Hello

Given 3 points:

(3,2) (-3,-3) (5,-1) are the vertices of a triangle. Find the area of the triangle.

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- Oct 21st 2007, 03:24 PMsubzero06find the area coordinate triangle
Hello

Given 3 points:

(3,2) (-3,-3) (5,-1) are the vertices of a triangle. Find the area of the triangle. - Oct 21st 2007, 03:58 PMgalactus
Do you have to use a particular method?.

Try using the distance formula to find the lengths of the sides, then use Heron's formula.

You could show off and find the equations of the lines that make up the triangles sides and then integrate.(Nerd)

$\displaystyle \int_{-3}^{3}\left[(\frac{5x}{6}-\frac{1}{2})-(\frac{x}{4}-\frac{9}{4})\right]dx+\int_{3}^{5}\left[(\frac{-3x}{2}+\frac{13}{2})-(\frac{x}{4}-\frac{9}{4})\right]dx$ - Oct 21st 2007, 04:16 PMsubzero06
i tried using ABS formula

Let (x1,y1)=(3,2) ; (x2,y2)=(-3,-3) ; (x3-y3) = (5,-1)

ABS(x2*y1-x1*y2+x3*y2-x2*y3+x1*y3-x3*y1)/2

=ABS[(-3)(2)-(3)(-3)+(5)(-3)-(-3)(-1)+(3)(-1)-(5)(2)]/2

=ABS(-6+9-15-3-3-10)/2

=ABS(-28)/2

=-14

for some reason i get -14...am i suppose to get a "-" for an area?? - Oct 21st 2007, 04:20 PMJhevon
i didn't check your calculations, but if you take the absolute value of a negative number, the answer is positive. so ABS(-28)/2 = 28/2 = 14

so provided your other calculations are correct, you would get +14 as the answer, which makes more sense than a negative answer - Oct 21st 2007, 04:31 PMsubzero06
oooh yea i get it now

thank you very much!