Answer
The given points are vertices of a paralelogram because they form two pairs of congruent opposite sides
Work Step by Step
If ABCD is a Parallelogram, than:
1. AB // CD
2. AD // BC
3. d(AB) = d(CD)
4. d(AD) = d(BC)
1.
AB // CD if m(AB) = m(CD)
m(AB) = $\frac{y_{b} - y_{a}}{x_{b} - x_{a}}$
m(AB) = $\frac{3}{6}$
m(CD) = $\frac{{y_{d}-y_{c}}}{x_{d}-x_{c}}$
m(CD) = $\frac{3}{6}$
m(AB) = m(CD)
2.
AD // BC if m(AD) = m(BC)
m(AD) = $\frac{y_{d} - y_{a}}{x_{d} - x_{a}}$
m(AD) = $\frac{-2}{6}$
m(BC) = $\frac{{y_{c} - y_{b}}}{x_{c}-x_{b}}$
m(BC) = $\frac{-2}{6}$
m(AD) = m(BC)
3.
d(AB) = d(CD)
d(AB) = $\sqrt {(x_{b}-x_{a})^{2} + (y_{b}-y_{a})^{2}}$
$\sqrt {36 + 9} $= $\sqrt {45}$
d(CD) = $\sqrt {(x_{d}-x_{c})^{2} + (y_{d}-y_{c})^{2}}$
$\sqrt {36 + 9} $= $\sqrt {45}$
d(AB) = d(CD)
4.
d(AD) = d(BC)
d(AD) = $\sqrt {(x_{d}-x_{a})^{2} + (y_{d}-y_{a})^{2}}$
$\sqrt {36 + 4} $= $\sqrt {40}$
d(BC) = $\sqrt {(x_{c}-x_{b})^{2} + (y_{c}-y_{b})^{2}}$
$\sqrt {36 + 4} $= $\sqrt {40}$
d(AD) = d(BC)
Thus, all conditions are met and ABCD is proved of being a parallelogram.
