Well it might be a simple question.  Given a set of 2D or 3D points, the question is:

How to find the center of geometry of an object?

According to the following figure, the center of geometry differs from the center of mass if it is calculated in the simplest form i.e., homogenous density of mass. The problem appears, indeed, in the computation of those.
Commonly, one approach is to average X coordinates and Y coordinates separately i.e., find an average position to the given points (here in 2D). This may be used as centroid for the set of points representing an object.
As shown, due to the extra vertex along the bottom edge, for a simple rectangle the resulting centroid is (0.5,0.4) while the correct answer is (0.5,0.5).
Note that the example given is too simple. The problem of interest however is for complex shapes in 2D and objects in 3D for which only coordinates of vertices are available.
BTW, an efficient computational way is of interest.

Update 1:
Just to mention that I have checked some web links such as Wikipedia's however my current problem is that there are group of 2D and 3D points wish to find a point as representative for those. Thus centroid became of interest. The points are given without any topological information. You may consider them as point cloud. The demonstration here provided to make it clear that the commonly known averaging of coordinates (see for example this) may be incorrect as shown in the example. Hope the point is clear now.  

  

Update 2:
Here are some implementations for comparison:


aa    = accepted answer below
chull = convex-hull of points i.e., the golden polygon
cent  = centroid proposed in Wikipedia and discussed in aa as the polygon centroid
centl = centroid of polyline as explained in aa


Visually, centl looks better representative for the geometry given compared to cent. Two others look promising here but usually they are too biased if the dispersion of the points were inhomogeneous as it is a usual case.
And also consider that although convex-hull makes problem reasonably simpler however it may generate too long and too short edges without any symmetrical positioning in the space, that is, awareness is necessary if you do simple averaging (i.e., without weighting) for both cases: entire points (green) or convex-hull polygon vertices (blue).

  

Update 3:
One application could be found in this post.


Question: Functional complete and xor
Answer: You have to consider an induction on the number of occurrences of connective in the formula.

Consider $\land$

Basis : in $p \land q$, when $p$ and $q$ have different truth values, then $p \land q$ is false.

Induction step : let $F_n^{\land}(p,q)$ a formula built with only the literals $p$ and $q$ with $n$ occurrences of $\land$, and assume as Induction hypo that it is identically false. 

If we consider now $F_{n+1}^{\land}(p,q)$ (i.e $F_n^{\land}(p,q) \land p$ or $F_n^{\land}(p,q) \land q$), again it is identically false.

Consider $xor$ (i.e. $\nLeftrightarrow$)

Basis : in $p \nLeftrightarrow q$, when $p$ and $q$ have different truth values, then $p \nLeftrightarrow q$ is true.

Induction step : let $F_n^{\nLeftrightarrow}(p,q)$ a formula built with only the literals $p$ and $q$ with $n$ occurrences of $\nLeftrightarrow $, ans assume as Induction hypo that it is not identically false. 

If we consider now $F_{n+1}^{\nLeftrightarrow}(p,q)$ (i.e $F_n^{\nLeftrightarrow}(p,q) \nLeftrightarrow p$ or $F_n^{\land}(p,q) \nLeftrightarrow q$), it can be true or false, and adding a new literal $p$ (or $q$), the result $F_{n+1}^{\nLeftrightarrow}$ will "flip" according to the value of $p$ (or $q$).

Thus, we have that it is not identically false.