I develop some program and I encountered to this problem:

![enter image description here](https://i.stack.imgur.com/5reAv.png)

We know point **P**, **X**, and angle **a**. Range of angle is <0°,360°> between lines **p** and **o**. How I can determine in which side of line **o** is point **X**? Line **p** is always perpendicular to x axe. I thought that I would use this equation: (**Bx - Ax) \* (Cy - Ay) - (By - Ay) \* (Cx - Ax)** (slope) but second point of line **o** is missing and I have no idea how do this with given angle.
['https://math.stackexchange.com/questions/969292', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/183624/']
You have two vectors: $X-P$ and $\langle \cos\left(\frac{\pi}{2}-\alpha\right), \sin\left(\frac{\pi}{2}-\alpha\right)\rangle$. $X$ lies below the line if and only if $(X-P) \times \langle \cos\left(\frac{\pi}{2}-\alpha\right), \sin\left(\frac{\pi}{2}-\alpha\right)\rangle$ (when both are regarded as vectors in $\mathbb{R}^3$ with zero third coordinate) points upwards (so that the three form a right-handed system).
Another way (renamed the angle as "A" to avoid confussion with the coordinate "a"): if (a-x)\*tan(A)>b-y then X is below the line.