SOLUTION: find the area of the triangle whose vertices are (1/3,8/3), (5/4, 1/4)and (67,-13/6)

Question 1103173: find the area of the triangle whose vertices are (1/3,8/3), (5/4, 1/4)and (67,-13/6)
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
find the area of the triangle whose vertices are A(1/3,8/3), B(5/4, 1/4) and
C(67,-13/6)
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  A      B      C      A
 1/3    5/4     67    1/3
 8/3    1/4   -13/6   8/3

Add the diagonal products starting upper left
1/12 - 65/24 + 536/3 = 4225/24
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Add the diagonal products starting lower left
20/3 + 67/4 - 13/18 = 817/36
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Find the difference between the 2 sums.
4225/24 - 817/36 = 11041/72
Area = 1/2 the difference
Area = 11041/144 sq units

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