + (y_d - y_c)}$
d(CD) = $\sqrt {(-10)^2 + (-2)^2}$
d(CD) = $\sqrt{104}$
d(AB) = d(CD)
2.
d(DA) = d(BC)
d(DA) = $\sqrt{(x_d - x_a) + (y_d - y_a)}$
d(DA) = $\sqrt {1^{2}+(-5)^2} $
d(DA) = $\sqrt {26}$
d(BC) = $\sqrt{(x_c - x_b) + (y_c - y_b)}$
d(BC) = $\sqrt {1^2 + (-5)^2}$
d(BC) = $\sqrt{26}$
d(DA) = d(BC)
3.
AB $\perp$ DA
$m_{AB}$ $*$ $m_{DA}$ = -1
$m_{AB}$ = $\frac{y_b - y_a}{x_b - x_a}$
$m_{AB}$ = $\frac{2}{10}$
$m_{DA}$ = $\frac{y_d - y_a}{x_d - x_a}$
$m_{DA}$ = $\frac{-5}{1}$
$\frac{2}{10}$ * $\frac{-5}{1}$ = -1
Thus, hence all the conditions are proven, ABCD is a rectangle
