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2002## Triangle On A Plane

Find the area of the triangle whose vertices are located at

A: (4,6)

B: (0,9)

C: (-5,-6)

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## Triangle On A Plane Solution

This problem has a *multitude* of possible solutions, many of which were submitted. Here is a sampling (note that, except the last one, none of these are 'complete' solutions; they are just summaries of the methods):

Using the distance formula, discover that the sides fit a^{2} + b^{2} = c^{2}, so it's a right triangle. Multiply the legs, divide by two.

Notice that two of the slopes of the sides are negative reciprocals, so the triangle is right. Ditto above.

Use Heron's formula. This is somewhat like using a sledgehammer to kill a mosquito, but it does work.

Draw a rectangle around triangle, and then subtract out the areas of the triangles which lie between the triangle and the rectangle.

And finally we have this solution, which was submitted by **CJ - EoB** and was used (in one form or another) by several solvers:

To find the area of figures defined by coordinates such as this, list out the coordinates in this way:

4 6

0 9

-5 -6

4 6

Then cross multiply on both diagonals and find the difference:

|(4*9 + 0*-6 + -5*6) - (6*0 + 9*-5 + -6*4)|

= |(36 - 30) - (-45 - 24)|

= |6- -69|

= 75

Then, divide this total by 2 to find the area:

**75/2 = 37.5**

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