Why does a line rotate clockwise when you substitute a counter-clockwise rotation? The point $(x, y)$ can be rotated by $\theta$ in a counter-clockwise direction about the origin $(0, 0)$. This point $(x', y')$ can be found by using the rotation matrix, as so:
$
\begin{bmatrix}
   cos(\theta) & -sin(\theta) \\
   sin(\theta) & cos(\theta)
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
x' \\ y'
\end{bmatrix}
$
That is, 
$
\begin{aligned}
x' & = x\cos(\theta) - y\sin(\theta) \\
y' & = x\sin(\theta) + y\cos(\theta)
\end{aligned}
$
Given the equation of a line
$Ax + By = C$
To rotate this line by $\theta$ in a counter-clockwise direction about the origin, one would think that you would simply substitute $(x', y')$ into the equation, giving you:
$A (x\cos(\theta) - y\sin(\theta)) + B (x\sin(\theta) + y\cos(\theta)) = C$
or with collected x and y terms:
$x(A\cos(\theta) + B\sin(\theta)) + y(B\cos(\theta) - A\sin(\theta) = C$
However, this is not the case, I now realise that this actually rotates it in a clockwise direction, but I don't understand why. 
Can anyone explain why this is the case? I feel that it has something to do with the form with collected x and y terms, due to it resembling the transpose of rotation matrix multiplied by $[A, B]^T$, but to me that doesn't make much sense.
 A: Your transformation expresses the new coordinates $x'$, $y'$ in terms of the old coordinates $x$, $y$.
On the other hand, in order to find the image of the rotated line in the new coordinate system, you must substitute to the old coordinates $x, y$ appearing in its equation their expressions in terms of the new coordinates $x'$, $y'$. 
In other words, you must apply the inverse of the tranformation you are considering.
A: Let $f : \mathbb{R}\to\mathbb{R}$. Think about how the graph of $f(x−c)$ is the graph of $f(x)$ moved to the right by $c$, not to the left. This is because $f(x−c)$ attains the value $f(x_0)$ at $x_0+c$:
$$f((x_0 + c) - c) = f(x_0)$$
so the entire function is displaced to the right.
Similarly here, by substituting $(x',y')$ into the line equation, the point $(x_0,y_0)$ satisfying the line equation is actually attained for $(x_0,y_0)$ rotated clockwise by $\phi$.
This is effectively rotating the entire line clockwise by $\phi$.
A: To explain Francesco Polizzi's comment in a symbolic way that I find more satisfying: let's say you have a set $S$ in the plane defined by the equation $f(x) = 0$, and you'd like to know its image $TS$ under the invertible transformation $T$. The image in question, in symbols, is $$\{T(x, y) | f(x, y) = 0\}.$$
This is equivalent (writing $(x', y') = T(x, y)$) to $$\{(x', y') | f( T^{-1}(x', y') ) = 0\}.$$
In your case, if $S$ is your original line $Ax + By = C$ and $TS$ is the line rotated counter-clockwise, the equation for $TS$ is such that if you apply a clockwise transformation to the coordinates in the equation (thereby taking $TS$ to $S$), you recover the equation for $S$.
