Problems from Axial and central symmetry

Given the triangle $A B C$ 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.

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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:

$A^{'} = (\begin{matrix} 2 \\ - 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) \cdot (\begin{matrix} 2 \\ 1 \end{matrix})$

$B^{'} = (\begin{matrix} 3 \\ - 5 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) \cdot (\begin{matrix} 3 \\ - 5 \end{matrix})$

$C^{'} = (\begin{matrix} 1 \\ - 4 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) \cdot (\begin{matrix} 1 \\ - 4 \end{matrix})$

Solution:

The homologous of the triangle $A B C$ is $A^{'} B^{'} C^{'}$ with coordinates $A^{'} = (2, - 1)$ , $B^{'} = (3, - 5)$ and $C^{'} = (1, - 4)$ .

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