Show algebraically that the points A(-3, 1), B(0, 4), C(0, 2) and D(-3, -1) form a parallelogram,
but not a rectangle.
+45 if the points were to form a rectangle, the slopes would have to be perpindicular, or be negative reciprocals of eachother. we find the slopes of all the lines:
\(\begin{align*} \overline{AB} &\Rightarrow \frac{4-1}{0+3} = \frac{3}{3} =1 \\ \overline{BC} &\Rightarrow \frac{4-2}{0-0} = \text{vertical line} \\ \overline{CD} &\Rightarrow \frac{2+1}{0+3} = \frac{3}{3} =1 \\ \overline{DA} &\Rightarrow \frac{-1-1}{-3+3} = \text{vertical line} \\ \end{align*}\)
in order for the lines to be perpindicular, the slopes of \(\overline{AB}\) and \(\overline{CD}\) would have to be 0, which would make a straight line, instead of 1, so the points do not make a rectangle. however, we have two pairs of parallel sides, so we know we have a parallelogram.
