The isometry f maps points (0,0), (1,0) and (0,1) into (3,3), (3,4) and (4,3).
Determine f in the form f = Ax+b
Are there any fixed points of f, and hence determine whether f is a translation, rotation, reflection or glide-reflection.
Attempt: I equated the transformed coordinates x' with transformation Ax+b to get $$ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ and $$ b = [3,3]^{T}$$ Is this correct?
Also I couldn't understand part 2, can you please explain it? Thank You.
Edit: For the second part, I deduced the transformation was glide- reflection. The transformation f first translates along line $y=x$ by 3 units and then reflects along the same line, hence it appears to be a glide - reflection. As for the fixed points of f, what does in this context? Are the common points among original and transformed coordinates called fixed points or is there any other criteria.
Your matrix $A$ is not correct. Correct is
$$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$
Hence $f(x)=x+(3,3)^T.$ This shows that $f$ has no fixed points.
Can you proceed ?