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This is the question:

Find the area of the rhombus $PQRS$ if the coordinates of point $P$, $Q$ and $R$ are $(6, 4)$, $(8, 7)$ & $(-6, 3)$ respectively.

Does any one know how to solve this?


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I don't think this is a rhombus - which has all 4 sides equal length... – Juan S Jun 6 '11 at 6:52
Compare -- if you knew all 4 vertices, a good first step in finding the area would be to forget one of them. So if you know only 3 vertices then you are already a step ahead. – David K Jul 1 '15 at 17:27

Wikipedia link to geometric meaning of the cross product of two vectors.

As that link shows, taking the magnitude of the cross product of two vectors also gives you the area of the parallelogram defined by those two vectors as the non-parallel sides. In your example, $Q$ is an end point of both vectors and as such we can find the vectors
$\vec{QP} = \langle 2,3 \rangle$
$\vec{QR}=\langle 14,4 \rangle$
and find their cross product and take its magnitude
$||\displaystyle \vec{QP} \times \vec{QR}|| = |2*4 - 14*4| = 48$

Is that the answer you wanted?

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