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Here are a few recommended readings before getting started with this lesson.
Like rotations and translations, reflections are rigid motions because they preserve the side lengths and angle measures. However, reflections can change the orientation of the preimage.
The principal of Jefferson High wants to build a physics lab by the chemistry lab. The plan, seen from the sky, is that the new building looks like a reflection of the chemistry lab through the walkway that connects the soccer field with the library.
Perform a reflection to the chemistry lab across the walkway in order to draw the physics lab.
To reflect the chemistry lab, reflect each corner of the building. The physics lab is the quadrilateral formed by the images. Remember that the image of a point that is on the line of reflection is the same point.
For simplicity, start by labeling each corner of the quadrilateral and the walkway.
To reflect ABCD across ℓ, a reflection can be performed on each vertex, one at a time. For example, to reflect A is a good start. To do so, follow the definition of reflections. First, draw a line perpendicular to ℓ passing through A.
Then A′ can be plotted as the point on line m where its distance to ℓ is the same as the distance from A to ℓ.
The same steps can be applied to reflect vertices B, C, and D. Notice that because both C and D are on the line ℓ, their images will maintain their same point locations, respectively.
Finally, the image of ABCD under a reflection across the line ℓ is the quadrilateral formed by A′, B′, C′, and D′. This quadrilateral represents the physics lab.
Reflections can be performed by hand with the help of a straightedge and a compass.
To reflect △ABC across the line ℓ, follow these three steps.
To reflect B and C, repeat the two previous steps.
The image of △ABC after the reflection is the triangle formed by A′, B′, and C′.
Up to this point, how to perform a reflection, when already given the line of reflection, has been understood. Now, consider a case if given a figure and its image under a reflection. How can the line of reflection be found? This question is answered in the following example.
While visiting a museum, Tearrik saw a painting containing the word MATH
and two pentagons. The picture caught his attention. There is some sort of reflection but he wants to know for sure.
Tearrik analyzed the painting. He determined that the picture was made by performing a reflection. Show how Tearrik figured that out and draw the line of reflection used in the picture.
The line of reflection is the perpendicular bisector of any segment connecting a point to its image.
When a point is reflected across a line, its image is such that the line of reflection is the perpendicular bisector of the segment connecting the point to its image. Therefore, to find the line of reflection of the painting, start by drawing a segment that connects a vertex and its image. For instance, draw CC′.
Next, construct the perpendicular bisector of CC′. That will represent the line of reflection used to make the painting.
Notice that drawing only one segment that connects a point to its image is enough to find the line of reflection. Before performing the reflection, the painting looked as follows.
In the following applet, there are two possible requests.
To reflect △ABC, place points A′, B′, and C′ where they should be after the reflection is applied. To draw the line of reflection, place the two points, so they lie on the line of reflection.
In previous lessons, the composition of rotations and translations were studied. Now it is time to learn about the composition of reflections. The first case to consider is when the lines of reflection are parallel.
Using a drone, Kevin took a photo of the roof of his house, Main Street, and Euclid Sreet — they are parallel streets.
Perform the following transformations to Kevin's house.
Draw both reflections over the original photo. Is there a single transformation that maps Kevin's house onto the final image?
To reflect Kevin's house across the midline of Main Street, start by drawing lines perpendicular to the line of reflection passing through each vertex of the house. Label each vertex and line for added clarity.
The image of A after the reflection is a point A′ on ℓ2 such that A′ and A are equidistant from the midline. The images of B, C, and D can be located in a similar way. By connecting A′, B′, C′ and D′, the image of Kevin's house after performing the first reflection can be obtained.
Applying a similar reasoning, the second reflection can be performed. After the second reflection is performed, the image of Kevin's house lies to the right of Euclid Street.
To determine whether there is a single transformation that maps Kevin's house to the final image, label the vertices of the initial polygon and its images.
From the previous diagram, the following conclusions can be drawn.
In the previous example, it was concluded that the composition of reflections in parallel lines gives the same result as a translation. This conclusion is not an isolated fact. Actually, there is a theorem that guarantees this result.
The composition of two reflections across parallel lines is a single translation. Furthermore, the translation vector is perpendicular to both parallel lines, and its magnitude is twice the distance between the parallel lines.
In the diagram, △ABC is first reflected across ℓ and then the image is reflected across m. Equivalently, the following statements hold true.
The proof will be developed focusing the attention on vertex C and its images, but the conclusions are true for all the vertices. First, start reflecting △ABC across ℓ. By definition of reflection, the line ℓ is the perpendicular bisector of CC′. Let P1 be the intersection point of ℓ and CC′.
Next, perform the reflection of △A′B′C′ across line m. This time, the line m is the perpendicular bisector of C′C′′. Let P2 be the intersection point of m and C′C′′.
Since ℓ and m are parallel lines and CC′ is perpendicular to ℓ, by the Perpendicular Transversal Theorem, CC′ is perpendicular to m. Also, C′C′′ is perpendicular to m. Consequently, CC′ and C′C′′ are parallel vectors with a common point. Therefore, these vectors belong to the same line.
The points C, C′, and C′′ are collinear.
CP1=P1C′, P2C′′=C′P2
Add terms
Factor out 2
Segment Addition Postulate
In the diagram above, △ABC was first reflected across m and then reflected across ℓ.
Additionally, there is also a theorem for the case where the lines of reflection intersect each other.
The composition of two reflections across intersecting lines is a single rotation. Additionally, the center of rotation is the point of intersection of the lines, and the angle of rotation is twice the measure of the acute or right angle formed by the lines.
In the diagram, △ABC is first reflected across ℓ and then the image is reflected across m. The same result is obtained when △ABC is rotated by an angle of 2α∘ about point P.
Let P be the intersection point between the lines ℓ and m. To prove that △A′′B′′C′′ is a rotation of △ABC, the following two facts will be proved.
By definition of reflection, the line ℓ is the perpendicular bisector of the segments AA′, BB′, and CC′.
Since P is on ℓ, the Perpendicular Bisector Theorem guarantees that P is equidistant from the endpoints of AA′, BB′, and CC′.In consequence, PA′=PA′′, PB′=PB′′, and PC′=PC′′. Finally, the Transitive Property of Equality can be used to obtain the first part of the proof.
That way, it has been shown that P is the same distance from a vertex of △ABC as it is from the corresponding vertex of △A′′B′′C′′.
Here, it will be shown that m∠APA′′ is 2α, where α is the acute angle formed by the lines. Let Q and T be the intersection points between AA′ and ℓ, and A′A′′ and m respectively.
Next, consider the right triangles PQA and PQA′. Notice that their hypotenuses PA and PA′ are congruent as well as their legs AQ and A′Q.
Therefore, △PQA and △PQA′ are congruent thanks to the Hypotenuse Leg Theorem. This congruence implies that ∠APQ and ∠QPA′ have the same measure.
m∠APQ=m∠QPA′(I)
Similarly, by the Hypotenuse Leg Theorem, △PTA′ and △PTA′′ are congruent. Consequently, ∠A′PT and ∠TPA′′ are congruent.
m∠A′PT=m∠TPA′′(II)
m∠AP