Rotating Lines
Problem:
Given a line of slope m in the Euclidean plane, what is the slope m’ of the line rotated (counterclockwise) by angle θ?
Solution:
Suppose we have an equation for the line of the form y = mx + b. We can ignore b as it is unrelated to the slope (in effect, we are working in an affine space).
So, y = mx for our purposes. Every point satisfying this equation is a multiple of
![\left[\begin{array}{c}1 \\ m\end{array}\right]](http://s.wordpress.com/latex.php?latex=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%20%5C%5C%20m%5Cend%7Barray%7D%5Cright%5D&bg=ffffff&fg=000000&s=0 "\left[\begin{array}{c}1 \\ m\end{array}\right]")
and, similarly, every point satisfying the equation y = m’x of the rotated line is a multiple of
![\left[\begin{array}{c}1 \\ m'\end{array}\right]](http://s.wordpress.com/latex.php?latex=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%20%5C%5C%20m%27%5Cend%7Barray%7D%5Cright%5D&bg=ffffff&fg=000000&s=0 "\left[\begin{array}{c}1 \\ m'\end{array}\right]")
Since the latter point is the image of the former after rotation by θ, the points are related by a rotation matrix, like so:
![\left[\begin{array}{c}1 \\ m'\end{array}\right] = \left[\begin{array}{cc}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{array}\right] \left[\begin{array}{c}1 \\ m\end{array}\right]](http://s.wordpress.com/latex.php?latex=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%20%5C%5C%20m%27%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ccos%5Ctheta%20%26%20-%5Csin%5Ctheta%20%5C%5C%20%5Csin%5Ctheta%20%26%20%5Ccos%5Ctheta%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%20%5C%5C%20m%5Cend%7Barray%7D%5Cright%5D&bg=ffffff&fg=000000&s=0 "\left[\begin{array}{c}1 \\ m'\end{array}\right] = \left[\begin{array}{cc}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{array}\right] \left[\begin{array}{c}1 \\ m\end{array}\right]")
Solving for m’ then yields
which, of course, is our solution in terms of m and θ.