Find points on parallel lines 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?
 A: 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. 
A: 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.

A: If the first line has direction $\vec{e}=(e_x,e_y)$ then the perpendicular has direction $\vec{n} = (-e_y,e_x)$. So starting from A the coordinates of B are
$$ (x_B,y_B)=(x_A,y_A) + d (-e_y,e_x) $$
