1 EXPLORATION: Points On A Perpendicular Bisector
Name Date9.2Perpendicular and Angle BisectorsFor use with Exploration 9.2Essential Question What conjectures can you make about a point onthe perpendicular bisector of a segment and a point on the bisector of anangle?1EXPLORATION: Points on a Perpendicular BisectorGo to BigIdeasMath.com for an interactive tool to investigate this exploration.Work with a partner. Use dynamic geometry software.a. Draw any segment and label it AB. Construct the perpendicular bisector ofAB.b. Label a point C that is on the perpendicular bisector of AB but is not on AB.c. Draw CA and CB and find their lengths. Then move point C to otherlocations on the perpendicular bisector and note the lengths of CA and CB.d. Repeat parts (a)–(c) with other segments. Describe any relationship(s) younotice.A3C21B00212345SamplePointsA(1, 3)B(2, 1)C(2.95, 2.73)SegmentsAB 2.24CA ?CB ?Line x 2 y 2.5EXPLORATION: Points on an Angle BisectorGo to BigIdeasMath.com for an interactive tool to investigate this exploration.Work with a partner. Use dynamic geometry software. a. Draw two rays AB and AC to form BAC. Construct the bisector of BAC.b. Label a point D on the bisector of BAC.289Copyright Big Ideas Learning, LLCAll rights reserved.
Name9.22DatePerpendicular and Angle Bisectors (continued)EXPLORATION: Points on an Angle Bisector (continued)c. Construct and find the lengths of the perpendicular segments from D to thesides of BAC. Move point D along the angle bisector and note how thelengths change.d. Repeat parts (a)–(c) with other angles. Describe any relationship(s) you notice.4E3B2A1DCF00123456SamplePointsA(1, 1)B(2, 2)C(2, 1)D(4, 2.24)RaysAB x y 0AC y 1Line 0.38 x 0.92 y 0.54Communicate Your Answer3. What conjectures can you make about a point on the perpendicular bisector of asegment and a point on the bisector of an angle? 4. In Exploration 2, what is the distance from point D to AB when the distance from D to AC is 5 units? Justify your answer.Copyright Big Ideas Learning, LLCAll rights reserved.290
9.29.2Notetaking with VocabularyNotetakingwithFor use after Lesson9.2 VocabularyFor use after Lesson 9.2In your own words, write the meaning of each vocabulary term.Name Dateequidistantequidistant9.2PracticeFor use after Lesson 9.2In your own words, write the meaning of each vocabulary term.TheoremsequidistantTheoremsPerpendicular Bisector TheoremPerpendicular Bisector TheoremIn a plane, if a point lies on the perpendicularIna plane,a point lieson itthebisectorof ifa segment,thenis s equidistantfrom the endpoints of the segment.from the endpoints of the segment.TheoremsCCAA CP is the bisectorof ABTheorem, then CA CB.IfPerpendicularBisectorIf CP is the bisector of AB, then CA CB.In a plane, if a point lies on the perpendicularNotes:bisector of a segment, then it is equidistantNotes:from the endpoints of the segment.BBPPCABP If CP is the bisector of AB, then CA CB.Converse of the Perpendicular Bisector TheoremConverseof the Perpendicular Bisector TheoremNotes:In a plane, if a point is equidistant from theIna plane,ofif aa segment,point is equidistanttheendpointsthen it liesfromon theendpointsof a bisectorsegment,ofthenit lies on theperpendicularthe segment.perpendicular bisector of the segment.If DA DB, then point D lies on the bisector of AB.ConversethepointPerpendicularBisectorTheorem, thenIf DA DBofD lies on the bisector ofAB.CCAABBPPIn a plane, if a point is equidistant from theCNotes:endpoints of a segment, then it lies on theNotes:Name D Dateperpendicular bisector of the segment.DNotetakingwith(continued)9.2 DB, then point DIf DAlies Vocabularyon the bisectorof AB.ANotes:Angle Bisector TheoremDBIf a point lies on the bisector of an angle, then it isequidistant from the two sides of the angle. BP If AD bisects BAC and DB AB and DC AC ,then DB DC .Notes:285285291DACCopyright Big Ideas Learning, LLCCopyright Big IdeasLearning,LLCAll rightsreserved.All rights reserved.Copyright Big Ideas Learning, LLCAll rights reserved.
X and Z? B is not collinear with X and Z. Because the twosegments containing points X and Z are congruent, B is theNotes:same distance from both X and Z , pointB is equidistant from—.X and Z, and point B is in the perpendicular bisector of XZNameDateMonitoring Progress and Modeling with MathematicsPractice (continued)9.23. GH 4.6; Because GK KJ and ⃖ ⃗HK ⃖ ⃗GJ , point H is on—. So, by the Perpendicularthe perpendicular bisector of GJConverseof theBisectorTheoremBisectorTheorem,GHAngle HJ 4.6.If a pointisBecausein the interiorangle andfromis equidistantAngleBisectorTheorem4. QR1.3;pointTofisanequidistantQ and S,—from thesidesof the angle, then it lieson bythethepointT istwoon theofthenQSIf a pointlieson perpendicularthe bisector ofbisectoran angle,it isbisector em.So, byequidistant from the two sides of the angle.definition ofsegment bisector,QR RS 1.3. DB DC,If DB AB and DC AC and ⃖ ⃗⃖ ⃗AB DandDC AC ,If ADbisects BACand5. AB 15;BecauseDB ACDBand pointis equidistant—fromand bisectsC,DCpointD is onthen ADB. BACAD. the perpendicular bisector of ACby the Converse of the Perpendicular Bisector Theorem. Bydefinition of segment bisector, AB BC.Notes:AB BCNotes:5x 4x 3 that is different is: Is point B collinear with2. ThequestionX andx Z?3 B is not collinear with X and Z. Because the twosegmentsAB 5 containing3 15 points X and Z are congruent, B is thesame distance from both X and Z, point B is equidistant from————,bisectorX andpoint VDB is inWDthe perpendicular6. UW Z,55;andBecauseand ⃖ ⃗UX VWpoint Uof XZ .—. So, by theison the perpendicularbisectorBisectorof VWConverseof the athematicsPerpendicularBisectorVUwith WU.If a point is in the interior of an angleandis⃖ ⃗ ⃖ ⃗ equidistantBBDDAACC 3. GHVU 4.6;Because GK KJ and HK GJ , point H is on UWfromthe two sidesbisectorofExamplesthe angle,the—.thenWorked-Outtheperpendicularof GJSo, itbyliestheonPerpendicular9x 1 7x 13bisectorthe angle.Bisector ofTheorem,GH HJ 4.6.2x 12Example #1 4. IfQRDB x1.3;Becausepoint T ACis equidistant 6ABDC, Q and S, and DCand DB from.—Findmeasure. Explainreasoning.point theT isindicatedon the perpendicularbisector yourof QSby the UWnition AD7 bisects6 13 55defiofsegmentbisector,QR RS L and1.3.M, point7. f thePerpendicularBisectorTheorem.So, by— by the ConverseN is on the perpendicularbisectorofLM5. AB 15; Because ⃖ ⃗DB ⃖ ⃗AC and point D is equidistantofthe Perpendicular Bisector Theorem. Because only one—Notes:from A and C, point D is on thebisector of AC—perpendicularline can be perpendicular to LMat point K, ⃗NK must be theby the Converse of the PerpendicularBisectorTheorem.By—, and P is on NK ⃗.perpendicular bisector of LMdefinitionof segmentbisector,Copyright BigIdeas Learning,LLC AB BC.AllABrightsreserved.286 BC8. no; Youwould need to know that either LN MN or5x MP.4x 3LPx 39. no; You would need to know that ⃖ ⃗PN ⃖ ⃗ML.AB 5 3 15 CDABA5x B 4x 3 C 10. yes; Because point Pfrom —L and M, point P—is equidistant— UX6. UW 55; Because VD WD and ⃖ ⃗, point U— VWis on the perpendicular bisector of LMthe Converse of—.byis on the perpendicular bisector of VWSo,—by the—thePerpendicularBisector Theorem. Also, LN MN, so ⃗PNExample#2BisectorPerpendicularTheorem,VU WU.—is a bisector of LM . Because P can only be on one of the—.UWis the perpendicularFindVUthe indicatedmeasure. Explainyour ⃗bisectors,PNbisectorof reasoning.LM9x 1 7x 1311. Because D is equidistant from ⃗BC and ⃗BA, ⃗BD bisects2x 12 ABC by the Converse of the Angle Bisector Theorem. So,x 6 m CBD 20 .m ABDUW 7 6 13 55—— , ⃗ ⃗S is an angle bisector of PQR PS ⃗QP , and SRQR.So,bytheAngleBisectorTheorem,PS RS 12.7. yes; Because point N is equidistant from L and M, pointCopyright Big Ideas Learning, LLC—NLMAll rights reserved. 292Copyright— Big Ideas ⃗ Learning, LLCline can be perpendicular to LM at point K, NK must be theADm ABD20 CB
Name Date9.2Practice (continued)PracticeAExtra PracticeIn Exercises 1–3, find the indicated measure. Explain your reasoning.1. AB2. EG3. SUGARH397E10107BD66CFS2x 24. Find the equation of the perpendicular bisector of AB.U3xTy2B 22 2xAIn Exercises 5–7, find the indicated measure. Explain your reasoning.5. m CAB6. DC7. BDACADBD30 30 20 B3x 1C55x – 1CBDA293Copyright Big Ideas Learning, LLCAll rights reserved.
Name Date6.2 BPractice BPracticeIn Exercises 1–3, tell whether the information in the diagram allows you toconclude that point P lies on the perpendicular bisector of RS, or on theangle bisector of DEF. Explain your reasoning.1.2.3.QEFDPFRPDPSQEIn Exercises 4–6, find the indicated measure. Explain your reasoning.5. m LNM4. AC x 25AN30 T(5x 24) LC60 (2x 3) (0, 5) Ky3(2x 8)B6. m UTW(52 , 52 )Mx (5, 0)U9V 9WD7. Write an equation of the perpendicular bisector of the segment with the endpointsG (3, 7) and H ( 1, 5).8. In the figure, line m is the perpendicularbisector of PR. Is point Q on line m? Ispoint S on line m? Explain your reasoning.P1310mQS12109. You are installing a fountain in the triangulargarden pond shown in thefigure. You want to placethe fountain the samedistance from each sideof the pond. Describe away to determinethe location of thefountain using anglebisectors.RCopyright Big Ideas Learning, LLCAll rights reserved.294
Angle Bisector Theorem If a point lies on the bisector of an angle, then it is equidistant from the two sides of the angle. If AD bisects BAC and DB AB and DC AC, then DB DC . AC Notes: 5 Converse of the Angle Bisector Theorem If a point is in the interior of an angle and is equidistant from the two sides of the .
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Graphite exploration - the importance of planning Graphite has become the focus for dozens of exploration companies since the mineral's investment boom of 2011-2012. Andrew Scogings, Industrial Minerals Consultant*, looks at the diferent exploration and testing methods and reporting conventions used by the graphite industry. IMX Resources Ltd
Table 1: Exploration Systems Development Organization-Managed Human Exploration Programs Are Baselined to Different Missions 17 Table 2: Change in Estimated Completion Date for Nine . Figure 6: Exploration Systems Development Organization's Integration Reviews 13 : Contents : Page ii GAO-18-28 Exploration Programs' Integration Approach .
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Points, Lines, Planes, & Angles Points & Lines: CLASSWORK 1. Name three collinear points on line q and on line s 2. Name 4 sets of non-collinear points 3. Name the opposite rays on line q and on line s 4. How many points are marked on line q? 5. How many points are there on line q? Points & Lines: HOMEWORK 6.
Exams (3 @ 100 points each) - 300 points Application Exercises (4 @ 5 points each) - 20 points Reading Synthesis Assignments (15 @ 3 points each) - 45 points HRM Issue Analysis Iterative Writing and Research Project (70 points) Issue Analysis Step 1: Individual preparatory assignment (10 points)
