Geometric Slerp - Calculating Points along an Arc I'm trying to understand how to use Geometric Slerp, as seen here.
Having looked at the following equation:

How can P0 and P1 be calculated in order to using this equation? Aren't P0 and P1 represented by 2 numbers? The 2 numbers being x and y coordinates? or have I miss understood the equation?
Below is what I'm trying to achieve; in a program, I have a camera following a car and when the car turns, the cameras position needs to update to stay behind it (I'm think using a Geometric Slerp is the way to go).
Below are two doodles to help you understand my description above. The first image shows the car and camera; the second shows the details:

Do I need to calculate P1 from P0's position to use this? Either way, I'm unsure how this can be implemented. Thanks. 
Edit:
I've tried to implement it using P0 and P1 as X Coordinates, but doesn't work as expected:
slerp = (((sin((1-t)*Omega))/(sin(Omega)))*p0)+(((sin(t*Omega))/(sin(Omega)))*p1)
 A: Your equation is a vector equation. So, yes, $P_0$ and $P_1$ are 2D points. You multiply these points by scalars, and then add together. 
A: Your equation is a vector equation. So, yes, $P_0$ and $P_1$ are 2D points that each have both $x$ and $y$ coordinates. You multiply these points by the given scalars, and then add together. In a computer implementation, either you do the $x$ and $y$ calculations separately, or you use a 2D point/vector class that supports addition and scalar multiplication.
To get $P_0$ and $P_1$, you'll need to know the position and orientation of the car at time $t=t_0$ and time $t=t_1$. Let's say at time $t=t_0$ the car is in position $Q_0$ and is heading in the direction of the vector $V_0$. Similarly, $Q_1$ and $V_1$ are the position and direction at time $t=t_1$. Then we have
$$
P_0 = Q_0 - L*V_0 \quad \text{and} \quad P_1 = Q_1 - L*V_1
$$
From your first picture, it appears that $V_0 = (0,1)$ -- the car is headed due north.
From your second picture, it looks like $V_1 = (-\sin\Omega, \cos\Omega)$.
If the car actually is standing still (only rotating, not moving forward) then we can write $Q_0 = Q_1 = (Q_x,Q_y)$, and we get the following
$$
\begin{align}
P_0 &= (Q_x, Q_y) - L*(0,1) = (Q_x, Q_y - L) \\
P_1 &= (Q_x, Q_y) - L*(-\sin\Omega, \cos\Omega) = (Q_x + L*\sin\Omega, Q_y - L*\cos\Omega)
\end{align}
$$
The camera position at time $t$ is then $C(t) = \text{slerp}(P_0, P_1, t)$. 
