# We can tell that f is not a translation or glide reflection (hence, it must be a rotation). How?

Let A=(0,1) B=(0,0) C=(1,0)

Suppose that f(A) = (0.4,1.8), f(B) = (1,1), and f(C) = (1.8,1.6).

How do we prove that if its not a translate or glide, then its a rotation?

Is it because since glide is a combination of translation and reflection, and this if its not a translation and not a glide, then its also not a reflection. So it has to be a rotation?

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Are we assuming that $f$ is an isometry? – Alex Becker May 15 '14 at 5:50
Yes f is an isometry – Instinct May 15 '14 at 5:51
Are you asking for a proof that all isometries of the plane are of these forms, or a proof that this particular one is a rotation? – Robert Israel May 15 '14 at 6:44

You can tell (e.g. by computing the signed area of the triangle) that the transformation $f$ preserves orientation. This rules out reflections and glide reflections. And you can tell it's no translation since the line $AB$ intersects the line $f(A)f(B)$ in a finite point, so $AB\not\Vert f(A)f(B)$ therefore it's no translation either.