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Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY Lesson 6: Segments That Meet at Right Angles Student Outcomes ๏ง Students generalize the criterion for perpendicularity of two segments that meet at a point to any two segments in the Cartesian plane. ๏ง Students apply the criterion to determine if two segments are perpendicular. Classwork Opening Exercise (2 minutes) As a class, present each problem, and have students determine an answer and justify it. Opening Exercise Carlos thinks that the segment having endpoints ๐จ(๐, ๐) and ๐ฉ(๐, ๐) is perpendicular to the segment with endpoints ๐จ(๐, ๐) and ๐ช(โ๐, ๐). Do you agree? Why or why not? No, the two segments are not perpendicular. If they were perpendicular, then ๐(โ๐) + ๐(๐) = ๐ would be true. MP.3 Working with a partner, given ๐จ(๐, ๐) and ๐ฉ(๐, โ๐), find the coordinates of a point ๐ช so that ฬ ฬ ฬ ฬ ๐จ๐ช โฅ ฬ ฬ ฬ ฬ ๐จ๐ฉ. ฬ ฬ ฬ ฬ โฅ ๐จ๐ฉ ฬ ฬ ฬ ฬ will be perpendicular to ๐จ๐ฉ ฬ ฬ ฬ ฬ , then ๐๐ + โ๐๐ = ๐. This means that ๐จ๐ช ฬ ฬ ฬ ฬ as Let the other endpoint be ๐ช(๐, ๐ ). If ๐จ๐ช ๐ ๐ long as we choose values for ๐ and ๐ that satisfy the equation ๐ = ๐. Answers may vary but may include (๐, ๐), (๐, ๐), and (๐, ๐) or any other coordinates that meet the requirement stated above. Scaffolding: ๏ง Graphing each exercise helps students picture the problem. ๏ง It is helpful to display posters of the previous dayโs findings around the room. Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 62 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY Example 1 (10 minutes) This example tries to get students to see that translating segments may help us determine whether the lines containing the segments are perpendicular. Example ฬ ฬ ฬ ฬ perpendicular? Are the lines containing the ฬ ฬ ฬ ฬ and ๐ช๐ซ Given points ๐จ(๐, ๐), ๐ฉ(๐๐, ๐๐), ๐ช(โ๐, ๐), and ๐ซ(๐, โ๐), are ๐จ๐ฉ segments perpendicular? Explain. One possible solution would be to translate ฬ ฬ ฬ ฬ ๐จ๐ฉ so that ๐จโฒ is on the origin (using the vector โฉโ๐, โ๐โช, or left ๐ units and down ๐ units) and to ฬ ฬ ฬ ฬ so that ๐ชโฒ is on the origin (using the vector โฉ๐, โ๐โช, translate ๐ช๐ซ right ๐ units and down ๐ unit): ๐จ(๐, ๐) โ ๐จโฒ(๐ โ ๐, ๐ โ ๐) โ ๐จโฒ(๐, ๐) ๐ฉ(๐๐, ๐๐) โ ๐ฉโฒ(๐๐ โ ๐, ๐๐ โ ๐) โ ๐ฉโฒ(๐, ๐๐) ๐ช(โ๐, ๐) โ ๐ชโฒ(โ๐ โ (โ๐), ๐ โ ๐) โ ๐ชโฒ(๐, ๐) ๐ซ(๐, โ๐) โ ๐ซโฒ(๐ โ (โ๐), โ๐ โ ๐) โ ๐ซโฒ(๐, โ๐) ฬ ฬ ฬ ฬ ๐จ๐ฉ will be perpendicular to ฬ ฬ ฬ ฬ ๐ช๐ซ if ฬ ฬ ฬ ฬ ฬ ฬ ๐จโฒ๐ฉโฒ is perpendicular to ฬ ฬ ฬ ฬ ฬ ฬ ๐ชโฒ๐ซโฒ. ฬ ฬ ฬ ฬ ฬ ฬ is perpendicular to ฬ ฬ ฬ ฬ ฬ ฬ ๐(๐) + ๐๐(โ๐) = ๐; therefore, ๐จโฒ๐ฉโฒ ๐ชโฒ ๐ซโฒ , which ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ means ๐จ๐ฉ โฅ ๐ช๐ซ. ๏ง Working in pairs, each student selects two points to be the endpoint of a segment. One partner identifies the coordinates of points ๐ด and ๐ต, and the other partner chooses coordinates for points ๐ถ and ๐ท. Although the likelihood of this happening is low, students may choose one of the same points, but not both. Students then plot the points on the same coordinate plane and construct ฬ ฬ ฬ ฬ ๐ด๐ต and ฬ ฬ ฬ ฬ ๐ถ๐ท . ๏ง Students must now determine whether the segments are perpendicular. ๏บ ๏ง How do your segments differ from those we worked with in Lesson 5? ๏บ ๏ง In that lesson, both segments had one endpoint located at the origin. Would translating your two segments change their orientation? ๏บ MP.1 & MP.3 Answers will vary. Some students will say no because they do not intersect. Students should recognize that, unless the segments are parallel, if we extend the segments, the lines containing the segments will intersect. No. If we translate the segments, thereby translating the lines containing the segments, the angles formed by the two lines will not change. If the lines were perpendicular, the images of the lines under the translation will also be perpendicular. ๏ง With your partner, translate each of your segments so that they have an endpoint at the origin. Then, use the results of yesterdayโs lesson to determine whether the two segments are perpendicular. ๏ง Switch papers with another pair of students, and confirm their results. ๏บ Answers will vary. Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 63 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY Exercise 1 (5 minutes) Have students try Exercise 1 using the example just completed as a template. The teacher may pull aside some groups for targeted instruction. Exercises Given ๐จ(๐๐ , ๐๐ ), ๐ฉ(๐๐ , ๐๐ ), ๐ช(๐๐ , ๐๐ ), and ๐ซ(๐ ๐ , ๐ ๐ ), find a general formula in terms of ๐๐ , ๐๐ , ๐๐ , ๐๐ , ๐๐ , ๐๐ , ๐ ๐ , and ๐ ๐ that will let us determine whether ฬ ฬ ฬ ฬ ๐จ๐ฉ and ฬ ฬ ฬ ฬ ๐ช๐ซ are perpendicular. 1. After translating the segments so that the image of points ๐จ and ๐ช lie on the origin, we get ๐จโฒ (๐, ๐), ๐ฉโฒ (๐๐ โ ๐๐ , ๐๐ โ ๐๐ ), ๐ชโฒ (๐, ๐), and ๐ซโฒ (๐ ๐ โ ๐๐ , ๐ ๐ โ ๐๐ ). ฬ ฬ ฬ ฬ ฬ ฬ ๐จโฒ๐ฉโฒ โฅ ฬ ฬ ฬ ฬ ฬ ฬ ๐ชโฒ ๐ซโฒ โบ (๐๐ โ ๐๐ )(๐ ๐ โ ๐๐ ) + (๐๐ โ ๐๐ )(๐ ๐ โ ๐๐ ) = ๐ Example 2 (7 minutes) In this example, students use the formula they derived in Exercise 1 to find the location of one endpoint of a segment given the other endpoint and the coordinates of the endpoints of a perpendicular segment. Because there are an infinite number of possible endpoints, students are asked only to determine one unknown coordinate given the other. ๏ง Given ฬ ฬ ฬ ฬ ๐ด๐ต โฅ ฬ ฬ ฬ ฬ ๐ถ๐ท and ๐ด(3, 2), ๐ต(7, 10), ๐ถ(โ2, โ3), and ๐ท(4, ๐2 ), find the value of ๐2 . While we will be using the formula that we derived in Exercise 1, it will be helpful if you explain each of the steps as you are solving the problem. ๏บ We translated ฬ ฬ ฬ ฬ ๐ด๐ต to ฬ ฬ ฬ ฬ ฬ ฬ ๐ดโฒ๐ตโฒ so that ๐ดโฒ is located at the origin. We moved three units to the left and down ๏บ ๏บ two units giving us ๐ตโฒ(7 โ 3, 10 โ 2). We translated ฬ ฬ ฬ ฬ ๐ถ๐ท to ฬ ฬ ฬ ฬ ฬ ฬ ๐ถโฒ๐ทโฒ so that ๐ถโฒ is located at the origin. We moved two units to the right and up three units, giving us ๐ทโฒ(4 + 2, ๐2 + 3). Using the formula we derived in the previous exercise, we can write the following equation and solve for ๐2 : (7 โ 3)(4 + 2) + (10 โ 2)(๐2 + 3) = 0 4(6) + 8(๐2 + 3) = 0 ๐2 + 3 = โ3 ๐2 = โ6 Teachers can assign different exercises to different students or groups and then bring the class back together to share. Some students may be able to complete all exercises, while others may need more guidance. Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 64 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY Exercises 2โ4 (14 minutes) 2. Recall the Opening Exercise of Lesson 4 in which a robot is traveling along a linear path given by the equation ๐ = ๐๐ โ ๐๐๐. The robot hears a ping from a homing beacon when it reaches the point ๐ญ(๐๐๐, ๐๐๐) and turns to ๐ ๐ travel along a linear path given by the equation ๐ โ ๐๐๐ = โ (๐ โ ๐๐๐). If the homing beacon lies on the ๐-axis, what is its exact location? (Use your own graph paper to visualize the scenario.) a. If point ๐ฌ is the ๐-intercept of the original equation, what are the coordinates of point ๐ฌ? ๐ฌ(๐, โ๐๐๐) b. What are the endpoints of the original segment of motion? ๐ฌ(๐, โ๐๐๐) and ๐ญ(๐๐๐, ๐๐๐) c. If the beacon lies on the ๐-axis, what is the ๐-value of this point, ๐ฎ? ๐ d. Translate point ๐ญ to the origin. What are the coordinates of ๐ฌโฒ, ๐ญโฒ, and ๐ฎโฒ? ๐ฌโฒ(โ๐๐๐, โ๐๐๐๐), ๐ญโฒ (๐, ๐), and ๐ฎโฒ(๐ โ ๐๐๐, โ๐๐๐) e. Use the formula derived in this lesson to determine the coordinates of point ๐ฎ. ฬ ฬ ฬ ฬ , so โ๐๐๐(๐ โ ๐๐๐) + (โ๐๐๐๐)(โ๐๐๐) = ๐ โบ ๐ = ๐๐๐๐. ฬ ฬ ฬ ฬ โฅ ๐ญ๐ฎ We know that ๐ฌ๐ญ Given ๐ฎ(๐, ๐) and we found ๐ = ๐๐๐๐, then the coordinates of point ๐ฎ are (๐๐๐๐, ๐). Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 65 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M4 GEOMETRY 3. A triangle in the coordinate plane has vertices ๐จ(๐, ๐๐), ๐ฉ(โ๐, ๐), and ๐ช(โ๐, ๐). Is it a right triangle? If so, at which vertex is the right angle? (Hint: Plot the points, and draw the triangle on a coordinate plane to help you determine which vertex is the best candidate for the right angle.) The most likely candidate appears to be vertex ๐ช. By translating the figure so that ๐ช is mapped to the origin, we can use the formula (โ๐ + ๐)(๐ + ๐) + (๐ โ ๐)(๐๐ โ ๐) = ๐ โ๐(๐) + ๐(๐) = ๐ ๐=๐ Yes, ฬ ฬ ฬ ฬ ๐ฉ๐ช and ฬ ฬ ฬ ฬ ๐จ๐ช are perpendicular. The right angle is โ ๐ช. 4. ฬ ฬ ฬ ฬ bisects ๐ฉ๐ซ ฬ ฬ ฬ ฬ ฬ , but ๐ฉ๐ซ ฬ ฬ ฬ ฬ ฬ does not ๐จ(โ๐, ๐), ๐ฉ(โ๐, ๐), ๐ช(๐, โ๐), and ๐ซ(โ๐, โ๐) are vertices of a quadrilateral. If ๐จ๐ช bisect ฬ ฬ ฬ ฬ ๐จ๐ช, determine whether ๐จ๐ฉ๐ช๐ซ is a kite. We can show ๐จ๐ฉ๐ช๐ซ is a kite if ฬ ฬ ฬ ฬ ๐จ๐ช โฅ ฬ ฬ ฬ ฬ ฬ ๐ฉ๐ซ. Translating ฬ ฬ ฬ ฬ ๐จ๐ช so that the image of point ๐จ lies on the origin and translating ฬ ฬ ฬ ฬ ฬ ๐ฉ๐ซ so that the image of point ๐ฉ lies at the origin lead to the equation: (๐ + ๐)(โ๐ + ๐) + (โ๐ โ ๐)(โ๐ โ ๐) = ๐ (๐๐)(โ๐) + (โ๐)(โ๐) = ๐ โ๐๐ + ๐๐ = ๐ This is a true statement; therefore, ฬ ฬ ฬ ฬ ๐จ๐ช โฅ ฬ ฬ ฬ ฬ ฬ ๐ฉ๐ซ, and ๐จ๐ฉ๐ช๐ซ is a kite. Closing (2 minutes) As a class, revisit the two theorems of todayโs lesson, and have students explain the difference between them. ๏ง Given points ๐(0, 0), ๐ด(๐1 , ๐2 ), ๐ต(๐1 , ๐2 ), ๐ถ(๐1 , ๐2 ), and ๐ท(๐1 , ๐2 ), ฬ ฬ ฬ ฬ โฅ ๐๐ต ฬ ฬ ฬ ฬ โบ (๐1 )(๐1 ) + (๐2 )(๐2 ) = 0; ๐๐ด ฬ ฬ ฬ ฬ ๐ด๐ต โฅ ฬ ฬ ฬ ฬ ๐ถ๐ท โบ (๐1 โ ๐1 )(๐1 โ ๐1 ) + (๐2 โ ๐2 )(๐2 โ ๐2 ) = 0. Exit Ticket (5 minutes) Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 66 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY Name Date Lesson 6: Segments That Meet at Right Angles Exit Ticket Given points ๐(2, 4), ๐(7, 6), ๐(โ3, โ4), and ๐(โ1, โ9): ฬ ฬ ฬ and ฬ ฬ ฬ ฬ a. Translate ฬ ๐๐ ๐๐ so that the image of each segment has an endpoint at the origin. ๐ป(๐, ๐) ๐บ(๐, ๐) ๐ผ(โ๐, โ๐) b. Are the segments perpendicular? Explain. ๐ฝ(โ๐, โ๐) c. Are the lines โก๐๐ and โก๐๐ perpendicular? Explain. Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 67 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY Exit Ticket Sample Solutions Given points ๐บ(๐, ๐), ๐ป(๐, ๐), ๐ผ(โ๐, โ๐), and ๐ฝ(โ๐, โ๐): a. Translate ฬ ฬ ฬ ฬ ๐บ๐ป and ฬ ฬ ฬ ฬ ๐ผ๐ฝ so that the image of each segment has an endpoint at the origin. Answers can vary slightly. Students have to choose one of the first two and one of the second two. If we translate ฬ ฬ ฬ ฬ ๐บ๐ป so that the image of ๐บ is at the origin, we get ๐บโฒ(๐, ๐), ๐ปโฒ(๐, ๐). If we translate ฬ ฬ ฬ ฬ ๐บ๐ป so that the image of ๐ป is at the origin, we get ๐บโฒ(โ๐, โ๐), ๐ปโฒ(๐, ๐). ฬ ฬ ฬ ฬ so that the image of ๐ผ is at the origin, we If we translate ๐ผ๐ฝ get ๐ผโฒ(๐, ๐), ๐ฝโฒ(๐, โ๐). ฬ ฬ ฬ ฬ so that the image of ๐ฝ is at the origin, we If we translate ๐ผ๐ฝ get ๐ฝโฒ(๐, ๐), ๐ผโฒ(โ๐, ๐). b. Are the segments perpendicular? Explain. โฒ ๐ฝโฒ , we determine whether the equation yields a true ฬ ฬ ฬ ฬ ฬ and ๐ผ ฬ ฬ ฬ ฬ ฬ ฬ Yes. By choosing any two of the translated ๐บโฒ๐ปโฒ statement: (๐๐ โ ๐๐ )(๐ ๐ โ ๐๐ ) + (๐๐ โ ๐๐ )(๐ ๐ โ ๐๐ ) = ๐. For example, using ๐บ(โ๐, โ๐), ๐ปโฒ(๐, ๐), and ๐ผโฒ(๐, ๐), ๐ฝโฒ(๐, โ๐): ฬ ฬ ฬ ฬ ฬ โฅ ๐ผโฒ๐ฝโฒ ฬ ฬ ฬ ฬ ฬ ฬ and ๐บ๐ป ฬ ฬ ฬ ฬ โฅ ๐ผ๐ฝ ฬ ฬ ฬ ฬ . โ๐(๐) + (โ๐)(โ๐) = ๐ is a true statement; therefore, ๐บโฒ๐ปโฒ c. Are the โก๐บ๐ป and โก๐ผ๐ฝ perpendicular? Explain. Yes, lines containing perpendicular segments are also perpendicular. Problem Set Sample Solutions 1. Scaffolding: Give students steps to follow to make problems more accessible. Are the segments through the origin and the points listed perpendicular? Explain. a. ๐จ(๐, ๐๐), ๐ฉ(๐๐, ๐) No ๐ โ ๐๐ + ๐๐ โ ๐ โ ๐ b. ๐ช(๐, ๐), ๐ซ(๐, โ๐) Yes ๐ โ ๐ + ๐ โ (โ๐) = ๐ 2. ฬ ฬ ฬ ฬ ฬ and ๐ด๐ต ฬ ฬ ฬ ฬ ฬ perpendicular? Translate ๐ด Given ๐ด(๐, ๐), ๐ต(๐, โ๐), and ๐ณ listed below, are ๐ณ๐ด to the origin, write the coordinates of the images of the points, and then explain without using slope. a. ๐ณ(โ๐, ๐) ๐ดโฒ(๐, ๐), ๐ตโฒ(โ๐, โ๐), ๐ณโฒ(โ๐, ๐) 1. Plot the points. 2. Find the common endpoint. 3. Translate that to (0, 0). 4. Translate the other points. 5. Use the formula (๐1 )(๐1 ) + (๐2 )(๐2 ) = 0 to determine perpendicularity. Yes (โ๐)(โ๐) + (โ๐)(๐) = ๐ Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 68 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 6 M4 GEOMETRY b. ๐ณ(๐๐, โ๐) ๐ดโฒ(๐, ๐), ๐ตโฒ(โ๐, โ๐), ๐ณโฒ(๐, โ๐) Yes (โ๐)(๐) + (โ๐)(โ๐) = ๐ c. ๐ณ(๐, ๐) ๐ดโฒ(๐, ๐), ๐ตโฒ(โ๐, โ๐), ๐ณโฒ(๐, ๐) No. (โ๐)(๐) + (โ๐)(๐) โ ๐ 3. Is triangle ๐ท๐ธ๐น, where ๐ท(โ๐, ๐), ๐ธ(โ๐, ๐), and ๐น(๐, โ๐), a right triangle? If so, which angle is the right angle? Justify your answer. Yes. If the points are translated to ๐ทโฒ(๐, ๐), ๐ธโฒ(๐, ๐), and ๐นโฒ(๐, โ๐), ๐(๐) + ๐(โ๐) = ๐, meaning the segments are perpendicular. The right angle is โ ๐ท. 4. A quadrilateral has vertices (๐ + โ๐, โ๐), (๐ + โ๐, ๐), (๐ + โ๐, ๐), and (โ๐, ๐). Prove that the quadrilateral is a rectangle. Answers will vary, but it is a rectangle because it has ๐ right angles. 5. Given points ๐ฎ(โ๐, ๐), ๐ฏ(๐, ๐), and ๐ฐ(โ๐, โ๐), find the ๐-coordinate of point ๐ฑ with ๐-coordinate ๐ so that the โก๐ฎ๐ฏ and โก๐ฐ๐ฑ are perpendicular. โ๐ 6. A robot begins at position (โ๐๐, ๐๐) and moves on a path to (๐๐๐, โ๐๐). It turns ๐๐° counterclockwise. a. What point with ๐-coordinate ๐๐๐ is on this path? (๐๐๐, ๐๐๐) b. Write an equation of the line after the turn. ๐ + ๐๐ = c. ๐๐ (๐ โ ๐๐๐) ๐ If it stops to charge on the ๐-axis, what is the location of the charger? (๐๐๐, ๐) 7. Determine the missing vertex of a right triangle with vertices (๐, ๐) and (๐, ๐) if the third vertex is on the ๐-axis. Verify your answer by graphing. (๐, 8. ๐๐ ) ๐ Determine the missing vertex for a rectangle with vertices (๐, โ๐), (๐, ๐), and (โ๐, ๐), and verify by graphing. Then, answer the questions that follow. (โ๐, ๐) a. What is the length of the diagonal? Approximately ๐. ๐๐ units Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 69 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY b. What is a point on both diagonals in the interior of the figure? ๐ (๐, ) ๐ 9. ฬ ฬ ฬ ฬ of right triangle ๐จ๐ฉ๐ช has endpoints ๐จ(๐, ๐) and ๐ฉ(๐, โ๐). Point ๐ช(๐, ๐) is located in Quadrant IV. Leg ๐จ๐ฉ a. Use the perpendicularity criterion to determine at which vertex the right angle is located. Explain your reasoning. Assume that the right angle is at ๐จ(๐, ๐). Then a translation ๐ unit left and ๐ units down maps point ๐จ to the origin. ๐จ(๐, ๐) โ ๐จโฒ (๐ โ ๐, ๐ โ ๐) โ ๐จโฒ (๐, ๐) ๐ฉ(๐, โ๐) โ ๐ฉโฒ(๐ โ ๐, โ๐ โ ๐) โ ๐ฉโฒ(๐, โ๐) ๐ช(๐, ๐) โ ๐ชโฒ(๐ โ ๐, ๐ โ ๐) By the criterion for perpendicularity, ๐(๐ โ ๐) + โ๐(๐ โ ๐) = ๐ ๐๐ โ ๐ โ ๐๐ + ๐๐ = ๐ ๐๐ โ ๐๐ + ๐ = ๐ ๐๐ = ๐๐ + ๐ ๐ ๐ ๐= ๐+ ๐ ๐ The solutions to this equation form a line that had a positive slope and a positive ๐-intercept so no point on this line lies in Quadrant IV. Therefore, the right angle cannot be at ๐จ(๐, ๐) and so must be at ๐ฉ(๐, โ๐). b. Determine the range of values that ๐ is limited to and why? If the right angle is at ๐ฉ(๐, โ๐), then a translation ๐ units left and ๐ unit up maps point ๐ฉ to the origin. ๐ฉ(๐, โ๐) โ ๐ฉโฒ(๐ โ ๐, โ๐ โ (โ๐)) โ ๐ฉโฒ(๐, ๐) ๐จ(๐, ๐) โ ๐จโฒ(๐ โ ๐, ๐ โ (โ๐)) โ ๐จโฒ(โ๐, ๐) ๐ช(๐, ๐) โ ๐ชโฒ(๐ โ ๐, ๐ โ (โ๐)) โ ๐ชโฒ(๐ โ ๐, ๐ + ๐) By the criterion for perpendicularity, โ๐(๐ โ ๐) + ๐(๐ + ๐) = ๐ โ๐๐ + ๐๐ + ๐๐ + ๐ = ๐ โ๐๐ + ๐๐ + ๐๐ = ๐ ๐๐ = ๐๐ โ ๐๐ ๐ ๐๐ ๐= ๐โ ๐ ๐ ๐ ๐ The point must lie on the graph of ๐ = ๐ โ (๐, โ ๐๐ and in Quadrant IV. The ๐-intercept of the graph is ๐ ๐๐ ), however since this point is not in the fourth quadrant, ๐ > ๐. Substituting ๐ for ๐ in the equation, ๐ it is determined that the ๐-intercept of the graph is ๐. ๐. This point does not lie in Quadrant IV so ๐ < ๐. ๐. The remaining vertex cannot coincide with another vertex, so ๐ โ ๐. Therefore, the value of the ๐-coordinate of the remaining vertex is limited to ๐ < ๐ < ๐. ๐ and ๐ โ ๐. Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 70 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 6 NYS COMMON CORE MATHEMATICS CURRICULUM M4 GEOMETRY c. Find the coordinates of point ๐ช if they are both integers. ๐ ๐ The value of ๐ must be ๐, ๐, ๐, ๐, or ๐. Substituting each into the equation ๐ = ๐ โ ๐๐ , the only value of ๐ ๐ that yields an integer value for ๐ is ๐ = ๐. ๐ ๐ ๐ = (๐) โ ๐= ๐๐ ๐ ๐ ๐๐ โ ๐ ๐ ๐ = โ๐ The coordinates of vertex ๐ช are (๐, โ๐). Lesson 6: Segments That Meet at Right Angles This work is derived from Eureka Math โข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M4-TE-1.3.0-09.2015 71 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.