# 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?

Jun 12, 2018

See below.

#### Explanation:

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

Using diagram.

We are looking for area of $\Delta A B C$ with co-ordinates:

$A = \left({x}_{1} , {y}_{1}\right) , B = \left({x}_{2} , {y}_{2}\right) , C = \left({x}_{3} , {y}_{3}\right)$

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

We then notice that:

$\text{Area of" DeltaABC= "area of trapezium " DBAE + "area of trapezium " EACF - "area of trapezium } D B C F$

$\therefore$

$\text{Area } \Delta A B C = \frac{1}{2} \left({y}_{2} + {y}_{1}\right) \left({x}_{1} - {x}_{2}\right) + \frac{1}{2} \left({y}_{1} + {y}_{3}\right) \left({x}_{3} - {x}_{1}\right) - \frac{1}{2} \left({y}_{2} + {y}_{3}\right) \left({x}_{3} - {x}_{2}\right)$

Factor out $\frac{1}{2}$ and expand:

$\frac{1}{2} \left[\left({y}_{2} + {y}_{1}\right) \left({x}_{1} - {x}_{2}\right) + \left({y}_{1} + {y}_{3}\right) \left({x}_{3} - {x}_{1}\right) - \left({y}_{2} + {y}_{3}\right) \left({x}_{3} - {x}_{2}\right)\right]$

$\frac{1}{2} \left[- {y}_{1} {x}_{2} + {y}_{2} {x}_{1} + {y}_{1} {x}_{3} - {y}_{3} {x}_{1} - {y}_{2} {x}_{3} + {y}_{3} {x}_{2}\right]$

Collect terms containing ${x}_{1}$

${y}_{2} {x}_{1} - {y}_{3} {x}_{1} \implies {x}_{1} \left({y}_{2} - {y}_{3}\right)$

Terms containing ${x}_{2}$

${y}_{3} {x}_{2} - {y}_{1} {x}_{2} \implies {x}_{2} \left({y}_{3} - {y}_{1}\right)$

Terms containing ${x}_{3}$

${y}_{1} {x}_{3} - {y}_{2} {x}_{3} \implies {x}_{3} \left({y}_{1} - {y}_{2}\right)$

We now have a formula:

$\frac{1}{2} | \left[{x}_{1} \left({y}_{2} - {y}_{3}\right) + {x}_{2} \left({y}_{3} - {y}_{1}\right) + {x}_{3} \left({y}_{1} - {y}_{2}\right)\right] |$

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.

Jun 17, 2018

A triangle with vertices $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$ has area $S$ given by the Shoelace formula,

$S = \frac{1}{2} | {x}_{1} {y}_{2} - {x}_{2} {y}_{1} + {x}_{2} {y}_{3} - {x}_{3} {y}_{2} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} |$

We can also write it as

$S = \frac{1}{2} | \left({x}_{1} - {x}_{3}\right) \left({y}_{2} - {y}_{3}\right) - \left({x}_{2} - {x}_{3}\right) \left({y}_{1} - {y}_{3}\right) |$

#### Explanation:

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

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

$S = a d - \frac{1}{2} a b - \frac{1}{2} c d - \frac{1}{2} \left(d - b\right) \left(a - c\right)$

$S = a d - \frac{1}{2} \left(a b + c d + a d - c d - a b + b c\right)$

$S = \frac{1}{2} \left(a d - b c\right)$

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.

$S = \frac{1}{2} | a d - b c |$

For an arbitrary triangle we translate $\left({x}_{3} , {y}_{3}\right)$ 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

$S = \frac{1}{2} | \left({x}_{1} - {x}_{3}\right) \left({y}_{2} - {y}_{3}\right) - \left({x}_{2} - {x}_{3}\right) \left({y}_{1} - {y}_{3}\right) |$

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

$S = \frac{1}{2} | {x}_{1} {y}_{2} - {x}_{2} {y}_{1} + {x}_{2} {y}_{3} - {x}_{3} {y}_{2} + {x}_{3} {y}_{1} - {x}_{1} {y}_{3} |$

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

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

Given vertices $\left({x}_{1} , {x}_{2} , \ldots , {x}_{n}\right) ,$ $\left({y}_{1} , {y}_{2} , \ldots , {y}_{n}\right) , \left({z}_{1} , {z}_{2} , \ldots , {z}_{n}\right)$ calculate

$A = {\left({x}_{1} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {y}_{2}\right)}^{2} + \ldots + {\left({x}_{n} - {y}_{n}\right)}^{2}$

$B = {\left({y}_{1} - {z}_{1}\right)}^{2} + {\left({y}_{2} - {z}_{2}\right)}^{2} + \ldots + {\left({y}_{n} - {z}_{n}\right)}^{2}$

$C = {\left({z}_{1} - {x}_{1}\right)}^{2} + {\left({z}_{2} - {x}_{2}\right)}^{2} + \ldots + {\left({z}_{n} - {x}_{n}\right)}^{2}$

$16 {S}^{2} = 4 A B - {\left(C - A - B\right)}^{2}$

but I won't derive that one now.