The $\langle u \; v \rangle$ coordinate system is displaced from the canonical system by $O_{uv}$.

What is the position of point $p$ with respect to the $O_{uv}$ origin?

$p' = p - O_{uv}$

Both $p$ and $O_{uv}$ are written in the canonical coordinate system.

The $\langle u v \rangle$ coordinate system is displaced and rotated from the canonical system.

What is the position of point $p$ in the $\langle u v \rangle$ coordinate system?

Do this in two steps:

1. Find $p' = p - O_{uv}$.
2. Project $p'$ onto the axes of $\langle u v \rangle$.

$p_{uv} = \begin{bmatrix} \cdots u \cdots \\ \cdots v \cdots \end{bmatrix} \left( p - O_{uv} \right)$

Scale each component of $(x, y)$ by separate amounts: $(s_x, s_y)$.

$\begin{bmatrix} s_x \; x \\ s_y \; y \end{bmatrix} = \begin{bmatrix} s_x & 0 \\ 0 & s_y \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

Problem: Find a transformation, $T$, that rotates points, $v$, on an object around that object's centre, $c$.

Since rotation is done around the origin, the centre must first be moved to the origin, then the rotation must be done, then the centre must be moved back to its original position:

So the final transformation is

$v''' = c+ R_\theta \; (v - c)$

This gets worse as more transformations are applied, which happens a lot in computer graphics.

The solution is to use Homogeneous Coordinates, describe in the next lecture.