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by Anurag@Gurome » Fri Jan 27, 2012 12:03 am
sud21 wrote:What is the area of the triangle with apexes (2,-2), (1,1), (0,0)?
Area of the triangle when coordinates of the vertices are (x1, y1), (x2, y2), (x3, y3) = (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]
So, area of the given triangle = (1/2)[2(1 - 0) + 1(0 + 2) + 0(-2 - 1)] = (1/2)[2 + 2 + 0] = 4/2 = 2 sq units
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by tomada » Sat Jan 28, 2012 3:33 pm
I did this somewhat less elegantly than Anurag did. I recognized that the slope from (0,0) to (2,-2) is -1. I also recognized that the slope from (0,0) to (1,1) is 1. Since these are the negative reciprocals of each other, I saw that the lines were perpendicular, indicating a right triangle.
Then I used the formula Area = (1/2)*base*height, using the line from (0,0) to (2.-2) as the base, and the line from (0,0) to (1,1) as the height.

Length of (0,0) to (1,1) = square root of 2.
Length of (0,0) to (2,-2) = 2 * square root of 2

(1/2)*base*height = (1/2)*(2 * square root of 2)* (square root of 2) = 2.
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