If you have 4 equation of the 4 sides of a parallelogram how do you find its area? lets suppose the equations as
$$A1x+B1y+C1=0 —–(1)$$
$$A1x+B1y+C2=0 —–(2)$$
$$A2x+B2y+D1=0 —–(3)$$
$$A2x+B2y+D2=0 —–(4)$$
My thought process started from
$$Area= b*h$$
we can easily find the height, say between (1) and (2) and take (3) or (4) as base
$$h= \frac{\left(C1-C2\right)}{\sqrt{A^2+B^2}}$$
And as for the base maybe we could find the intersection of (1) and (2) with (3) taking (3) as the base
using the distance formula we should be able to find the length of the base and hence the area of the parallelogram
But is there any elegant way? I hope this isn't a repost as there are a similar couple of question but based on vectors (i am alright with vectors being used, but introduce me like a 4-year-old, I have only a basic understanding, that too from physics.)
Best Answer
You have a good start.
To find the base, you can take the perpendicular distance between the other set of sides, and divide by the sine of the corner angle (which you can find by a dot product between the side normals, one of them turned by 90°).
If you keep track of everything symbolically, I believe the square roots will even cancel out at the end.
(Your area formula is wrong, by the way -- the factor of $\frac12$ is for a triangle, not a parallelogram).