02 Sep

What do you need for a rotation?

This question is a bit more difficult to answer than for a translation or a reflection.

We need an image. And we need an angle of rotation. But an angle of rotation isn’t enough – we need direction. In our high school geometry course, we begin to let the sign of the angle indicate the direction of the rotation. And that still isn’t enough. We need a center of rotation.

Are the two triangles congruent?


Yes. The two triangles are congruent because there is a rigid motion that maps one triangle onto the other. A rotation is a rigid motion. A rotation is an isometry.

What does a rotation buy us mathematically? What has to be congruent in the diagram? Of course the triangles are congruent. And their corresponding parts are congruent (angles, segments, etc.). Their perimeters and areas are equal. But what else? OB=OB’, OA=OA’, and OC=OC’. m∠BOB’=m∠COC’=m∠AOA’=angle of rotation.

How does a 90° rotation compare with a –90° rotation? Which one is clockwise? Which one is counter-clockwise? In the past, I would have told students that a positive rotation is clockwise and a negative rotation is counterclockwise. Instead, we let students look for regularity in repeated reasoning as they explored rotations in the coordinate plane.

Image     Image  

They determined which was which and recorded information about different angle rotations in a table so that they could reason abstractly and quantitatively.


So it’s the next day. Students don’t have out their chart – because while we used the chart to organize our information about rotations and generalize the rotations, we don’t want to rely on having to memorize it.

I sent a Quick Poll to assess student understanding of a rotation. Rotate the point (–4,5) –90° about the origin.


I monitored the results as they came in – and because only 9 students had it correct, I unchecked “Show Correct Answer” before I showed them the results.


I asked them to get up, find someone in the room who had answered differently, and convince that person that their answer was correct. I sent the poll again.


While we are down to two responses, only 10 students had it correct. (Not good. Usually at least a majority has it correct the second time around.) I still didn’t show them the correct answer. We went back to the drawing board (the Graphs page). If (4,5) is the answer, then what has to be true? From our earlier exploration, we knew that AO must equal OA’. Necessary. But not sufficient..because we also knew that m∠AOA’=90°. An angle measure of 77° isn’t good enough.


What happens when A’ is at (5,4)? AO=AO’ and the angle measure is 90°.


The students asked for another Quick Poll. This time, they rotated (4,2) 90° about the origin. 73% correct (up from 33% correct in the previous poll). What mistake did those who answered incorrectly make? They rotated clockwise instead of counterclockwise. An easier mistake to correct.


So it hit me while I was driving home that afternoon that while we had learned some about rotations, we had missed a huge opportunity to look for and make use of structure.

I started class the next day with another rotation question. 84% of the students got it correct.


We went back to the Graphs page. What has to be true?


AO=AO’. Yes. What else has to be true?

m∠AOA’=90°. Yes. What else has to be true?

The rotation is clockwise. Yes. What else has to be true?

A’ is in Quadrant II. Yes. What else has to be true?

Someone suggested that the lines containing segments AO and AO’ have to be perpendicular. Really? Yes. Perpendicular lines intersect to form right angles.

What else has to be true? Oh! The slopes of the perpendicular lines have to be negative reciprocals of each other.

Really? How do you know? Because my teacher told me last year. (They didn’t really say that part…I did.)

So this year we will actually prove that the product of the slopes of perpendicular lines is –1. But we are not ready for that yet. For now, we will go on what your Algebra I teacher told you. The slopes of perpendicular lines are negative reciprocals of each other.

How do you calculate slope? Inevitably, someone answered “rise over run”. Someone else answered “the change in y over the change in x”. What is the slope of line OA? 2/3. What is the slope of line OA’? -3/2. The lines are perpendicular.

We looked back at the problem from the day before with which students had difficulty. Can the image of (–4,5) rotated –90° about the origin be (4,5)? No! The line containing the pre-image and the center of rotation has a slope of –5/4. The line containing the other point and the center of rotation has a slope of 5/4. But an image of (5,4) does work. The slope of the line containing (5,4) and (0,0) is 4/5.

And so the journey continues, finding opportunities every day to enter into the Standards of Mathematical Practice with my students….

Note: We used the Transformations – Rotations lesson from Geometry Nspired as a guide for part of our exploration in this lesson.


Posted by on September 2, 2013 in Coordinate Geometry, Geometry, Rigid Motions


Tags: , , , , , , ,

3 responses to “Rotations

  1. Travis

    September 2, 2013 at 8:21 pm

    The ‘seek and convince’ strategy is excellent for various reasons.
    Ever try using the Shapes>Rectangle tool for a visual approach? Which will be interesting with a split screen!


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