Question

Given two parallel lines, explain how to transform one onto another using: A translation. Describe all possible translations.

Answer #1

Let *A*=(*X*1,*Y*1),
*B*=(*X*2,*Y*2),
*C*=(*X*3,*Y*3) and
*D*=(*X*4,*Y*4). We have that
*A**C* ∥ *B**D* and that you want to
transform *A* to *B* and *C* to *D* (of
course the same approach can be used to map *A* to
*D* and *C* to *B*). The notation
|*P**Q*| means the length of *P**Q*.
First of all, there are infinitely many transformations that would
map *A**C* to *B**D* (without any
constraints, if you assume that you want your map to be of some
specific class, there may be just a few or even none), and I will
just give you an example of such transformation.

What I will construct is called usually homothety or simply scaling, and this is one of the basic geometry transformations (along with rotation, translation and symmetry).

To start we need the center, let call it
*S*=(*S**x*,*S**y*). However, we
know that *S* is a point on the line passing through
*A* and *B*, and also on the line passing through
*C* and *D*. Therefore, *S* is the
intersection of those two lines (with the special case I will deal
later). To apply the transformation to the point say *P*,
just draw a line *S**P* and then set *P*′ on
the same line, but with
|*S**P*′|=|*B**S*||*A**S*||*S**P*|=|*D**S*||*C**S*||*S**P*|
(where positive sign means *X*′ is on the same side as
*X*, and the negative sign denotes opposite side). Please,
compare it with Intercept theorem (in my country this is known as
Thales' theorem). To give algebraic formula:

*T*(*P*)=|*B**S*||*A**S*|(*P*−*S*)+*S*

or written more explicitly:

*T*(*P**x*,*P**y*)=(|*B**S*||*A**S*|(*P**x*−*S**x*)+*S**x*,|*B**S*||*A**S*|(*P**y*−*S**y*)+*S**y*)

As you can see, if *S*=(0,0) then this becomes a simple
scaling. There is a special case when *A**B* ∥
*C**D*, i.e. when *S* does not exists, or as
some like to say *S* is in infinity. But in this degenerate
case homothety becomes just translation, so

*T*(*P*)=*P*+*A**B*

or

*T*(*P**x*,*P**y*)=(*P**x*+*B**x*−*A**x*,*P**y*+*B**y*−*A**y*)

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