I compute eigenvector on matrix \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} and find eigenvector of eigenvalue $1 , (\sqrt{2}+1 ,1 ).$ I think that axis of symmetry pass point $(1,2) ,$ (Because $T(0.0)=(2,4)$) so,I found axis of symmetry .
$y=(\sqrt{2}-1)x + 3-\sqrt{2} \quad(1)$
My question.. I try other method.
By $T(0,0)=(2,4),$ Axis of symmetry has slope $\frac{-1}{2}$ ans pass $(1,2)$ and get
$y=\frac{-1}{2}x+\frac{5}{2} \quad (2)$
but it's different $(1).$
I don't know why two formula is different
As a word of caution, the map $T$ is the composition of a reflection followed by a translation. Generally, a composition like this is not expressible as a single reflection. For general knowledge $T$ is called a glide reflection.
You may want to know the fixed points of $T$. Solve $T(x,y) = (x,y)$ and you get $y = (\sqrt{2} - 1)x - 2\sqrt2$. But $T$ is by no means a reflection above this line.
You may want to know only about the reflection part. Since you calculated the eigenvector for this matrix, you can deduce the reflection is in the line $$y = \frac{x}{\sqrt2 +1}$$