Math Help -> Geometry and Trigonometry -> Polygons and Triangles -> Area -> Vector area

Contents of the "Vector Area" section:

Determinant Area
Vector cross product area
Vector cross product area 2
Determinant area of polygon

The area of a triangle formed by vectors

Start with the vectors (a,b) and (c,d) and draw them so that they both start from the origin (0,0). The end of the vectors are joined to form a triangle. What is the area of this triangle?

K = |(ad-bc)|/2

Here's why. Consider this triangle:

Draw a rectangle around this triangle, so that the rectangle is divided into four regions:

The area of the triangle is the area of the rectangle minus the three new triangles.

triangle = rectangle - yellow - pink - blue:

bc - ab/2 - cd/2 - (a-c)(d-b)/2
bc - ab/2 - cd/2 - (ad-ab-cd+bc)/2
bc - ab/2 - cd/2 - ad/2 + ab/2 + cd/2 - bc/2
bc/2 + bc/2 - ab/2 - cd/2 - ad/2 + ab/2 + cd/2 - bc/2
bc/2 - ad/2
(bc - ad)/2

The drawing, above, shows b>d and c>a. What if these relationships don't exist? Let's consider the case where d>b and c>a.

We will color this diagram the same way as the previous one, but it's a little more complicated because the colored regions will overlap. First, we make the rectangle the same way -- its height is b, and its width is c. Then we color the green and yellow regions as before.

Then we color the blue region as before, but this time we will add it to the area of the rectangle because it is outside the rectangle. The blue region partially overlaps the green triangle. The overlap is shown as a darker blue.

Then we color the pink region as before, showing the partial overlap with the blue region as a darker pink:

triangle = rectangle - yellow - pink + blue

bc - ab/2 - cd/2 + (c-a)(d-b)/2
bc - ab/2 - cd/2 + (cd-bc-ad+ab)/2
bc - ab/2 - cd/2 + cd/2 - bc/2 - ad/2 + ab/2
bc/2 + bc/2 - ab/2 - cd/2 + cd/2 - bc/2 - ad/2 + ab/2
bc/2 - ad/2
(bc - ad)/2

So either way, the area is the same: (bc-ad)/2. Sometimes this ends up being a negative number, and the area can't be negative, so one final adjustment is needed:

K = |(bc-ad)|/2

which is the same as

K = |(ad-bc)|/2

Related Pages in this website

Triangle Area using Determinant

Cross Product, near the bottom of the page, explains why the magnitude of the cross product of two vectors equals twice the area of the triangle formed by the vectors.

Triangle Area using Vectors Cross Products

Heron's formula for the area of a triangle, if all you know is the lengths of its sides.

Polygon Area using Determinant

The webmaster and author of the Math Help site is Graeme McRae.
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