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I need to find a translation that has the same effect as a translation on a point $P\in\mathbb{E}^2$, the full question:

Let $\theta$ be a nonzero angle and $\textbf{b}$ a translation vector in the plane. Give a geometric construction for a point $P\in\mathbb{E}^2$ such that

$$\text{Rot}(O,\theta)(P)=\text{Trans}(-\textbf{b})(P).$$

Am I supposed to find a rotation that has the same effect as translating by $-\textbf{b}$ for any $\theta$? Because the geometric interpretation I came up with only works for one theta:

Draw $\textbf{b}$ from the origin. Draw a line perpendicular to $\textbf{b}$ through the origin. Draw another line perpendicular to $\textbf{b}$ through $\frac{1}{2}\textbf{b}$ (a perpendicular bisector of $\textbf{b}$). Choose a point $P$ on this second line such that the angle between the vector $OP$ and the first line we drew through the origin is $\frac{1}{2}\theta$. Then drawing $-\textbf{b}$ from this point $P$ gives us a point $Q$ that satisfies that the angle $POQ=\theta$, because the line through the origin is a perpendicular bisector of the $-\textbf{b}$ vector we just drew. Then a rotation of $\theta$ through the origin moves $P$ to $Q$ and so does a translation by $-\textbf{b}$.

I tried to draw this to make it more clear: Rotation that's equivalent to translation

But this only works for a given theta, is this what the question has in mind? If not, how would I approach this (and also the linear algebra equivalent down below)?

I am then asked the same thing, but using Linear Algebra. Find $x$ and $y$ such that

$$ A \begin{pmatrix} x_1 \\x_2 \end{pmatrix} + \begin{pmatrix} b_1 \\b_2 \end{pmatrix} = \begin{pmatrix} x_1 \\x_2 \end{pmatrix} , \text{where} A= \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{pmatrix}. $$

If I understand this correctly this says: For what vector $\textbf(x)$ is first rotating by $\theta$ and then translating by $\textbf{b}$ equal to doing nothing.

For this again I'm not sure if I have to solve this so it is correct for all $\theta$. Just using Gaussian elimination gave me a not so satisfying answer that I'm having trouble to interpret.

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    $\begingroup$ The way I read the problem statement, you’re given a specific angle $\theta$ and vector $\mathbf b$. $\endgroup$ – amd Feb 14 at 20:15

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