1 / 10

Translations

AB and DA. B and D. Translations. Lesson 9-1. Lesson Quiz. 1. Is the transformation below an isometry? Explain. No; the angles are not congruent. For Exercises 2 and 3, ABCD is an image of KLMN. 2. Name the images of  L and  N.

axelle
Télécharger la présentation

Translations

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. AB and DA B and D Translations Lesson 9-1 Lesson Quiz 1. Is the transformation below an isometry? Explain. No; the angles are not congruent. For Exercises 2 and 3, ABCD is an image of KLMN. 2. Name the images of L and N. 3. Name the sides that correspond to KL and NK. Use the diagram below. 4. Find the image of MNV under the translation (x, y)  (x – 2, y + 5). M(–5, 4), N(–4, 6), V(–1, 5) 5. Write a rule to describe the translation MNV WZP. (x, y)  (x + 4, y + 3) 9-2

  2. Reflections Lesson 9-2 Check Skills You’ll Need (For help, go to Lesson 3-7.) Write an equation for the line through point A that is perpendicular to the given line. 1. 2. 3. Check Skills You’ll Need 9-2

  3. Reflections Lesson 9-2 Check Skills You’ll Need Solutions 1. A line perpendicular to a vertical line is a horizontal line. The equation of the horizontal line through (1, –2) is y = –2. 2. A line perpendicular to a horizontal line is a vertical line. The equation of the vertical line through (–1, –1) is x = –1. 3. The given line has slope 1. A line perpendicular to a line with slope 1 has a slope of –1. The equation of a lien with slope –1 through (–1, 2) is y = –x + 1. 9-2

  4. Reflections Lesson 9-2 A reflection (or flip) is an isometry in which a figure and its image have opposite orientations. 9-2

  5. Reflections Lesson 9-2 You can use the following two rules to reflect a figure across a line r. • If a point A is on line r, then the image of A is A itself (that is, A’ = A). • If a point B is not on line r, then ris the perpendicular bisector of . 9-2

  6. Reflections Lesson 9-2 In other words, a point and its reflection image are equidistant from the line of reflection. 9-2

  7. Point P is 2 units right of the reflection line, the y-axis. Therefore, the image of P' is 2 units left of the reflection line. Reflections Lesson 9-2 Additional Examples Finding a Reflection Image If point P(2, –1), is reflected across the y-axis, what are the coordinates of its reflection image? So the coordinates of P' are (–2, –1). Quick Check 9-2

  8. First locate vertices X, Y, and Z and draw XYZ in a coordinate plane. Locate points X , Y , and Z such that the line of reflection x = 4 is the perpendicular bisector of XX , YY , and ZZ . Draw the reflection image X Y Z . Reflections Lesson 9-2 Additional Examples Drawing Reflection Images XYZ has vertices X(0, 3), Y(2, 0), and Z(4, 2). Draw XYZ and its reflection image in the line x = 4. Quick Check 9-2

  9. Because a reflection is an isometry, PD and PD form congruent angles with line . PD and PW also form congruent angles with line because vertical angles are congruent. Therefore, PD and PW form congruent angles with line by the Transitive Property. Reflections Lesson 9-2 Additional Examples Real-World Connection Show that PD and PW form congruent angles with line . Quick Check 9-2

  10. XYZ has vertices X(–2, 3), Y(1, 1), and Z(2, 4). Draw XYZ and its reflection image in each line. 4. the x-axis 5. the line x = 5 Reflections Lesson 9-2 Lesson Quiz Find the coordinates of the image point for each given point and reflection line. 1. R(4, –5) across x = –2 2. S(–11, 2) across y = 1 3. T(0, 5) across x-axis R'(–8, –5) S'(–11, 0) T'(0, 5) 9-2

More Related