MHB How Can You Calculate the Area of a Polygon Using Coordinates?

[cavalieregi]

I have been given various corners which are x and y coordinates for a shape. The coordinates are listed in a vector e.g. xpoints = [x1, x2, x3, …, xn, x1] and ypoints = [y1, y2, y3, …, yn, y1] so that corner 1 would be (x1,y2) and corner n would be (xn, yn). I have listed the first point last so that I can calculate the distance between each point in the shape when using the distance equation. These distances added up will give the perimeter. These points could be anything. The lines made between each point are straight. I was wondering is there a way where I could calculate the total area roughly. I don’t think I could integrate it as I don’t have a function for the shape. I was thinking maybe since all of the lines are straight maybe triangles could be used? Note this is for a programming assignment but is there a mathematical method anyone could think of which I could use to approach this. The picture shows examples of possible areas. Also note the points can be negative to. Also is there any way some could think of only calculating the shaded area. See picture for reference. It include possible shapes.

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[I like Serena]

I have been given various corners which are x and y coordinates for a shape. The coordinates are listed in a vector e.g. xpoints = [x1, x2, x3, …, xn, x1] and ypoints = [y1, y2, y3, …, yn, y1] so that corner 1 would be (x1,y2) and corner n would be (xn, yn). I have listed the first point last so that I can calculate the distance between each point in the shape when using the distance equation. These distances added up will give the perimeter. These points could be anything. The lines made between each point are straight. I was wondering is there a way where I could calculate the total area roughly. I don’t think I could integrate it as I don’t have a function for the shape. I was thinking maybe since all of the lines are straight maybe triangles could be used? Note this is for a programming assignment but is there a mathematical method anyone could think of which I could use to approach this. The picture shows examples of possible areas. Also note the points can be negative to. Also is there any way some could think of only calculating the shaded area. See picture for reference. It include possible shapes.

Hi cavalieregi! Welcome to MHB! :)

Yes. There is a formula to calculate the area.

We can divide the area in a series of triangles.
Suppose $\mathbf p_i$ is the vector pointing to point i, which has coordinates (xi,yi).

Then the first triangle is between $\mathbf p_1$ , $\mathbf p_2$ and $\mathbf p_3$. The second triangle is between $\mathbf p_1$, $\mathbf p_3$ and $\mathbf p_4$, and so on. That is, all triangles have $\mathbf p_1$ as a common point.

The formula is:
$$\text{Area} = \frac 1 2 \left| \sum_{k=2}^{n-1} (\mathbf p_k - \mathbf p_1) \times (\mathbf p_{k+1} - \mathbf p_1)\right|$$

The operator $\times$ indicates the cross product between vectors.
In this case the cross product between the 2 vectors (a,b) and (c,d) can be represented as (ad - bc).

Calculating it like this, means that a hole such as you have, will indeed not be counted.
More specifically, a shape that is oriented counterclockwise is counted positive, while a shape that is oriented clockwise is counted negative.
Taking the absolute value of the result ensures you get a positive area.

 

[cavalieregi]

Thanks for that could you please give me an example with a small intersecting polygon. I don't know how I would apply this formula to any area. Thanks.

 

[I like Serena]

Thanks for that could you please give me an example with a small intersecting polygon. I don't know how I would apply this formula to any area. Thanks.

Let's start with a simple polygon.
Say ((1,1), (2,1), (2,2), (1,2)), which is a square of size 1, which is oriented counterclockwise (aka positive).

Btw, I forgot a factor 1/2, which I've added to my original post just now.

The (new) formula says that:
\begin{aligned}\text{Area}
&= \frac 1 2 \left|\ \Big((2,1) - (1,1)\Big) \times \Big((2,2) - (1,1)\Big) + \Big((2,2) - (1,1)\Big) \times \Big((1,2) - (1,1)\Big)\ \right| \\
&= \frac 1 2 \left|\ (1,0) \times (1,1) + (1,1) \times (0,1)\ \right| \\
&= \frac 1 2 \left|\ (1\cdot 1 - 0 \cdot 1) + (1 \cdot 1 - 1 \cdot 0)\ \right| \\
&= 1\end{aligned}

Note that the symbol $\times$ in this context denotes the so called cross product of vectors as opposed to normal multiplication.