Question

critique & explain
two students are trying to determine whether compositions of rigid motions are commutative. paula translates a triangle and then reflects it across a line. when she reflects and then translates, she gets the same image. she concludes that compositions of rigid motions are commutative.
translate. then reflect. reflect. then translate.
keenan rotates a triangle and then reflects it. when he changes the order of the rigid motions, he gets a different image. he concludes that compositions of rigid motions are not commutative.
rotate. then reflect. reflect. then rotate.
a. should paula have used grid paper? explain.
b. communicate precisely do you agree with paula or with keenan? explain.
habits of mind
communicate precisely what should you look for to determine whether two given rigid motions are commutative?

Explanation:

A.

Grid paper provides a coordinate system. It helps in precisely measuring distances (for translations) and accurately reflecting over lines (by having clear reference points for coordinates). Without it, visual estimations might lead to incorrect conclusions about whether the compositions result in the - same image.

Keenan is correct. Rigid motions (translations \(T\), reflections \(R\), rotations \(M\)) are not always commutative. The composition of rigid motions \(f\circ g\) (where \(f\) and \(g\) are rigid motions) is not the same as \(g\circ f\) in general. Paula's case might be a special case (e.g., if the translation is parallel to the line of reflection in a particular way), but Keenan's example shows a non - commutative case. Since a mathematical property (commutativity) must hold in all cases for the general statement to be true, and Keenan has shown a counter - example, compositions of rigid motions are not commutative.

To determine if two rigid motions \(f\) and \(g\) are commutative (\(f\circ g=g\circ f\)), we need to check the position of multiple points (not just the overall shape) in the plane. Apply \(f\circ g\) and \(g\circ f\) to at least three non - collinear points (e.g., the vertices of a triangle). If for all these points \(f(g(P))=g(f(P))\) (where \(P\) is a point in the plane), then the rigid motions are commutative for that particular pair. If there exists at least one point \(Q\) such that \(f(g(Q))
eq g(f(Q))\), then the rigid motions are not commutative.

Answer:

Yes, Paula should have used grid paper. Grid - paper allows for precise coordinate - based analysis of translations and reflections. It helps in accurately determining if the two compositions (translate - then - reflect and reflect - then - translate) result in the exact same image. Without grid paper, it is more prone to visual misjudgment.

B.