12.5: Composition of Transformations
Learning Objectives
- Perform a glide reflection.
- Perform a reflection over parallel lines and the axes.
- Perform a double rotation with the same center of rotation.
- Determine a single transformation that is equivalent to a composite of two transformations.
Review Queue
- Reflect \begin{align*}ABCD\end{align*}
ABCD over the \begin{align*}x-\end{align*}x− axis. Find the coordinates of \begin{align*}A'B'C'D'\end{align*}. - Translate \begin{align*}A'B'C'D'\end{align*} such that \begin{align*}(x,y) \rightarrow (x+4,y)\end{align*}. Find the coordinates of \begin{align*}A''B''C''D''\end{align*}.
- Now, start over. Translate \begin{align*}ABCD\end{align*} such that \begin{align*}(x,y) \rightarrow (x+4,y)\end{align*}. Find the coordinates of \begin{align*}A'B'C'D'\end{align*}.
- Reflect \begin{align*}A'B'C'D'\end{align*} from #3 over the \begin{align*}x-\end{align*}axis. Find the coordinates of \begin{align*}A''B''C''D''\end{align*}. Are they the same as #2?
Know What? An example of a glide reflection is your own footprint. The equations to find your average footprint are in the diagram below. Determine your average footprint and write the rule for one stride. You may assume your stride starts at (0, 0).
Glide Reflections
Now that we have learned all our rigid transformations, or isometries, we can perform more than one on the same figure. In your homework last night you actually performed a composition of two reflections. And, in the Review Queue above, you performed a composition of a reflection and a translation.
Composition (of transformations): To perform more than one rigid transformation on a figure.
Glide Reflection: A composition of a reflection and a translation. The translation is in a direction parallel to the line of reflection.
So, in the Review Queue above, you performed a glide reflection on \begin{align*}ABCD\end{align*}. Hopefully, in #4, you noticed that the order in which you reflect or translate does not matter. It is important to note that the translation for any glide reflection will always be in one direction. So, if you reflect over a vertical line, the translation can be up or down, and if you reflect over a horizontal line, the translation will be to the left or right.
Example 1: Reflect \begin{align*}\triangle ABC\end{align*} over the \begin{align*}y-\end{align*}axis and then translate the image 8 units down.
Solution: The green image below is the final answer.
\begin{align*}A(8,8) & \rightarrow A''(-8,0)\\ B(2,4) & \rightarrow B''(-2,-4)\\ C(10,2) & \rightarrow C''(-10,-6)\end{align*}
One of the interesting things about compositions is that they can always be written as one rule. What this means is, you don’t necessarily have to perform one transformation followed by the next. You can write a rule and perform them at the same time.
Example 2: Write a single rule for \begin{align*}\triangle ABC\end{align*} to \begin{align*}\triangle A''B''C''\end{align*} from Example 1.
Solution: Looking at the coordinates of \begin{align*}A\end{align*} to \begin{align*}A''\end{align*}, the \begin{align*}x-\end{align*}value is the opposite sign and the \begin{align*}y-\end{align*}value is \begin{align*}y - 8\end{align*}. Therefore the rule would be \begin{align*}(x,y) \rightarrow (-x,y-8)\end{align*}.
Notice that this follows the rules we have learned in previous sections about a reflection over the \begin{align*}y-\end{align*}axis and translations.
Reflections over Parallel Lines
The next composition we will discuss is a double reflection over parallel lines. For this composition, we will only use horizontal or vertical lines.
Example 3: Reflect \begin{align*}\triangle ABC\end{align*} over \begin{align*}y = 3\end{align*} and \begin{align*}y = -5\end{align*}.
Solution: Unlike a glide reflection, order matters. Therefore, you would reflect over \begin{align*}y = 3\end{align*} first, followed by a reflection of this image (red triangle) over \begin{align*}y = -5\end{align*}. Your answer would be the green triangle in the graph below.
Example 4: Write a single rule for \begin{align*}\triangle ABC\end{align*} to \begin{align*}\triangle A''B''C''\end{align*} from Example 3.
Solution: Looking at the graph below, we see that the two lines are 8 units apart and the figures are 16 units apart. Therefore, the double reflection is the same as a single translation that is double the distance between the two lines.
\begin{align*}(x,y) \rightarrow (x,y-16)\end{align*}
Reflections over Parallel Lines Theorem: If you compose two reflections over parallel lines that are \begin{align*}h\end{align*} units apart, it is the same as a single translation of \begin{align*}2h\end{align*} units.
Be careful with this theorem. Notice, it does not say which direction the translation is in. So, to apply this theorem, you would still need to visualize, or even do, the reflections to see in which direction the translation would be.
Example 5: \begin{align*}\triangle DEF\end{align*} has vertices \begin{align*}D(3, -1), E(8, -3),\end{align*} and \begin{align*}F(6, 4)\end{align*}. Reflect \begin{align*}\triangle DEF\end{align*} over \begin{align*}x = -5\end{align*} and \begin{align*}x = 1\end{align*}. This double reflection would be the same as which one translation?
Solution: From the Reflections over Parallel Lines Theorem, we know that this double reflection is going to be the same as a single translation of \begin{align*}2(1 -(-5))\end{align*} or 12 units. Now, we need to determine if it is to the right or to the left. Because we first reflect over a line that is further away from \begin{align*}\triangle DEF\end{align*}, to the left, \begin{align*}\triangle D''E''F''\end{align*} will be on the right of \begin{align*}\triangle DEF\end{align*}. So, it would be the same as a translation of 12 units to the right. If the lines of reflection were switched and we reflected the triangle over \begin{align*}x = 1\end{align*} followed by \begin{align*}x = -5\end{align*}, then it would have been the same as a translation of 12 units to the left.
Reflections over the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} Axes
You can also reflect over intersecting lines. First, we will reflect over the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} axes.
Example 6: Reflect \begin{align*}\triangle DEF\end{align*} from Example 5 over the \begin{align*}x-\end{align*}axis, followed by the \begin{align*}y-\end{align*}axis. Determine the coordinates of \begin{align*}\triangle D''E''F''\end{align*} and what one transformation this double reflection would be the same as.
Solution: \begin{align*}\triangle D''E''F''\end{align*} is the green triangle in the graph below. If we compare the coordinates of it to \begin{align*}\triangle DEF\end{align*}, we have:
\begin{align*}D(3,-1) & \rightarrow D'(-3,1)\\ E(8,-3) & \rightarrow E' (-8,3)\\ F(6,4) & \rightarrow F'(-6,-4)\end{align*}
If you recall the rules of rotations from the previous section, this is the same as a rotation of \begin{align*}180^\circ\end{align*}.
Reflection over the Axes Theorem: If you compose two reflections over each axis, then the final image is a rotation of \begin{align*}180^\circ\end{align*} of the original.
With this particular composition, order does not matter. Let’s look at the angle of intersection for these lines. We know that the axes are perpendicular, which means they intersect at a \begin{align*}90^\circ\end{align*} angle. The final answer was a rotation of \begin{align*}180^\circ\end{align*}, which is double \begin{align*}90^\circ\end{align*}. Therefore, we could say that the composition of the reflections over each axis is a rotation of double their angle of intersection.
Reflections over Intersecting Lines
Now, we will take the concept we were just discussing and apply it to any pair of intersecting lines. For this composition, we are going to take it out of the coordinate plane. Then, we will apply the idea to a few lines in the coordinate plane, where the point of intersection will always be the origin.
Example 7: Copy the figure below and reflect it over \begin{align*}l\end{align*}, followed by \begin{align*}m\end{align*}.
Solution: The easiest way to reflect the triangle is to fold your paper on each line of reflection and draw the image. It should look like this:
The green triangle would be the final answer.
Investigation 12-2: Double Reflection over Intersecting Lines
Tools Needed: Example 7, protractor, ruler, pencil
- Take your answer from Example 7 and measure the angle of intersection for lines \begin{align*}l\end{align*} and \begin{align*}m\end{align*}. If you copied it exactly from the text, it should be about \begin{align*}55^\circ\end{align*}.
- Draw lines from two corresponding points on the blue triangle and the green triangle. These are the dotted lines in the diagram below.
- Measure this angle using your protractor. How does it related to \begin{align*}55^\circ\end{align*}?
Again, if you copied the image exactly from the text, the angle should be \begin{align*}110^\circ\end{align*}.
From this investigation, we see that the double reflection over two lines that intersect at a \begin{align*}55^\circ\end{align*} angle is the same as a rotation of \begin{align*}110^\circ\end{align*} counterclockwise, where the point of intersection is the center of rotation. Notice that order would matter in this composition. If we had reflected the blue triangle over \begin{align*}m\end{align*} followed by \begin{align*}l\end{align*}, then the green triangle would be rotated \begin{align*}110^\circ\end{align*} clockwise.
Reflection over Intersecting Lines Theorem: If you compose two reflections over lines that intersect at \begin{align*}x^\circ\end{align*}, then the resulting image is a rotation of \begin{align*}2x^\circ\end{align*}, where the center of rotation is the point of intersection.
Notice that the Reflection over the Axes Theorem is a specific case of this one.
Example 8: Reflect the square over \begin{align*}y = x\end{align*}, followed by a reflection over the \begin{align*}x-\end{align*}axis.
Solution: First, reflect the square over \begin{align*}y = x\end{align*}. The answer is the red square in the graph above. Second, reflect the red square over the \begin{align*}x-\end{align*}axis. The answer is the green square below.
Example 9: Determine the one rotation that is the same as the double reflection from Example 8.
Solution: Let’s use the theorem above. First, we need to figure out what the angle of intersection is for \begin{align*}y = x\end{align*} and the \begin{align*}x-\end{align*}axis. \begin{align*}y = x\end{align*} is halfway between the two axes, which are perpendicular, so is \begin{align*}45^\circ\end{align*} from the \begin{align*}x-\end{align*}axis. Therefore, the angle of rotation is \begin{align*}90^\circ\end{align*} clockwise or \begin{align*}270^\circ\end{align*} counterclockwise. The correct answer is \begin{align*}270^\circ\end{align*} counterclockwise because we always measure angle of rotation in the coordinate plane in a counterclockwise direction. From the diagram, we could have also said the two lines are \begin{align*}135^\circ\end{align*} apart, which is supplementary to \begin{align*}45^\circ\end{align*}.
Know What? Revisited The average 6 foot tall man has a \begin{align*}0.415 \times 6 = 2.5\end{align*} foot stride. Therefore, the transformation rule for this person would be \begin{align*}(x,y) \rightarrow (-x,y+2.5)\end{align*}.
Review Questions
- Explain why the composition of two or more isometries must also be an isometry.
- What one transformation is equivalent to a reflection over two parallel lines?
- What one transformation is equivalent to a reflection over two intersecting lines?
Use the graph of the square below to answer questions 4-7.
- Perform a glide reflection over the \begin{align*}x-\end{align*}axis and to the right 6 units. Write the new coordinates.
- What is the rule for this glide reflection?
- What glide reflection would move the image back to the preimage?
- Start over. Would the coordinates of a glide reflection where you move the square 6 units to the right and then reflect over the \begin{align*}x-\end{align*}axis be any different than #4? Why or why not?
Use the graph of the triangle below to answer questions 8-10.
- Perform a glide reflection over the \begin{align*}y-\end{align*}axis and down 5 units. Write the new coordinates.
- What is the rule for this glide reflection?
- What glide reflection would move the image back to the preimage?
Use the graph of the triangle below to answer questions 11-15.
- Reflect the preimage over \begin{align*}y = -1\end{align*} followed by \begin{align*}y = -7\end{align*}. Write the new coordinates.
- What one transformation is this double reflection the same as?
- What one translation would move the image back to the preimage?
- Start over. Reflect the preimage over \begin{align*}y = -7\end{align*}, then \begin{align*}y = -1\end{align*}. How is this different from #11?
- Write the rules for #11 and #14. How do they differ?
Use the graph of the trapezoid below to answer questions 16-20.
- Reflect the preimage over \begin{align*}y = -x\end{align*} then the \begin{align*}y-\end{align*}axis. Write the new coordinates.
- What one transformation is this double reflection the same as?
- What one transformation would move the image back to the preimage?
- Start over. Reflect the preimage over the \begin{align*}y-\end{align*}axis, then \begin{align*}y = -x\end{align*}. How is this different from #16?
- Write the rules for #16 and #19. How do they differ?
Fill in the blanks or answer the questions below.
- Two parallel lines are 7 units apart. If you reflect a figure over both how far apart with the preimage and final image be?
- After a double reflection over parallel lines, a preimage and its image are 28 units apart. How far apart are the parallel lines?
- A double reflection over the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} axes is the same as a _________ of ________ \begin{align*}^\circ\end{align*}.
- What is the center of rotation for #23?
- Two lines intersect at an \begin{align*}83^\circ\end{align*} angle. If a figure is reflected over both lines, how far apart will the preimage and image be?
- A preimage and its image are \begin{align*}244^\circ\end{align*} apart. If the preimage was reflected over two intersected lines, at what angle did they intersect?
- A rotation of \begin{align*}45^\circ\end{align*} clockwise is the same as a rotation of ________ \begin{align*}^\circ\end{align*} counterclockwise.
- After a double reflection over parallel lines, a preimage and its image are 62 units apart. How far apart are the parallel lines?
- A figure is to the left of \begin{align*}x = a\end{align*}. If it is reflected over \begin{align*}x = a\end{align*} followed by \begin{align*}x = b\end{align*} and \begin{align*}b > a\end{align*}, then the preimage and image are _________ units apart and the image is to the _________ of the preimage.
- A figure is to the left of \begin{align*}x = a\end{align*}. If it is reflected over \begin{align*}x = b\end{align*} followed by \begin{align*}x = a\end{align*} and \begin{align*}b > a\end{align*}, then the preimage and image are _________ units apart and the image is to the _________ of the preimage.
Review Queue Answers
- \begin{align*}A'(-2, -8), B'(4, -5), C'(-4, -1), D'(-6, -6)\end{align*}
- \begin{align*}A''(2, -8), B''(8, -5), C''(0, -1), D''(-2, -6)\end{align*}
- \begin{align*}A'(2, 8), B'(8, 5), C''(0, 1), D''(-2, 6)\end{align*}
- The coordinates are the same as #2.