Question

How do you find the area of a triangle from the co-ordinates of its vertices, without having to calculate the length of the triangles sides?

Answer 1

See below.

Explanation:

Finding the area of a triangle from its co-ordinates.

Using diagram.

We are looking for area of `Delta ABC` with co-ordinates:

`A=(x_1,y_1), B= (x_2 ,y_2), C=(x_3 , y_3)`

We form perpendiculars BD, AD and CF to the x axis.

We then notice that:

`"Area of" DeltaABC= "area of trapezium " DBAE + "area of trapezium " EACF - "area of trapezium " DBCF`

`:.`

`"Area " Delta ABC=1/2(y_2+y_1)(x_1-x_2) + 1/2(y_1+y_3)(x_3-x_1)-1/2(y_2+y_3)(x_3-x_2)`

Factor out `1/2` and expand:

`1/2[(y_2+y_1)(x_1-x_2) + (y_1+y_3)(x_3-x_1)-(y_2+y_3)(x_3-x_2)]`

`1/2[-y_1x_2+y_2x_1+y_1x_3-y_3x_1-y_2x_3+y_3x_2]`

Collect terms containing `x_1`

`y_2x_1-y_3x_1=>x_1(y_2-y_3)`

Terms containing `x_2`

`y_3x_2-y_1x_2=>x_2(y_3-y_1)`

Terms containing `x_3`

`y_1x_3-y_2x_3=>x_3(y_1-y_2)`

We now have a formula:

`1/2|[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]|`

The absolute bars ensure that the area will be positive, whatever order we take them in.

This is a much better method than calculating the length of the sides, which invariably will be in surd form, making further calculations difficult. There isn't a lot of information on this and similar methods, and the reason for answering my own question was to show this method. I have to give credit to contributors Dean R and Binyaka C for pointing me in the direction of this and similar methods.

Answer 2

A triangle with vertices `(x_1, y_1), (x_2, y_2)` and `(x_3,y_3)` has area `S` given by the Shoelace formula,

`S = 1/2 | x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3 |`

We can also write it as

`S = 1/2 | (x_1 - x_3)(y_2 - y_3) - (x_2 - x_3)(y_1-y_3) |`

Explanation:

Let's imagine a triangle with vertices `(0,0), (a,b), (c,d).` For now think of `(a,b)` and `(c,d)` in the first quadrant, `0< c < a`, `0 < b < d.`

The triangle's area `S` is the `a \times d` rectangle less the three right triangles.

`S = ad - 1/2 ab - 1/2 cd - 1/2 (d-b)(a-c)`

`S = ad - 1/2(ab + cd + ad -cd - ab + bc)`

`S = 1/2 (ad - bc)`

The area is half the cross product. This is the signed area; if we had taken the points in the other order we'd get negative the area.

`S = 1/2 | ad - bc |`

For an arbitrary triangle we translate `(x_3, y_3)` to the origin,

We substitute `a = x_1 - x_3, b=y_1-y_3,` `c=x_2-x_3, d=y_2-y_3` giving

`S = 1/2 | (x_1 - x_3)(y_2 - y_3) - (x_2-x_3)(y_1-y_3) |`

That's one of our formulas. Multiplying it out, we get the standard form of the Shoelace Formula,

`S = 1/2 | x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3 |`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

When we have three or more dimensions for the vertices we should do it this way:

Given vertices `(x_1, x_2, ..., x_n),` `(y_1, y_2, ..., y_n), (z_1, z_2, ..., z_n)` calculate

`A = (x_1 - y_1)^2 + (x_2 - y_2)^2 + ... + (x_n - y_n)^2`

`B = (y_1 - z_1)^2 + (y_2 - z_2)^2 + ... + (y_n - z_n)^2`

`C = (z_1 - x_1)^2 + (z_2 - x_2)^2 + ... + (z_n - x_n)^2`

`16S^2 = 4AB - (C-A-B)^2`

but I won't derive that one now.