Let's consider 2 functions, $f(x)$ and $g(x)$ defiened :
$f(x)=ax+b$
$g(x)=cx+d$
Let's also consider a function $m(x)$ defiened as $m(x)=ex+h$ that verifie this property :
$f(x)$ is symmetrical to $g(x)$ in relation to $m(x)$. ($m(x)$ is the equation of the of the line which is the axis of symmetry between $f(x)$ and $g(x)$)
Here $f(x)=2x$ and $g(x) =0.5x$. Then $m(x)=1x$
My question is how can we find what is the expression of $m(x)$ with the expression of $f(x)$ and $g(x)$?
I already solved that for $e$ (in the equation $m(x)=ex+h$) :
$e=\text{tan}(\frac{\text{tan}^{-1}(a)+\text{tan}^{-1}(c)}{2})$
But I can't solve it for $h$.
Could you help me?
Note that all three lines meet at the same point $(x_i,y_i)$: $$ax_i+b=cx_i+d\\(a-c)x_i=d-b\\x_i=\frac{d-b}{a-c}$$ Then use $$ax_i+b=ex_i+h\\h=(a-e)x_i+b$$