I am trying to remember how to compute the symmetric point to an $(x=a,y=b)$ point with respect to a line, i.e. $y=mx +b$, without luck.
Is there a closed form equation for this type of transformation?
As an example, we can consider $(5,10)$ w.r.t the line $y=-x$.
Yes. Let $p$ be the point, $v$ be the vector of the line's direction, and $c$ be any point on the line.
First, take the vector $p-c$ and compute its component $n$ orthogonal to the line:
$$n = (p-c) - \frac{(p-c)\cdot v}{\|v\|^2}v$$
Now reflect $p$ about the line by subtracting twice $n$ from it, to get the reflected point $r$:
$$r = p - 2n.$$
Addendum: in the case your line is in slope-intercept form, you can take $c = (0,b)$ and $v = (1,m)$.
For your example we have $b=(0,0)$ and $p-b = (5,10)$. Since $v=(1,-1)$ we get $$n = (5,10) - \frac{-5}{2}(1,-1) = (15/2, 15/2).$$
Finally $$p-2n = (-10, -5).$$