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How do people come up with transformations??

It can't be a coincidence when the textbook has you turn a weird shape into a perfect rectangle!!

My question: What is the procedure for transforming a parallelogram into a rectangle?

I believe that it should be very easy, and involves some type of $x=\frac{1}{2}(au+bv)$ and $y=\frac{1}{2}(au-bv)$.

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Hint:

Let the parallelogram be defined by the vectors $(p,q)$ and $(r,s)$. You want to turn it to a "perfect rectangle", i.e. presumably map to the vectors $(w,0)$ and $(0,h)$.

Write

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}p&r\\q&s\end{bmatrix}=\begin{bmatrix}w&0\\0&h\end{bmatrix}$$

expand and solve for the unknown coefficients $a,b,c,d$. This is a system of four equations in four unknowns, but that splits in two systems of two equations.

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Let the parallelogram have the vertices $ABCD$ and the rectangle the vertices $A'B'C'D'$.

Assuming the transformation is the net result of linear and affine transformations, you could stick to homogeneous coordinates and try to solve $$ T(OA, OB, OC, OD) = (OA', OB', OC', OD') $$ where $OX$ is the vector from the origin to point $X$.

You should provide the dimensions (2D, 3D, nD?) and additional constraints (e.g. particular sides oriented parallel to some coordinate axis) that might simplify the problem.

An example would help.

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I'm not very sure if this answers your question, but if you define a linear isomorphism $T: \mathbb{R^2} \to \mathbb{R^2}$ such that $T(0,1) = v$ and $T(1,0) = w$, the image of the square $[0,1] \times [0,1]$ gets mapped to a parallelogram, with a vertex in $(0,0)$ and such that the segments $\vec{0v}$ and $\vec{0w}$ are two of its sides. Now you can take the inverse function to map the parallelogram to $[0,1] \times [0,1]$. (You can calculate $T^{-1}$ since $T$ is an isomorphism,it all just comes down to inverting a 2 by 2 matrix).

You can further generalize this by considering linear isomorpshisms in $\mathbb{R}^k$, or more in general a special family of functions which are injective "almost everywhere".

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