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Give a equation of rotation the plane around the point $(1,1)$, rotated by an angle $\frac{\pi}{4}$ ?

Idea: I think this should be deposition rotation and translation. Points under rotation and translation:

$(1,1) \mapsto (1,1)$

$(2,2) \mapsto (0,2)$

Rotation matrix: \begin{pmatrix} \cos(\pi/4) & -\sin(\pi/4) \\ \sin(\pi/4) & \cos(\pi/4) \end{pmatrix}

I have problem how to write equation connecting this two facts. Any ideas?

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  • $\begingroup$ What is a "lineral equation"? Do you mean a linear equation? Your sentence, "I think this should be deposition rotation and translation." doesn't parse in English. Please clarify what you mean. $\endgroup$ – Adrian Keister Jul 5 at 16:52
  • $\begingroup$ As a general rule, you need to come up with a matrix that does the following: translate a point to be rotated as centered about the origin, then rotate according to your (correct) rotation matrix, then un-translate what you did before. $\endgroup$ – Adrian Keister Jul 5 at 16:53
  • $\begingroup$ I am not sure how to define linear equation. I think is better remove them. I consider only equation. $\endgroup$ – Martin Inf1n1ty Jul 5 at 16:55
  • $\begingroup$ @AdrianKeister your idea is clear for my. But I don't know how I will describe that equation. $\endgroup$ – Martin Inf1n1ty Jul 5 at 16:58
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    $\begingroup$ The point $(2,2)$ does not go to $(2,0)$. Since rotation preserves distance, $(2,2)$ gets sent to the point north of $(1,1)$ at a distance of $\sqrt{2}$. $\endgroup$ – John Douma Jul 5 at 17:50
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First, translate the plane so that $(1,1)$ goes to the origin.

This gives $$\begin{pmatrix} x-1 \\ y-1 \end{pmatrix}$$

Then rotate by $\frac{\pi}{4}$ to get $$\begin{pmatrix}\frac{1}{\sqrt2}\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix} x-1 \\ y-1 \end{pmatrix}=\begin{pmatrix} \frac{x-y}{\sqrt{2}} \\ \frac{x+y-2}{\sqrt{2}}\end{pmatrix}$$

Finally, do the inverse translation, i.e. add $1$ to each coordinate to get $$\begin{pmatrix} \frac{x-y+\sqrt{2}}{\sqrt{2}} \\ \frac{x+y+\sqrt{2}-2}{\sqrt{2}}\end{pmatrix}$$

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