How do I rotate a line segment in a specific point on the line? If I have two points $A$ and $B$, which $AB$ is a line segment, how can I rotate the $AB$ in another Point $C$ which is a point on the line $AB$. 
Thank you in advance.
 A: Lets just say you have a point $A=(1, 1)$ and $B=(1, 3)$. Suppose you want to
rotate $\overline{AB}$ about the point $C=(1, 2)$ by $\theta = \frac{\pi}{2}$ counter-clockwise, which will give you a segment from $(0, 2)$ to $(2, 2)$.  
Step 1.  You need to translate $C$ to the origin, i.e. apply the linear map  $Ax = x-C$. 
Step 2. Rotate the segment by using the rotational matrix
\begin{align}
R(\theta) =
\begin{pmatrix}
\cos\theta & -\sin \theta\\
\sin\theta & \cos\theta
\end{pmatrix}.
\end{align}
Step 3. Translate back, i.e. $A^{-1}x= x+C$.
Thus, the entire process becomes $A^{-1}R(\theta)A$, a conjugation. Any how, lets see this in our example. 
Take the point $(1, 1)$ which gets map to $(1,1)-(1, 2) = (0, -1)$. Then the rotation by $R(\pi/2)$ gives
\begin{align}
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0\\
-1
\end{pmatrix}
=
\begin{pmatrix}
1\\
0
\end{pmatrix}.
\end{align} 
Lastly, translate back gives $(1, 0) +(1, 2) = (2, 2)$. 
A: Using complex numbers:
In the complex, multiplying by $e^{i\theta}$ amounts to a rotation around the origin by the angle $\theta$. To rotate around another point, translate so that the center moves to the origin, rotate and translate back.
Hence, for any point $a$ in the plane
$$a'=(a-c)e^{i\theta}+c.$$
This expands as
$$a'_x=(a_x-c_x)\cos\theta-(a_y-c_y)\sin\theta+c_x,\\
a'_y=(a_x-c_x)\sin\theta+(a_y-c_y)\cos\theta+c_y.
$$
