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I have two points $p_1$ and $p_2$ on a 2-dimensional graph, each having an $x$-coordinate and a $y$-coordinate. I want to rotate $p_2$ by $60^\circ$ around $p_1$, such that $p_1$ is fixed in its position. So, how to know the new $p_2$ coordinates (after rotation)? I think there's some relation between the line that connects $p_1$ to $p_2$ and the angle $60^\circ$, but can't figure what is.

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Write the points in column vectors: $\begin{pmatrix} x\\y \end{pmatrix}$. Then

$P_2':=\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{pmatrix}\cdot (P_2-P_1) + P_1$

where choose $\varphi=\pm 60^\circ$ according which direction you want to rotate.

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    Yes, matrix multiplication, I could have expanded it. First reduced the problem where $P_1$ is the origo (shifted by $-P_1$ before rotation and $+P_1$ after), and the matrix multiplication gives the rotated vector.2012-09-30
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In the complex number plane, a rotation of a point, $z$, through an angle, $\theta$, is accomplished by $z \to z e^{i \theta}$.

If $z = x + iy$, where $x,y \in \mathbb R$, and $\theta = 60^\circ$, then

$x + iy \to (x + i y)(\cos 60^\circ + i \sin 60^\circ) \to (x + iy)(1 + i\sqrt 3)/2 \to (x-\frac 12y\sqrt 3) + i(y +\frac 12x\sqrt 3)$

In $\mathbb R^2$, this translates into $(x,y) \to (x-\frac 12y\sqrt 3, \frac 12x\sqrt 3 + y)$