# 1. Is the transformation below an isometry? Explain.

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1. Is the transformation below an isometry? Explain.
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. B and D 3. Name the sides that correspond to KL and NK. AB and DA 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

Write an equation for the line through point A that is perpendicular
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

horizontal line through (1, –2) is y = –2.
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

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

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 r is the perpendicular bisector of 9-2

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

Finding a Reflection Image
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? 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. So the coordinates of P' are (–2, –1). Quick Check 9-2

Drawing Reflection Images
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. First locate vertices X, Y, and Z and draw XYZ in a coordinate plane. Draw the reflection image X Y Z . Locate points X , Y , and Z such that the line of reflection x = 4 is the perpendicular bisector of XX , YY , and ZZ . Quick Check 9-2

Real-World Connection
Reflections Lesson 9-2 Additional Examples Real-World Connection Show that PD and PW form congruent angles with line . 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. Quick Check 9-2

Find the coordinates of the image point for each given point and
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) 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 9-2