affine transformation of triangle

Question

$(a,b,c),(a',b',c') \in (\mathbb{R}^2)^3$ triples of pw different points which are not colinear.

Show: There's an affine transformation $x \mapsto Ax+s$ with $Aa+s=a',Ab+s=b',Ac+s=c'$ mapping one "triangle" to the other.

Hints? Where and how does non-colinearity come into play?

Answer 1

Hints?

Assume $a$ and $a'$ both are the origin. Then the other two coordinates will describe $A$. When $a$ and $a'$ are not the origin, you can use $s$ to make them agree, and use difference vectors to determine $A$.

Where and how does non-colinearity come into play?

An affine transformation preserves collinearity. So you can't find an affine transformation which turns collinear points into non-collinear ones.

Answer 2

By the condition of non-colinearity we have that both pairs of vectors $(b-a,\ c-a)$ and $(b'-a',\ c'-a')$ form bases in $\Bbb R^2$.

So, there is a linear transformation $A$ which takes one basis to the other one, in particular now $A(b-a)=b'-a'$ and $A(c-a)=c'-a'$. Then choose $s:=a'-Aa$.