m3equals333 wrote:
I just calculated the slope of the two apparent legs (AB and BC), confirmed they were negative inverses (perpendicular), then used those #'s to calculate distances and finally 1/2bh
I think this was, what I was originally asking about. And you are right. It actually takes only seconds to calculate the slopes of AB and BC.
In case they they are perpendicular, they should be negative inverses (reciprocals) as you say. So
Slope: (y2-y1)/(x2-x1)
Slope BC: (3-(-3)) / ((4-(-4)) = 6/8 = 3/4
Slope AB: 3-7/4-1 = -4/3
I also realized the following:
From point C, if you move three points up, four points to the right. You get to zero. If you move again three points up, four to the right you get to point B.
From point B, if you move four points up, and three points backwards you get to point A.
Four points up, three points left is the negative reciprocal of three points up and four points right. So it has to be perpendicular....(in practice this is the same as calculating the slopes but it may be a quicker way of seeing it in this particular case).