LinkBack URL
About LinkBacksDo 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.
![\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](http://latex.codecogs.com/png.latex?\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??
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 = 14Originally Posted by subzero06
so provided your other calculations are correct, you would get +14 as the answer, which makes more sense than a negative answer
  |
