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Suppose that a linear transformation $M:R^2 \rightarrow R^2$ maps a triangle $ABC$ to a congruent triangle $A'B'C'$

($\{A, B, O\}, \{B, C, O\},\{C, A, O\}$ are not colinear, and $A,B,C\neq O$)

Is it true that the linear transformation always conserves length?

Or can there possibly be a counterexample?

(I'm talking about transformations on $R^2$ that can be represented as a 2 by 2 matrix)

Thanks in advance

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    $\begingroup$ use translations $\endgroup$
    – user126154
    Mar 10, 2014 at 15:46
  • $\begingroup$ it is a 'linear' map... so the origin must map onto the origin $\endgroup$ Mar 15, 2014 at 1:34
  • $\begingroup$ well, after the edit my comment looses sense. By the way, why don't you accept the zyx answer? In my opinion it is the "right" answer. Or, if you are not happy with that, could you explain better what exactly you need? $\endgroup$
    – user126154
    Mar 20, 2014 at 12:51

2 Answers 2

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It is true, whether or not "linear" means homogeneous-linear taking $0$ to $0$.

Triangle $ABC$, if it is nondegenerate (nonzero area), determines a linear coordinate system in which $A=(0,0)$, $B=(0,1)$ and $C=(1,0)$, and coordinate $(x,y)$ is assigned to the point $A + x(B-A) + y(C-A)$. The distances between pairs of points are fully and uniquely determined by the coordinates of the points and the lengths $AB, BC$ and $CA$. Therefore, the unique invertible affine (linear inhomogeneous) transformation

"for all $(x,y)$, send the point with $ABC$-coordinates $(x,y)$ to the point with $A'B'C'$-coordinates $(x,y)$"

is an isometry if $ABC$ is congruent to $A'B'C'$ and both triangles are non-degenerate. It is also the unique affine-linear transformation sending $ABC$ to $A'B'C'$, so that any linear transformation with that property is an isometry.

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By definition, all congruent triangles have the same size. So you can map the sides $AB$, $BC$, $CA$ to $A'B'$, $B'C'$ and $C'A'$, respectively. Now let's consider if this mapping is an isometry. Clearly, $|AB|=|A'B'|$, as we have defined by congruency. So the vector $AB$ is mapped to another vector of the same length. The same happens to the vector $BC$ mapped to $B'C'$. These vectors however are linearly independend, as you defined. Now you use the fact that a linear mapping is a isometry iff the mapping of each basis vector keeps the original length constant.

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