Hello
Given 3 points:
(3,2) (-3,-3) (5,-1) are the vertices of a triangle. Find the area of the triangle.
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.
$\displaystyle \int_{-3}^{3}\left[(\frac{5x}{6}-\frac{1}{2})-(\frac{x}{4}-\frac{9}{4})\right]dx+\int_{3}^{5}\left[(\frac{-3x}{2}+\frac{13}{2})-(\frac{x}{4}-\frac{9}{4})\right]dx$
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??