What is the area of a triangle with vertices (x_1, y_1), (x_2, y_2), (x_3, y_3) ?

2 Answers
Oct 18, 2017

1/2 abs(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)

Explanation:

Given three vertices: (x_1, y_1), (x_2, y_2), (x_3, y_3)

Start by assuming:

x_1 < x_3 < x_2

y_1 < y_2 < y_3

The triangle can be drawn in a rectangle with vertices:

(x_1, y_1), (x_2, y_1), (x_2, y_3), (x_1, y_3)

dividing it into 4 triangles, the other 3 of which are:

  • (x_1, y_1), (x_2, y_1), (x_2, y_2) with area: 1/2 (x_2-x_1)(y_2-y_1)

  • (x_2, y_2), (x_2, y_3), (x_3, y_3) with area: 1/2 (x_2-x_3)(y_3-y_2)

  • (x_3, y_3), (x_1, y_3), (x_1, y_1) with area: 1/2 (x_3-x_1)(y_3-y_1)

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So the area of the given triangle is:

(x_2-x_1)(y_3-y_1) - 1/2 (x_2-x_1)(y_2-y_1) - 1/2 (x_2-x_3)(y_3-y_2) - 1/2 (x_3-x_1)(y_3-y_1)

=1/2 (x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)

Note the symmetry of the final expression. It is symmetric in (x_1, y_1), (x_2, y_2), (x_3, y_3) except for its sign.

If we are only interested in the unsigned area of the triangle then we can ignore our initial assumptions and write the universal formula:

"Area" = 1/2 abs(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)

Jun 20, 2018

Just a small comment to the answer from George C.

Explanation:

The answer from George C. is 100% right, even for triangles with such vertices, that they do not fit on a rectangle. An example would be a triangle with the vertices
(0,0), (2,3), (4,4)
GeoGebraGeoGebra

Even if the rectangle does not fit, we can still apply the formular and get the solution.

A=1/2(|0*3+2*4+4*0-0*4-4*3-2*0|)
A=1/2(|8-12|)
A=1/2(|-4|)
A=4/2=2" units" (confirmed by GeoGebra)