- #1

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With Vertices of (x1,y1) (x2,y2) (x3,y3)

Area= 1/2 [(y2-y1)(x1-x3)-(x2-x1)(y1-y3)]

There was no description with the formula. Where did it come from and why does it work?

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- Thread starter your
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- #1

- 1

- 0

With Vertices of (x1,y1) (x2,y2) (x3,y3)

Area= 1/2 [(y2-y1)(x1-x3)-(x2-x1)(y1-y3)]

There was no description with the formula. Where did it come from and why does it work?

- #2

- 606

- 1

With Vertices of (x1,y1) (x2,y2) (x3,y3)

Area= 1/2 [(y2-y1)(x1-x3)-(x2-x1)(y1-y3)]

There was no description with the formula. Where did it come from and why does it work?

Look here http://en.wikipedia.org/wiki/Triangle#Using_Heron.27s_formula , under "using coordinates", the third "T" there.

DonAntonio

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