# find the area coordinate triangle

• Oct 21st 2007, 03:24 PM
subzero06
find 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 PM
galactus
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 PM
subzero06
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 PM
Jhevon
Quote:

Originally Posted by subzero06
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??

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 PM
subzero06
oooh yea i get it now
thank you very much!