First, find the x and y components of the line from (a,b) to (x,y).
If we draw a horizontal line from (x,y), we can see that the angle between the extension of the (a,b) to (x,y) line and the horizontal is equal to angle Q because they are alternate interior angles. Since we're given angle R, the angle between the line from (x,y) to (c,d) and the horizontal is Q - R.
Now that we have that angle, we can find the x and y components of the second line segment. The x component is $l \cos(R-Q)$ and the y component is $l \sin(Q-r)$.
Adding components, we find that if (a,b) is the origin, (c,d) is at: $$(x+l \cos(R-Q), y+l \sin(Q-R))$$
If (a,b) is not at the origin, we simply shift (c,d):
$$(c,d) = (a+x+l \cos(R-Q), b+y+l \sin(Q-R))$$
Note: $\angle Q = Tan^{-1}(y/x)$
!Triangles