0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Joey Zou Jul 6 '16 at 22:08
  • $\begingroup$ You are correct that was the problem. The answer is $|-12|$ $\endgroup$ – Yusha Jul 6 '16 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.