Three of the vertices of a parallelogram are $(0,1),(1,2)$, and $(2,1)$. The area of the parallelogram is
(A) 1
(B) 2
(C) $\sqrt{2}$
(D) $2 \sqrt{2}$
(E) 4
Solution

There are three possible locations for the fourth vertex, but each resulting parallelogram has the same area.

Locating the fourth vertex at $(3,2)$ yields a parallelogram with base 2 and height 1 . Its area is $(2)(1)=2$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-313.jpg?height=450&width=529&top_left_y=1190&top_left_x=1359)

ANSWER: (B)

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