# Problems from 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.

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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:

$$A'= \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

$$B'= \begin{pmatrix} 3 \\ -5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -5 \end{pmatrix}$$

$$C'= \begin{pmatrix} 1 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -4 \end{pmatrix}$$

### Solution:

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

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