Get angle of unknown end point of a line - c++

Question

Short Version:
Basically, I'm trying to figure out the X and Y values of an unknown Point C using the ending point of Line AB ( Point B ).
I tried using atan2( B.y, B.x) but it gives me the angle respective to the X-Axis which leads to wrong coordinates. I figured that I needed to get the angle formed by Line BC and the imaginary horizontal axis in which Point B lies ( Angle P ). Is there a way to obtain that angle? If getting that angle would not do what I want to do, what would?
Long Version:
I'm trying to do some math using C++ and I got stuck with this certain problem.
I have two vectors, Vector A and Vector B ( illustrated by Point A and Point B and they form the line AB ). Now, I want to get Vector C ( Point C ), which is L units away from Vector B ( illustrated by Line BC ).
I searched around, read some books, and got this formula to get the point that I need.
/* Let:
* L = length from B to C
* ( B.x, B.y ) = start point
* ( C.x, C.y ) = end point
* theta = angle respective to X-Axis ( using atan2( B.y, B.x ) )
*/
// To get C.x use formula cos( theta ) = ( C.x - B.x ) / L then derive
C.x = ( L * cos( theta ) ) + B.x;
// To get C.y use formula sin( theta ) = ( C.y - B.y ) / L then derive
C.y = ( L * sin( theta ) ) + B.y;
However, this does not result into Point C as illustrated above because the resulting C will be respective to the X-Axis. After reading some more, I figured out that I need to get the angle ( illustrated by purple Angle P ) between the Line BC ( pale red line ) and the imaginary horizontal axis ( teal line ) in which Point B lies.
Another example:
In this case, using the formula above would result to C being in the wrong place ( above ( X , 5 ) )
I am not sure if getting Angle P would do what I want, if you know a much more appropriate answer, please do so.

Answer 1

You don't need trig for this. It's simple proportions. I handy rule I found is that if you find yourself using sin, cos and atan2 then you're probably doing it wrong.
Here's how to do it without trig
dx = B.x - A.x;
dy = B.y - A.y;
dist = sqrt(dx*dx + dy*dy);
C.x = B.x + L*(dx/dist);
C.y = B.y + L*(dy/dist);

Answer 2

The first thing to do to is to create a tempB as if A was the origin(so you can find angle of B to A and so forth).
Subtract A's values from B.
tempB.x = B.x - A.x
tempB.y = B.y - A.y
Then you can use atan2 along with sin/cos (your formulas) to obtain C.x and C.y.
theta = atan2(tempB.y, tempB.x)
C.x = ( L * cos( theta ) ) + B.x;
C.y = ( L * sin( theta ) ) + B.y;
Another alternative is to obtain the unit vector of tempB.
magnitude = sqrt( pow(tempB.x,2) + pow(tempB.y,2))
unit.x = tempB.x/magnitude
unit.y = tempB.y/magnitude
Multiplying this unit vector by L gives you the offset from B.
C.x = B.x + unit.x * L
C.y = B.y + unit.y * L

Answer 3

What you need (in vector terms) is C = B + L*normalize(B-A). If you will be doing a lot more of this sort of calculation it would be worth using a vector library so that you can write your code exactly like that. Otherwise, You should use John's answer, which breaks down the same calculation into its elementary steps.
Edit: Here's a library you could use quite easily (it's just header files): GLM