Axial and central symmetry
Given the triangle $ABC$ defined by the following points $A = (2,1)$, $B = (3, 5)$ and $C = (1,4)$, calculate its homologous if its symmetry axis is the $x$-coordinate axis.
From the formula given in the paragraph of axial symmetries, we can calculate the points homologous from the apexes of the triangle. This way, we get:
$$ A'= \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \end{pmatrix} $$
$$ B'= \begin{pmatrix} 3 \\ -5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -5 \end{pmatrix} $$
$$ C'= \begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \end{pmatrix} $$
The homologous of the triangle $ABC$ is $A'B'C'$ with coordinates $A' = (2, -1)$, $B '= (3, -5)$ and $C' = (1, -4)$.