I have two parallel lines, I know the coordinates of one point (the orange one) on one line, I know also the distance and I have to find the point that is on the other line, the green one.
Is it possible to calculate the coordinates of the other point?
On
From the statement “I have two parallel lines”, I am assuming that L: y = mx + c is the equation of the parallel line that passes through B with m and c known.
Form the circle (C) centered at A and radius = 4. That is, $C: (x – 5)^2 + (y – 2)^2 = 4^2$.
Combining L and C together, we get a quadratic equation in x (or in y) in the form $H: Ax^2 + Bx + C = 0$.
Because L is tangent to C, H should have equal roots and that root is $\dfrac {-B}{2A}$.
simpler Method
Let N be the normal to L and N passes through A. The equation of N can be found.
Solving L and N, we get the required.
Obviously you need to know the common slope, $m=\tan \theta$, of the lines. Let the first one be$$y-mx+c=0$$ From the point $A=(5,2)$ in this line you get $c=5m-2$. Let the other line be $$y-mx+d=0$$ where $$d=c+c'=c+\frac {d}{\cos \theta}=c+4\sqrt{1+m^2}=5m-2+4\sqrt{1+m^2}$$ (see the figure below)
You get the point $B=(x,y)$ from the system $$\begin{cases}y-mx+5m-2+4\sqrt{1+m^2}=0\\\frac{y-mx+5m-2}{\sqrt{1+m^2}}=4\end{cases}$$ where the first equation is the point $B=(x,y)$ belonging to the second line and the second equation is the formula of the distance from $B$ to the first line.