The signed area of a polygon is given by
|A = (1/2)||
| = (1/2) (
x1y2 + x2y3 + ... + xny1 - x2y1 - x3y2 - ... - x1yn )
The proof of this can be obtained by induction. We
the formula is true for a triangle (n=3). Let's assume it's true for any
n, and show it's true for n+1.
Here is our polygon with n points, labeled counterclockwise, P1, P2, ..., Pn.
Its area is given by the formula above,
An = 1/2(
Remember, this is the "signed area", which is positive if the vertices are given in counterclockwise order. When we add the n+1st point, we will change the area of the polygon, and the amount (and sign) of the change is exactly the area of the newly formed triangle
The area of the yellow triangle, At, is given by the same formula:
At = 1/2(xnyn+1 + xn+1y1 + x1yn - xn+1yn - x1yn+1 - xny1)
Here is the new polygon with n+1 points, labeled counterclockwise, P1, P2, ..., Pn, Pn+1.
Its area, An+1, is the sum of the areas of the original n-gon, An, and the new triangle, At.
An+1 = An + At
An+1 = 1/2(
An+1 = 1/2 (x1y2 + x2y3 + ... + xnyn+1 + xn+1y1 - x2y1 - x3y2 - ... - xn+1yn - x1yn+1),
which proves, by induction, that the formula is right.
What if the new point is inside the original polygon, you might ask. In that case, the new triangle, Pn, Pn+1, P1, has its vertices named in clockwise order, not counterclockwise. So the area of the new triangle, as calculated by the formula, is negative. The algebraic addition of the area of the triangle to the area of the original n-gon results in the new, smaller, area of the n+1-gon.
Triangle Area using Determinant
Triangle Area using Vectors
Introduction to Matrices
Points and Triangles -- answers the question: is the point inside the triangle?
Points and Lines -- answers questions like finding the equation of a line given a point and a slope, or another point
Heron's formula for the area of a triangle, if all you know is the lengths of its sides.
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