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4 More functions. One can have many more functions, formulas, and images. 4). We compute the images of A under g and h, leaving the other three vertices to you: g(A) = g(2, 1) = (2× 2 − 1, 2+3× 1) = (3, 5); h(A) = h(2, 1) = (3× 2+1, 2 − 1 2 +4) = (7, 5). ) Fig. 5 Distortion and preservation. Looking at the three functions f, g, and h we have considered so far, we notice a progressive ‘deterioration’: f simply failed to preserve distances (mapping ABCD to a bigger rectangle), g failed to preserve right angles (but at least sent parallel lines to parallel lines), while h did not even preserve straight lines (it mapped AB and CD to curvy lines).
We simply draw KP, measure it either with a ruler or with a compass, then ‘build’ a 70 0 angle ‘to the left hand’ of KP with the help of a protractor, and finally pick a point P ′ on the angle’s ‘new’ leg so that |KP ′ | = |KP|. That’s all! Fig. 4 It’s an isometry! 2 and prove that every rotation is indeed an isometry. We return to our watch example and prove that |LS| = |L′ S ′ |, which says that the distance between the two images L′ , S′ is equal to the distance between the two original points L, S; the general case is proven in exactly the same way.
Fig. 4 It’s an isometry! 2 and prove that every rotation is indeed an isometry. We return to our watch example and prove that |LS| = |L′ S ′ |, which says that the distance between the two images L′ , S′ is equal to the distance between the two original points L, S; the general case is proven in exactly the same way. 19 Fig. 22): they have two pairs of equal sides as |OS| = |OS′ | (short hands) and |OL| = |OL ′ | (long hands). If we show the in-between angles ∠ LOS and ∠ L ′OS ′ to be equal, then the two triangles are congruent and, of course, |LS| = |L′S ′|.