I think the following way works, but I'd like to know whether there is a better one ( maybe more direct).
Desmos construction illustrating the way I've followed : https://www.desmos.com/calculator/ijv5dxivfa
My idea is to treat the line $ y = \tan (\alpha)\space x$ as the $y=x$ line in a new referential the X-coordinate of which would be $ y = \tan (\alpha - \pi/4) \space x$ and the Y-coordinate of which would be : $ y = -\tan (\alpha - \pi/4)^{-1} \space x$. This new coordinate system is therefore the standard one rotated $\beta = \alpha - \pi/4$ radians counterclockwise.
After that , one has to find the new coordinates $(X_N, Y_N)$ of point $P$ in the new coordinate system, applying the " rotation by $\beta$ radians clockwise" matrix :
$$ X_N = X \cos(\beta) + Y \sin (\beta)$$
and
$$ Y_N= Y \cos(\beta) - X \sin (\beta)$$.
Because symmetry by the $ y=x$ line amounts to switching the coordinates, the coordinates of $P'$ ( image of P under axial symmetry w.r.t the "new" $ y=x$ line ) in the new coordinate system are $P'= ( Y_N , X_N)$.
Applying to this last expression the " rotation by $\beta$ radians counterclockwise" matrix , one obtains finally the coordinates of $P'$ in the old coordinate system:
$$P' = \bigg {(}Y_N \cos(\beta) - X_N \sin (\beta) , Y_N \cos(\beta) + X_N \sin (\beta) \bigg {)} $$