# Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3).

Find the area of the parallelogram with vertices at (−1,1), (−5,5), (−2,−1), and (−6,3).

I can't figure out what I'm doing wrong..

I let:

$a = |DC| = (-2,-1) - (-6,3) = (4,-4)$

$b = |AD| = (-1,1) - (-5,5) = (4,-4)$

so:

$\begin{bmatrix} 4&-4\\4&-4\end{bmatrix} \implies$ Area $= -16 + 16 = 0$ ?

The determinant is the area of the parallelogram is it not?

• Not sure how you're labelling vertices, but $a$ appears to be $|CD|$, while $b$ appears to be $|AB|$. You get a result of zero because those lines are parallel. – Joey Zou Jul 6 '16 at 22:08
• You are correct that was the problem. The answer is $|-12|$ – Yusha Jul 6 '16 at 22:11