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Find the area of triangle ABC that has the coordinates A(1, 3, 0), B(0, 2, 5), and C(-1, 0, 2).

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One way to do that is to find the lengths of ABC sides and use Heron's formula. Not pretty nymbers!

You can find the distance h between C and AB.

Area A(ABC) = 1/2 AB h.

I noticed that ABC "split in half" by yz-plane.

Midpoint M(0,3/2,1) of AC is in yz-plane. ABC's median BM is in yz-plane.

This fact for some reason fascinated me, and, I guess, made me find area of AMB first and then double it to get area of ABC.

BM = sqrt(65)/2.

BM equation

(0)x + (8)y + (-1)z + (-11) = 0,

distance between A and BM is

H=|(0)(1)+(8)(3)+(-10)(0)+(-11)| /sqrt (0^2+8^2+(-1)^2))

H=13/sqrt(65)

A(ABC) = BM*H = (sqrt(65)/2)*(13/sqrt(65)) = 13 / 2 = 6.5

The answer seems to be too good to require such a long solution.

I also think that the fact that vertices A, B, and C are respectively in xy-plane, yz-plane, and xz-plane, should provide a shortcut.

Any suggestions?

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# Homework Help: Area of a triangle in 3-D

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