1 / 9

SOLUTION

The endpoints of RS are R (1, –3) and S (2, –6). Graph the image of RS after the composition. STEP 1. Graph RS. STEP 2. Reflect RS in the y -axis. R ′ S ′ has endpoints R ′(–1, –3) and S ′(–2, –6). EXAMPLE 2. Find the image of a composition. Reflection: in the y -axis.

hart
Download Presentation

SOLUTION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. The endpoints of RSare R(1, –3) and S(2, –6). Graph the image of RSafter the composition. STEP 1 Graph RS STEP 2 Reflect RSin the y-axis. R′S′ has endpoints R′(–1, –3) and S′(–2, –6). EXAMPLE 2 Find the image of a composition Reflection: in the y-axis Rotation: 90° about the origin SOLUTION

  2. EXAMPLE 2 Find the image of a composition o Rotate R′S′ 90 about the origin. R′′S′′ has endpoints R′′(3, –1) andS′′(6, –2). STEP 3

  3. In the diagram, a reflection in line kmaps GH to G′H′. A reflection in line m maps G′H′to G′′H′′. Also, HB = 9 andDH′′ = 4. a. Name any segments congruent to each segment: HG, HB, and GA ~ ~ ~ ~ a. HG H′G′ ,andHG H′′G′′. HB H′B. GA G′A . EXAMPLE 3 Use Theorem 9.5 SOLUTION

  4. In the diagram, a reflection in line kmaps GH to G′H′. A reflection in line m maps G′H′to G′′H′′. Also, HB = 9 andDH′′ = 4. b. Yes, AC = BDbecause GG′′ and HH′′ are perpendicular to both kand m,so BDand ACare opposite sides of a rectangle. EXAMPLE 3 Use Theorem 9.5 b. Does AC = BD? Explain. SOLUTION

  5. In the diagram, a reflection in line kmaps GH to G′H′. A reflection in line m maps G′H′to G′′H′′. Also, HB = 9 andDH′′ = 4. c.What is the length of GG′′ ? c. By the properties of reflections, H′B = 9 and H′D = 4. Theorem 9.5 implies that GG′′ = HH′′ = 2 BD, so the length of GG′′ is 2(9 + 4), or 26 units. EXAMPLE 3 Use Theorem 9.5 SOLUTION

  6. 3.Graph RSfrom Example 2. Do the rotation first, followed by the reflection. Does the order of the transformations matter? Explain. Yes; the resulting segment R′′ S ′′is not the same. for Examples 2 and 3 GUIDED PRACTICE SOLUTION

  7. for Examples 2 and 3 GUIDED PRACTICE 4.In Example 3, part (c), explain how you know that GG′′ = HH′′. SOLUTION They are opposite sides of a parallelogram.

  8. ANSWER 5. The preimage is reflected in line k , then in line m. Describe a single transformation that maps the blue figure to the green figure. Translation for Examples 2 and 3 GUIDED PRACTICE Use the figure below for Exercises 5 and 6. The distance between line k and line mis 1.6 centimeters.

  9. 6. What is the distance between Pand P′′? If you draw PP′ , what is its relationship with line k? Explain. ANSWER 3.2 cm; They are perpendicular. for Examples 2 and 3 GUIDED PRACTICE

More Related