Calculating coordinates OK, I have a picture which will hopefully make my explanation a bit clearer.

I have a line $(a, b)$ to $(x, y)$ and all I know is the end points of the line.
I am trying to draw a line from the end point of the original line.
It will have an angle of $\theta$ from the original line and a length of $l$.
Based on these two values I need to work out the coordinates of the end point of this new line.
Length l can be any positive value.
Angle theta can lie anywhere from $-80^\circ$ to $+80^\circ$.
Thanks for any help you can provide!
::Gets out pen and paper and starts scribbling::
 A: 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: $\angle Q = Tan^{-1}(y/x)$

A: OK, so after a lot of scribbling...
I came up with this...

The dotted lines represent lines parallel to the x and y axes.
OK, so.
m = x - a
n = y - b
alpha = inverse tan (n/m)
so...
beta = alpha - theta
(oops, I forgot length l of the new line)
p = l cos(beta)
q = l sin(beta)
SOOOOO...
c = x + p
d = y + q
Phew!
