integrate?
http://mathworld.wolfram.com/TriangleArea.html
How do I use three vertices to calculate the area of a triangle?
The three vertices are (0,0), (2,1), and (-1,6).
I've figured out that the equations of the lines that make up the triangle are y = -6x, y = (1/2)x, and y = (-5/3)x + 4.34ish.
Now I have to integrate to find out the area somehow and am confused as how to do this. Any help would be great! Thanks!
3 answers
i tried but the answer i received was not 6.5, which is the correct answer.
Just integrate
a = ∫[-1,0] (13-5x)/3 - (-6x) dx
+ ∫[0,2] (13-5x)/3 - x/2 dx
= 13/6 + 13/3 = 13/2
Or
The distance from (0,0) to (1,2) is √5. Let that be the base of the triangle.
The altitude is the distance from (-1,6) to y=x/2 (or -x+2y+0=0) is
|1+12|/√5 = 13/√5
Thus, the area is 1/2 bh = (1/2)(√5)(13/√5) = 13/2
a = ∫[-1,0] (13-5x)/3 - (-6x) dx
+ ∫[0,2] (13-5x)/3 - x/2 dx
= 13/6 + 13/3 = 13/2
Or
The distance from (0,0) to (1,2) is √5. Let that be the base of the triangle.
The altitude is the distance from (-1,6) to y=x/2 (or -x+2y+0=0) is
|1+12|/√5 = 13/√5
Thus, the area is 1/2 bh = (1/2)(√5)(13/√5) = 13/2