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.
Development:
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:
$$ A'= \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \end{pmatrix} $$
$$ B'= \begin{pmatrix} 3 \\ -5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -5 \end{pmatrix} $$
$$ C'= \begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \end{pmatrix} $$
Solution:
The homologous of the triangle $$ABC$$ is $$A'B'C'$$ with coordinates $$A' = (2, -1)$$, $$B '= (3, -5)$$ and $$C' = (1, -4)$$.