Find the area of the parallelogram whose vertices are at `A(0,8),` `B(-2,6),` `C(-4,-6)` and `D(-2,-4).`
This figure is really a parallelogram, for example because the opposite sides
have the same length: `|AB| = |CD| = 2sqrt(2)` and `|BC| = |AD| = 2sqrt(37).` Or we can check that the opposite sides have the same slope.
The area of a parallelogram `ABCD` is twice the area of the triangle `ABC` (or `BCD,` or `CDA,` or `DAB` ). Therefore it is `|AB|*|BC|*|sin(B)|.`
The simplest way to compute this for the points with known coordinates is to note that this expression is the absolute value of the cross product:
`A = |vec(BA) xx vec(BC)| = |lt2,2gt xx lt-2,-12gt| = `
`= |2*(-12)-2*(-2)| = |-24 + 4| = |-20| = 20.`
This is the answer. If you don't know the cross product, you can use Heron's formula for any mentioned triangle (and multiply by 2).
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