# Axial and central symmetry

An axial symmetry is a transformation, therefore each point $P$ on the plane maps another point $P'$ also of the plane, so that the axis $e$ would be the perpendicular bisector of the segment $PP'$. The axial symmetries are inverse isometries, because they preserve the distances between its points and its homologous, but its orientation is inverse. The axial symmetry not only appears between an object and its reflection, since many figures that can break in two sections by means of a line are symmetrical with regard to the line. These objects have one (or more) symmetry axes.

The axial symmetry happens when the points of a figure coincide with the points of another one, taking as a reference a line that is known by the name of axis of symmetry. In the axial symmetry we find the same phenomenon as in an image reflected in the mirror.

We call the points that belong to the symmetrical figure homologous points, that is to say, $A’$ is homologous of $A$, $B’$ is homologous of $B$, and $C’$ is homologous of $C$. Besides, the existing distances between the points of the original figure are equal to the distances between the points of the symmetrical figure. In this case: The axial symmetry can also happen in an object with regard to one or more symmetry axes.

If we bend the figure on the traced symmetry axis, we might observe clearly that the points of the opposite parts coincide, that is to say, both parts correspond.

Next, we are going to study the expression in coordinates of the axial symmetries.

Let $P = (x, y)$ and $P '= (x', y ')$ be two points of the plane let's give its expression in coordinates according to the position of its axis:

• The symmetry axis is the y-coordinate axis:

In this case the algebraic representation of the transformation can be done by means of the following system:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$

Next, we are going to calculate the symmetrical of the point $P$ by means of a symmetry which axis is they-coordinate axis. Let $P = (2,2)$ be a point of the plane, then its symmetrical is calculated by means of the following equations system:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \end{pmatrix} \Rightarrow \left\{ \begin{array}{l} x'=-2 \\ y'=2 \end{array} \right.$$

Therefore, the symmetrical point regarding the y-coordinate axis is the point $P'=(-2,2)$.

• The symmetry axis is the x-coordinate axis:

In this case the algebraic representation of the transformation can be done by means of the following system: $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$

We continue with the previous example, let's remember that we had a point P of coordinates $(2,2)$ and in the previous example we had calculated its symmetrical regarding the $y$-coordinate axis. Now we are going to calculate its symmetrical with regard to the $x$-coordinate axis, and we will call this new point $P"$.We calcute its coordinates by means of the following equations system:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \end{pmatrix} \Rightarrow \left\{ \begin{array}{l} x'=2 \\ y'=-2 \end{array} \right.$$

Therefore, the symmetrical point with regard to the x-coordinate axis is the $P''=(2,-2)$.

To finish, of axial symmetries, we are going to study what happens with the composition of axial symmetries:

• The composition of two symmetries with the parallel axes $e$ and $e'$ is translation, which vector has the length twice the distance between the axes, the direction is perpendicular to the axes and its sense is the one that goes from $e$ to $e'$.

• The composition of two symmetries with the perpendicular axes $e$ and $e'$ is a central symmetry with regard to the point where two axes of symmetry meet.

We take again the point $P = (2,2)$ and we are going to apply a symmetry to it regarding the y-coordinate axis and then a symmetry regarding the $x$-coordinate axis. By the last example, the symmetrical point for the $y$-coordinate axis was the point $P' = (-2,2)$. Then, to calculate the symmetrical regarding the $x$-coordinate axis, we solve the following equations system:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 2 \end{pmatrix} \Rightarrow \left\{ \begin{array}{l} x'=-2 \\ y'=-2 \end{array} \right.$$

If first we do the symmetry regarding the x-coordinate axis and then the symmetry concerning the y-coordinates axis, we will get to the same point $Q=(-2, -2)$. As we will see in a while, this transformation is called an axial symmetry.

A central symmetry, centered at point $O$, is a movement of the plane where every point $P$ of the plane has to map to another point $P'$, being $O$ the average point of the segment of endpoints $P$ and $P'$. Note that a central symmetry is equivalent to a rotation of $180^\circ$.

A point is a centre of symmetry of a figure if it defines a central symmetry.

Next we are going to see the expression in coordinates of a central symmetry changing the center of symmetry.

• Coordinates by means of a center symmetry $O=(0,0)$:

In the following image we see how a central symmetry behaves being the centre the origin of coordinates of a point:

Next, a triangle and its homologous are seen by means of a symmetry:

In both cases, the transformation has the following system associate: $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}$$

Given the segment $AB$ formed by the points $A = (1,0)$ and $B = (2,3)$, we are going to calculate its symmetry regarding the centre of coordinates. To do it, we will calculate the symmetry of the points $A$ and $B$. Precisely, the symmetrical of $A$ is $A'$:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} \Rightarrow \left\{ \begin{array}{l} x'=-1 \\ y'=0 \end{array} \right.$$

Therefore $A '= (-1,0)$ . The symmetrical of the point $B$ is $B'$:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \end{pmatrix} \Rightarrow \left\{ \begin{array}{l} x'=-2 \\ y'=-3 \end{array} \right.$$

Therefore, the symmetry of the segment $AB$ is the segment $A'B'$ that passes through the points $A' = (-1,0)$ and $B '= (-2, -3)$.

• Coordinates by means of a center symmetry $O=(a, b)$:

A point $P'$ homologous to a point $P=(x, y)$ by means of a central symmetry of center $O=(a, b)$:

And the figure homologous to a triangle has this form:

Therefore, its associated system, is: $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} 2a \\ 2b \end{pmatrix}$$

where we remember that the values $(a, b)$ are the coordinates of the centre of the symmetry.

We are going to consider the center of the central symmetry $O = (1,2)$ and we want to calculate the symmetrical regarding $O$ of the point $A = (3,7)$. Then, for the system of equations associated with the central symmetry with origin $O=(a, b)$, the coordinates of the symmetrical point are:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 7 \end{pmatrix} + \begin{pmatrix} 2\cdot 1 \\ 2\cdot 2 \end{pmatrix} \Rightarrow \left\{ \begin{array}{l} x'=-3+2=-1 \\ y'=-7+4=-3 \end{array} \right.$$

Therefore, the symmetrical of the point $A$ regarding the centre $O = (1,2)$ it is the point $A'= (-1, -3)$.

To finish this section about axial symmetries, we are going to see what happens when we compose more than one central symmetry simultaneously:

• Symmetries composition with the same centre: As a center symmetry $O$ is equivalent to a rotation with center $O$ and range $180^\circ$, when applying another transformation, the angle will be $360^\circ$, so the same figure is obtained, that is called a regression. It is an involutive transformation.

• Symmetries composition with different center: The composition of two central symmetries with different center is a translation.