If the images of the points \(\mathbf{A}(1,3), \mathbf{B}(3,1)\) and \(\mathbf{C}(2,4)\) in the line \(x+2 y=4\) are D, …

The mirror line is \(x+2 y-4=0\)
image of \(\mathrm{A}(1,3)\) is \(\frac{x-1}{1}=\frac{y-3}{2}=-2\left(\frac{1+6-4}{5}\right)\)
image of \(\mathrm{B}(3,1)\) is \(\frac{x-3}{1}=\frac{y-1}{2}=-2\left(\frac{3+2-4}{5}\right)\)
image of \((2,4)\) is \(\frac{x-2}{1}=\frac{y-4}{2}=-2\left(\frac{2+8-4}{5}\right)\)

So \(D=\left(-\frac{1}{5}, \frac{3}{5}\right), E=\left(\frac{13}{5}, \frac{1}{5}\right), F=\left(\frac{-2}{5}, \frac{-4}{5}\right)\)
Centroid \(=\left(\frac{\frac{-1}{5}+\frac{13}{5} -\frac{2}{5}}{3}, \frac{\frac{3}{5}+\frac{1}{5}-\frac{4}{5}}{3}\right)\)
\(=\left(\frac{10}{15}, 0\right)=\left(\frac{2}{3}, 0\right)\)