For angle $\theta$ and vector $b$, construct the point whose image under rotation by $\theta$ about $O$ matches its image under translation by $-b$

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:

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.

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