Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM 8•2 Lesson 3: Translating Lines Student Outcomes ο§ Students learn that when lines are translated, they are either parallel to the given line or they coincide. ο§ Students learn that translations map parallel lines to parallel lines. Classwork Exercise 1 (3 minutes) Students complete Exercise 1 independently in preparation for the discussion that follows. Exercises Draw a line passing through point π· that is parallel to line π³. Draw a second line passing through point π· that is parallel to line π³ and that is distinct (i.e., different) from the first one. What do you notice? 1. Students should realize that they can only draw one line through point π· that is parallel to π³. Discussion (3 minutes) Bring out a fundamental assumption about the plane (as observed in Exercise 1): ο§ Given a line πΏ and a point π not lying on πΏ, there is at most one line passing through π and parallel to πΏ. – Based on what we have learned up to now, we cannot prove or explain this, so we have to simply agree that this is one of the starting points in the study of the plane. – This idea plays a key role in everything we do in the plane. A first consequence is that given a line πΏ and a point π not lying on πΏ, we can now refer to the line (because we agree there is only one) passing through π and parallel to πΏ. Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 31 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM 8•2 Exercises 2–4 (9 minutes) Students complete Exercises 2–4 independently in preparation for the discussion that follows. ββββββ . What do you notice about π³ and its image, π³′? Translate line π³ along the vector π¨π© 2. Scaffolding: Refer to Exercises 2–4 throughout the discussion and in the summary of findings about translating lines. π³ and π³′ coincide. π³ = π³′. ββββββ . Translate line π³ along vector π¨π© ββββββ . What do you notice about π³ and its image, π³′? Line π³ is parallel to vector π¨π© 3. π³ and π³′ coincide, again. π³ = π³′. ββββββ . What do you notice about π³ and its image, π³′? Translate line π³ along the vector π¨π© 4. π³ β₯ π³′ Discussion (15 minutes) ο§ ο§ Now we examine the effect of a translation on a line. Thus, let line πΏ be given. Again, let the translation be along a given βββββββ π΄π΅, and let πΏ′ denote the image line of the translated πΏ. We want to know what πΏ′ is relative to π΄π΅ and line πΏ. If πΏ = πΏπ΄π΅ , or πΏ β₯ πΏπ΄π΅ , then πΏ′ = πΏ. – If πΏ = πΏπ΄π΅ , then this conclusion follows directly from the work in Lesson 2, which says if πΆ is on πΏπ΄π΅ , then so is πΆ′; therefore, πΏ′ = πΏπ΄π΅ and πΏ = πΏ′ (Exercise 2). – If πΏ β₯ πΏπ΄π΅ and πΆ is on πΏ, then it follows from the work in Lesson 2, which says that πΆ′ lies on the line π passing through πΆ and parallel to πΏπ΄π΅ . However, πΏ is given as a line passing through πΆ and parallel to πΏπ΄π΅ , so the fundamental assumption that there is just one line passing through a point, parallel to a line (Exercise 1), implies π = πΏ. Therefore, πΆ′ lies on πΏ after all, and the translation maps every point of πΏ to a point of πΏ′ . Therefore, πΏ = πΏ′ again (Exercise 3). Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 Note to Teacher: The notation πππππ πππ‘πππ(πΏ) is used as a precursor to the notation students encounter in high school Geometry (i.e., π(πΏ)). We want to make clear the basic rigid motion that is being performed, so the notation πππππ πππ‘πππ(πΏ) is written to mean the translation of πΏ along the specified vector. 32 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 8•2 – Caution: One must not over-interpret the equality πππππ πππ‘πππ(πΏ) = πΏ (which is the same as πΏ = πΏ′). – All the equality says is that the two lines πΏ and πΏ′ coincide completely. It is easy (but wrong) to infer from the equality πππππ πππ‘πππ(πΏ) = πΏ that for any point π on πΏ, πππππ πππ‘πππ(π) = π. Suppose the vector βββββ π΄π΅ lying on πΏ is not the zero vector (i.e., assume π΄ ≠ π΅). Trace the line πΏ on a transparency to obtain a red line πΏ, and now slide the transparency along βββββ π΄π΅ . Then, the red line, as a line, coincides with the original πΏ, but clearly every point on πΏ has been moved by the slide (the translation). Indeed, as we saw in Example 2 of Lesson 2, πππππ πππ‘πππ(π΄) = π΅ ≠ π΄. Therefore, the equality πΏ′ = πΏ only says that for any point πΆ on πΏ, πππππ πππ‘πππ(πΆ) is also a point on πΏ, but as long as βββββ π΄π΅ is not a zero vector, πππππ πππ‘πππ(πΆ) ≠ πΆ. Strictly speaking, we have not completely proved πΏ = πΏ′ in either case. To explain this, let us define what it means for two geometric figures πΉ and πΊ to be equal, that is, πΉ = πΊ: it means each point of πΉ is also a point of πΊ; conversely, each MP.6 point of πΊ is also a point of πΉ. In this light, all we have shown above is that if every point πΆ′ of πΏ′ belongs to πΏ, then π is also a point of πΏ′. To show the latter, we have to show that this π is equal to πππππ πππ‘πππ(π) for some π on πΏ. This then completes the reasoning. However, at this point of students’ education in geometry, it may be prudent not to bring up such a sticky point because they are already challenged with all of the new ideas and definitions. Simply allow the preceding reasoning to stand for now, and clarify later in the school year when students are more comfortable with the geometric environment. ο§ Next, if πΏ is neither πΏπ΄π΅ nor parallel to πΏπ΄π΅ , then πΏ′ β₯ πΏ. ο§ If we use a transparency to see this translational image of πΏ by the stated translation, then the pictorial evidence is clear: the line πΏ moves in a parallel manner along βββββββ π΄π΅, and a typical point πΆ of πΏ is translated to a point πΆ′ of πΏ′. The fact that πΏ′ β₯ πΏ is unmistakable, as shown. In the classroom, students should be convinced by the pictorial evidence. If so, leave it at that (Exercise 4). Here is a simple proof, but to present it in class, begin by asking students how they would prove that two lines are parallel. Ensure students understand that they have no tools in their possession to accomplish this goal. It is only then MP.2 that they see the need for invoking a proof by contradiction (see discussion above). If there are no obvious ways to do & MP.7 something, then you just have to do the best you can by trying to see what happens if you assume the opposite is true. Thus, if πΏ′ is not parallel to πΏ, then they intersect at a point πΆ′. Since πΆ′ lies on πΏ′, it follows from the definition of πΏ′ (as the image of πΏ under the translation π) that there is a point πΆ on πΏ so that πππππ πππ‘πππ(πΆ) = πΆ′. It follows from Lesson 2 that πΏπΆπΆ′ β₯ πΏπ΄π΅ . However, both πΆ and πΆ′ lie on πΏ, so πΏπΆπΆ′ β₯ πΏ, and we get πΏ β₯ πΏπ΄π΅ . This contradicts the assumption that πΏ is not parallel to πΏπ΄π΅ , so πΏ could not possibly intersect πΏ′. Therefore, πΏ′ β₯ πΏ. ο§ Note that a translation maps parallel lines to parallel lines. More precisely, consider a translation π along a vector βββββββ π΄π΅. Then: If πΏ1 and πΏ2 are parallel lines, so are πππππ πππ‘πππ(πΏ1 ) and πππππ πππ‘πππ(πΏ2 ). Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 33 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM 8•2 ο§ The reasoning is the same as before: Copy πΏ1 and πΏ2 onto a transparency, and then translate the transparency along βββββββ π΄π΅. If πΏ1 and πΏ2 do not intersect, then their red replicas on the transparency will not intersect either, no matter what βββββββ π΄π΅ is used. So, πππππ πππ‘πππ(πΏ1 ) and πππππ πππ‘πππ(πΏ2 ) are parallel. ο§ These findings are summarized as follows: Given a translation π along a vector βββββββ π΄π΅, let L be a line, and let πΏ′ denote the image of πΏ by π. MP.6 – If πΏ β₯ πΏπ΄π΅ or πΏ = πΏπ΄π΅ , then πΏ′ β₯ πΏ. – If πΏ is neither parallel to πΏπ΄π΅ nor equal to πΏπ΄π΅ , then πΏ′ β₯ πΏ. Exercises 5–6 (5 minutes) Students complete Exercises 5 and 6 in pairs or small groups. 5. Line π³ has been translated along vector ββββββ π¨π©, resulting in π³′. What do you know about lines π³ and π³′? π³ β₯ π»(π³) 6. Translate π³π and π³π along vector ββββββ π«π¬. Label the images of the lines. If lines π³π and π³π are parallel, what do you know about their translated images? Since π³π β₯ π³π , then (π³π )′ β₯ (π³π )′. Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 34 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 8•2 Closing (5 minutes) Summarize, or have students summarize, the lesson. ο§ We know that there exists just one line, parallel to a given line and through a given point not on the line. ο§ We know that translations map parallel lines to parallel lines. ο§ We know that when lines are translated, they are either parallel to the given line or they coincide. Lesson Summary ο§ Two lines in the plane are parallel if they do not intersect. ο§ Translations map parallel lines to parallel lines. ο§ Given a line π³ and a point π· not lying on π³, there is at most one line passing through π· and parallel to π³. Exit Ticket (5 minutes) Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 35 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM Name 8•2 Date Lesson 3: Translating Lines Exit Ticket 1. Translate point π along vector βββββ π΄π΅ . What do you know about the line containing vector βββββ π΄π΅ and the line formed when you connect π to its image π′? 2. Using the above diagram, what do you know about the lengths of segments ππ′ and π΄π΅? 3. Let points π΄ and π΅ be on line πΏ and the vector βββββ π΄πΆ be given, as shown below. Translate line πΏ along vector βββββ π΄πΆ . What do you know about line πΏ and its image, πΏ′? How many other lines can you draw through point πΆ that have the same relationship as πΏ and πΏ′? How do you know? Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 36 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 3 NYS COMMON CORE MATHEMATICS CURRICULUM 8•2 Exit Ticket Sample Solutions 1. Translate point π along vector ββββββ π¨π©. What do you know about the line containing vector ββββββ π¨π© and the line formed when you connect π to its image π′? ββββββ and ππ′ is parallel. The line containing vector π¨π© 2. Using the above diagram, what do you know about the lengths of segment ππ′ and segment π¨π©? The lengths are equal: |ππ′| = |π¨π©|. 3. βββββ be given, as shown below. Translate line π³ along vector π¨πͺ βββββ . Let points π¨ and π© be on line π³ and the vector π¨πͺ What do you know about line π³ and its image, π³′? How many other lines can you draw through point πͺ that have the same relationship as π³ and π³′? How do you know? π³ and π³′ are parallel. There is only one line parallel to line π³ that goes through point πͺ. The fact that there is only one line through a point parallel to a given line guarantees it. Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 37 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 8•2 Problem Set Sample Solutions 1. βββββ . Sketch the images, and label all points Translate ∠πΏππ, point π¨, point π©, and rectangle π―π°π±π² along vector π¬π using prime notation. 2. What is the measure of the translated image of ∠πΏππ? How do you know? The measure is ππ°. Translations preserve angle measure. 3. Connect π© to π©′. What do you know about the line that contains the segment formed by π©π©′ and the line βββββ ? containing the vector π¬π β‘ββββββ π©π©′ β₯ β‘ββββ π¬π. 4. Connect π¨ to π¨′. What do you know about the line that contains the segment formed by π¨π¨′ and the line containing βββββ ? the vector π¬π β‘βββββ β‘ββββ coincide. π¨π¨′ and π¬π 5. Given that figure π―π°π±π² is a rectangle, what do you know about lines that contain segments π―π° and π±π² and their translated images? Explain. β‘βββ β₯ β‘βββ Since π―π°π±π² is a rectangle, I know that π―π° π±π². Since translations map parallel lines to parallel lines, then β‘ββββββ π―′π°′ β₯ β‘ββββββ π±′π²′. Lesson 3: Translating Lines This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from G8-M2-TE-1.3.0-08.2015 38 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.