Elementary coordinate geometry

Transcription

1 CHAPTER 2 Elementary coordinate geometry Thepageonwhichyoudrawmay,forallpracticalpurposes,beconsideredasawindowontoaplaneextending uniformly to infinity. We shall not look too closely at the assumptions made in this statement, but instead rely strongly on intuition depending on visual experience to deduce important facts about this plane. Using computers to draw requires translating from geometry to numbers i. e. to a coordinate system and back again.thereareafewbasicformulasthatareusedoverandoveragain.itisbesttomemorizethem.calculating the distance between points whose coordinates are given requires Pythagoras Theorem, which we recall almost at the beginning of this chapter. Before that, however, comes a discussion of the areas of parallelograms; and even before that comes a short note about distinguishing points from vectors. Towards the end of the chapter we shall look at a number of results related principally to projections. Formanyreaders,theresultspresentedinthischapterwillbewellknown.Evenforthem,however,theuseof visual reasoning might be interesting and, in some aspects, novel Points and vectors It is important to distinguish points from vectors, even though a coordinate system assigns a pair of numbers toeitherapointoravector. Pointsare...well,points. Theypossessnoattributeotherthan position,andin particulartheyare(inspiteofhowtheyaredrawn!)withoutdimensionorsizeorcolororsmellor...anything other than position. Vectors, on the other hand, have magnitude and direction. They measure relative position. It is very important to keep in mind that both points and vectors are objects independent of which coordinate system is being used. Vectorscanbeaddedtoeachother,andtheycanbemultipliedbyconstants. Thereisalsoakindoflimited algebrainvolvingpoints. If P = (x P, y P )and Q = (x Q, y Q )aretwopointsthenthereisauniquevectorwith tailat Pandheadat Qwhosecoordinatesare x Q x P and y Q y P,describingtherelativepositionofthetwo points.itiswrittenas Q P.Onereasonthatitiscommontoconfusepointswithvectorsisthattoeachpoint P correspondsthevector P Ofromtheoriginto P.However,ifthecoordinatesystemchanges,theoriginmay change. The points themselves won t change, but the vectors they correspond to will likely do so. Ifwearegivenacoordinatesystem,thevectorwithcoordinates x, ywillbe [x, y]andthepointwiththose coordinateswillbe (x, y). The (x, y)correspondstothevector [x, y]fromtheorigintoitself butirepeatthat this point and this vector are not the same geometrical object. If Pisapointand vavectoritmakessensetoconsider P + vasapoint itisthepoint Qsuchthat Q P = v. Itisthepoint P displacedor translatedby v. If tisarealnumberbetween 0and 1thenthepoint tofthe wayfrom Pto Qisequalto P + t(q P)withcoordinates (1 t)x P + tx Q, (1 t)y P + ty Q ). Iwriteitas (1 t)p +tq.itisakindofweightedaverageof Pand Q.Forexample,thepointpointmidwaybetween Pand Qis (1/2)P + (1/2)Q.Asweshallseelater,wecanalsotakeweightedaveragesofcollectionsofseveralpoints. Insummary:wecansubtracttwopointstogetavector;orcalculateaweightedaveragetoobtainanotherpoint; butthesumoftwopointsorascalarmultipleofapointmakesnointrinsicsense.

2 Chapter 2. Elementary coordinate geometry Areas of parallelograms Area is a somewhat sophisticated concept, not easily analyzed in complete rigor. We are used to thinking of it asanumber,butofcoursethenumberinvolveddependsontheunitsinvolved itisreallyaratioofthearea ofaregiontothatofaunitsquare.soareaseemstobeafundamental,geometricalcharacteristicofaregion.it isinterestingthateuclidstartsoffbookiofthe Elementswithpropertiesofareathatareencapsulatedinafew particularly simple axioms. One of these is that congruent regions have the same area. Recall that one region is said to be congruent to another if it is obtained from it by translation, rotation, or reflection, without altering the relative distances between points of the region. Another basic principle is the additive principle of areas: If two regionshavethesameareaandcongruentregionsareaddedtoeach,thenthenewregionsalsohavethesame area.thisleadstothefollowingmoregeneralcriterion,whichisveryclosetotheoneusedimplicitlybyeuclid in his treatment of area: (Euclid sfirstcriterionforareas)tworegionshavethesameareaiftheycanbechoppedintosmallerpieces which are congruent. Thisdoesnotallowforatreatmentofareaswithcurvedboundaries,butitdoesallowustoseethat Aparallelogramhasthesameareaasarectanglewiththesamebaseandheight. Why is this true? A proof according to Euclid s criterion must show how to decompose the parallelogram and the rectangle into congruent pieces. In some circumstances, this is simple. The complexity of the decomposition involveddependsonhowskewedtheparallelogramis,orhowfarremoveditisfromtherectangleitistobe comparedto.ifitnottooskewed,thenwecanlopoffatriangleatoneendoftheparallelogramandpasteitinat theothertomakearectangle. But this means exactly that in these circumstances we can decompose the rectangle and the parallelogram into congruent pieces. Iftheparallelogramisveryskewed,however,thenwhatwelopoffatoneendisnotatriangle,andthisargument fails. The first, simple argument works when the parallelogram is mildly skewed i. e. when the piece chopped off one end is indeed a triangle. This happens when the entire parallelogram fits into the region shown in this figure: Just about all proofs of the result are the same for mildly skewed parallelograms. There are lots of different ways toproceedfortherest.hereareafew: Proof 1.Wecangetanideaofapossiblewaytoproceedifweagaintranslatetheloppedoffregiontotheleftand glueiton,justasifitwereatriangle.

3 Chapter 2. Elementary coordinate geometry 3 Thenaturalthingtodonowistolopoffthebitoftriangleatthefarrightandshiftitbackagaintofillinarectangle. This gives us finally a way to chop up both the rectangles and the original parallelogram into congruent regions. As the parallelogram gets more and more skewed, the number of pieces the parallelogram gets chopped up into increases,butthereisadefinitepatterntothewaythingsgo.hereareacoupleofpicturestoshowwhathappens: Exercise 2.1. Define the skew of a parallelogram to be the length of the perpendicular projection of its upper leftcornerontoitsbaseline,dividedbythelengthofthebase.countnegativelytotheleft.aparallelogramisa rectangleifandonlyifitsskewis 0.Theargumentaboveshowsthatiftheskew ssatisfies 0 < s 1,thenthe simpledecompositionwillprovetheclaim.explainbyapicturewhathappensif 1 s < 0. Exercise 2.2. Explain the argument in the previous exercise by producing figures in PostScript. Exercise 2.3. Thesecondgroupofpicturesshowswhathappensif 1 < s 2. Whatabout 2 s < 1? 2 < s 3? Exercise 2.4. Iftheskew ssatisfies n < s n + 1(npositive),whatistheleastnumberofpiecesinthe decomposition of the parallelogram and rectangle into congruent pieces suggested by the above reasoning? Exercise 2.5. The reasoning above has just shown how the decomposition of rectangle and parallelogram works inafewcases,andtheexercisesabovehaveshownhowtoincludeafewmorecases.writeoutindetailarecipe for making congruent decompositions of rectangle and parallelogram that will prove the claim when the skew s satisfies 0 < n < s n + 1. Proof 2. The transformation of a rectangle into a parallelogram with the same base and height is called a shear. Theresultweareprovingamountstothis:

4 Chapter 2. Elementary coordinate geometry 4 Shears preserve area. Ashearcanbevisualizedasacontinuoussequenceofslidingmotions,ifyouthinkoftheoriginalrectangleas madeupofverythinstripspiledontopofeachother.likeaslidingdeckofcards. In this way, preservation of area under shear becomes intuitive you can think of the rectangle as an infinite number of horizontal strips piled on top of one another. Shearing it just translates each of these, not changing itsarea,hencenotchangingtheareaofthetotalfigureasitissheared. Thissortofreasoningisnotalways dependable, but it is valuable nonetheless. Historically, it played an important role in the development of calculus long before the nature of limits was understood clearly. Here, however, it suggests an entirely valid and perhapsthebestmotivatedproofoftheresult.wedon thavetochopuptherectangleintoaninfinitenumberof horizontal strips, but just enough strips so that each one becomes only a mildly skewed parallelogram when it is sheared. Proof 3.ThattwoparallelogramswiththesamebaseandheighthavethesameareaisPropositionI.35inEuclid s Elements. But his proof of it depends on the subtractive principle of areas: If congruent regions are taken away from two regions of equal area, then the remaining regions have equal area. Thesimplestofallproofsdependsonthisprinciple,butitisnotthesameasEuclid s. Itcanbeexplainedina single pair of diagrams: Exercise 2.6. Analyze Euclid s own proof of I.35 by breaking it up into a sequence of pictures. Exercise 2.7.Neither the result nor any of these proofs depends on interpreting area as a number, nor even how tocomparetheareaoftwodistinctrectangles.makeafirststepinthisdirectionbyexplaininginyourownwords howtoconstructgeometrically,foranyrectanglewithbase bandheight a,asquareofthesamearea. (Thisis II.14 of Euclid s Elements).

5 Chapter 2. Elementary coordinate geometry Lengths The principal result concerning lengths is Pythagoras Theorem. Foraright angledtrianglewithshortsides aand bandlongside c, c 2 = a 2 + b 2. Thisresult,asalsotheoneintheprevioussection,canbephrasedintermsofequalityofareas.Weerectsquares oneachofthesidesofthegiventriangle.thetheoremassertsneithermorenorlessthanthat Theareaofthelargestsquareisthesumoftheareasoftheothertwo. There are many ways to prove Pythagoras Theorem. There is even a book which purports to contain 365 different proofs,oneforeachdayoftheyear(andincludesafewextra). TheproofgivenhereisclosetoEuclid sown (Pythagoras TheoremisI.47inthe Elements).IfirstsawitinabookbyHowardEves,butitprobablyderives originally from the proof of a generalization of Pythagoras Theorem due to the later Alexandrian mathematician Pappus. It exhibits a decomposition of the larger square(the square on the hypotenuse ) into rectangles whose areas match the smaller squares(the squares on the sides ). The proof proceeds by a sequence of shears and translations, which we know to preserve areas, transforming the rectangles in the large square into the squares on the sides.

6 Chapter 2. Elementary coordinate geometry 6 Exercise 2.8. Thisisveryelegant,butiflookedatcloselythereappeartobeafewgaps.Findthemandfillthem in Vector projections If visavectorintheplane,thenanyothervector ucanbeexpressedasthesumofavector u 0 parallelto vanda vector u perpendiculartoit. u u = u 0 + u v u 0 We ask the following question:

7 Chapter 2. Elementary coordinate geometry 7 If v = [a, b]and u = [x, y],howdowecalculate u 0? Theprojection u 0 willbeascalarmultipleof v,say u 0 = cv,andourproblemistocalculate c.thelengthofthe projection will be u 0 = c v. Soifweknowthelength u 0,wecancalculate c = u 0 / v.inordertogetthesignof c,weintroducethe notion of signed length. If the ordinary length of the projection is s, then its signed length(relative to the vector v)isjust siftheprojectionisinthesamedirectionas v,but sifintheoppositedirection. v positive length negative length Wenowneedtofindaformulaforthesignedlengthof u 0. Thefirstobservationisthattheparallelprojectionisanadditivefunctionof u,whichmeansthatif u = u 1 + u 2 thentheprojectionof uisthesumoftheprojectionsof u 1 and u 2. u u 1 u u 2 Since uisequaltothesumofitsprojectionsontothe xand yaxes,itisonlynecessarytofindthesignedlengths oftheprojectionsof [x, 0]and [0, y]andaddthemtogether. Let slookattheprojectionof [x, 0]. v = [a, b] s x [x, 0] [a, 0]

8 Chapter 2. Elementary coordinate geometry 8 Let s x bethesignedlengthoftheprojectionof [x, 0],andlet v = [a, b].thetwotrianglesinthefigurearesimilar, soweseethat s x x = a v, s x = ax v. Similarlyif s y isthatof [0, y]then s y = by v. u The figure deals with positive lengths, but the final result remains valid for negative ones as well. Hence the signedlengthof u = [x, y]is ax + by s = s x + s y =. v If u = [x, y]and v = [a, b]arevectors,theprojectionof uontoalineparallelto vis ( ) ax + by [a, b] ax + by a2 + b 2 a2 + b = 2 a 2 + b 2 [a, b]. Thereisanotherformulaforthesignedlength.If θistheanglebetween uand vthen s = u cosθ. u v θ u 0 signed length of u 0 = u cosθ Ifwecomparethetwoformulasweseethattheangle θbetweenthevectors v = [a, b]and u = [x, y]canbefound from this identity: ax + by cosθ = u v. If u = [a, b]and v = [x, y]thenthenumeratoroftheformulaaboveiscalledtheir dot product: u v = ax + by.

9 Chapter 2. Elementary coordinate geometry 9 Thustheformulaforthecosineoftheanglebetweenthembecomes cosθ = u v u v. The dot product satisfies a number of simple formal algebraic rules: cx y = c(x y) x cy = c(x y) (x + y) z = x z + y z x x = x 2 where x isthelengthofthevector x,thedistanceofitsheadfromitstail Rotations Wenowlookatanewproblem:Supposewestartwiththevector v = [x, y]androtateitaroundtheoriginby angle θ.whatarethecoordinatesofthenewvector? θ Rotation by θ Theanswertothisdependsdirectlyontheanswertothesimplestcase,thatwheretheangleis 90. If v = [x, y]thenrotating vcounter clockwiseby 90 givesus v = [ y, x]. Thiscanbeseeneasilyinthispicture: Butnowweareingoodshape,sincefromthisfigure Rotationby 90 v θ v we can deduce: vrotatedby θis (cosθ)v + (sin θ)v

10 Chapter 2. Elementary coordinate geometry 10 If v = [x, y]thenrotating vby θgivesus (cosθ) v + (sin θ) v = [x cosθ y sin θ, x sin θ + y cosθ]. Thisexpressioncanbecalculatedbyamatrix.Rotationby θtakesthevector [ x y ]to [ [ x y ] cosθ sinθ ] sin θ cosθ Inthisbook,vectorswillusuallyberowvectors,andmatriceswillmultiplythemontheright.Thisisacommon convention in computer graphics, as opposed to that in mathematics, and makes especially good sense in dealing with PostScript calculations, as we shall see. If vistheunitvector [cosα, sin α]and V is vrotatedby β,weobtainontheonehandthevectorcorresponding toangle α + β,andontheother,accordingtotheformulaforrotationsihavejustderived,thevector Thisgivesusthecosineandsinesumrules: [ [cos(α + β), sin(α + β)] = [cosα, sin α] cosβ sinβ cos(α + β) = cosαcosβ sin α sin β sin(α + β) = sin α cosβ + cosαsin β ] sinβ. cosβ 2.6. The cosine rule The cosine rule is a generalization of Pythagoras Theorem that applies to triangles which are not necessarily right angled. c a C b (Cosinerule)Inatrianglewithsides a, b, c,andangle Copposite c c 2 = a 2 + b 2 2ab cosc. I sketch three proofs, the first a mixture of algebra and geometry, the second almost purely geometric, and the third almost entirely algebraic.

11 Chapter 2. Elementary coordinate geometry 11 c 2 c 1 a C 2 C 1 b The first one applies the cosine sum formula and Pythagoras Theorem to the diagram above. C The second generalizes Euclid s proof of Pythagoras Theorem. It begins by showing that the two rectangular areas in the diagram above have equal areas, and then finally applies the definition of cosine. Asforthethird,itusesthedot productin2d.recall: Iftheanglebetweentwovectors uand vis θthen cosθ = u v u v. Nowstartbywritingthecosineruleintermsofvectors.Wewanttoshowthat cosθ = u v 2 u 2 v 2 2 u v

12 Chapter 2. Elementary coordinate geometry 12 where uand varevectorsalongthesidesofthetrianglewithlengths aand b,andthereforethethirdsideofthe triangleis u v.followingtheequationabovewereduce (u v) (u v) u u v v 2 u v u u 2 u v + v v u u v v = 2 u v = u v u v = cos θ sure enough. Exercise 2.9. Finish both the first two proofs. For the second, add several diagrams illustrating the argument Dot products in higher dimensions Thedotproductoftwovectorsinanynumber nofdimensionsgreaterthantwoorthreeisbydefinitionthesum of the products of their coordinates: (x 1, x 2,...,x n ) (y 1, y 2,..., y n ) = x 1 y 1 + x 2 y x n y n. Theformalruleswehaveseentobetrueintwodimensionsholdalsointhreedimensionsandmore. Forvectors uand vin 2or 3dimensions where θistheanglebetweenthem. u v = u v cosθ Theproofofthisisjustthereverseofthethirdargumentinthelastsection. In particular: Thedotproductoftwovectorsis 0preciselywhentheyareperpendiculartoeachother. Ofcourseitisonlyin 2or 3dimensionsthatwehaveageometricdefinitionoftheanglebetweentwovectors.In higher dimensions, this formula is used to define that angle algebraically Lines Onewayofrepresentinglinesintheplaneisbymeansofanequation y = mx + b. Butthiscannotrepresentverticallines,thoseparalleltothe y axis,whichhaveanequation x = a.theuniform waytorepresentalllinesisbymeansofanequation Ax + By + C = 0. Forexample, y mx b = 0or x a = 0. Lineswhicharenotverticalarethosewith B 0,inwhichcase wecansolvefor y.theproblemwiththisschemeisthatif Ax + By + C = 0istheequationofalineand cis anon zeroconstantthen cax + cby + cc = (ca)x + (cb)y + (cc) = 0isalsotheequationofthesameline. This means that the coordinates [A, B, C] of a line are homogeneous only determined up to multiplication by a non zero scalar. This is the first place in which homogeneous coordinates occur in this book. They will play an extremely important role later on, especially when we come to 3D graphics, and also in understanding how PostScript handles coordinates in 2D. Intheequation y = mx + bboth mand bhaveageometricalinterpretation mistheslopeand bthe y intercept. Whatisthegeometricalsignificanceof A, B,and Cintheequation Ax + By + C = 0?

13 Chapter 2. Elementary coordinate geometry 13 Suppose C = 0.Theequationis Ax + By = 0,whichcanberewritten [A, B] [x, y] = 0. Butthisistheconditionthat [x, y]beperpendicularto [A, B].Inotherwords, [A, B]isthedirectionperpendicular totheline Ax + By = 0.Inotherwords,theline Ax + By = 0istheuniquelinethatis(1)perpendiculartothe vector [A, B] and(2) passing through the origin. Nowlookatthegeneralcase Ax + By + C = 0.If P = (x P, y P )and Q = (x Q, y Q )aretwopointsonthisline, then Ax P + By P + C = 0 Ax Q + By Q + C = 0 A(x Q x P ) + B(y Q y P ) = 0 whichsaysthatthevector Q Pisperpendicularto [A, B].Inotherwords,wehavethefollowingpicture: P Ax + By + C = 0 [A, B] Ax + By = 0 Q Thevector [A, B]isperpendiculartotheline Ax + By + C = 0. Whatisthemeaningof C?If C = 0thelinepassesthroughtheorigin.Thismakesplausible: The quantity C A2 + B 2 isthesigneddistanceoftheline Ax + By + C = 0fromtheline Ax + By = 0. Exercise Explain why this is true.(hint: use projections.) Exercise Givenaline Ax + By + C = 0andapoint P,findaformulafortheperpendicularprojectionof Pontotheline. Exercise Giventwolines A 1 x + B 1 y + C 1 = 0and A 2 x + B 2 y + C 2 = 0,findaformulaforthepointof intersection. Exercise Given two points P and Q, find a formula for the line containing them. Exercise Givenaline Ax + By + C = 0,findaformulaforthelineobtainedbyrotatingitby 90 around the origin. Theline Ax+By +C = 0separatestheplaneintotwohalves,onewhere Ax+By +C > 0andtheotherwhere itisnegative.whichsideiswhich?

14 Chapter 2. Elementary coordinate geometry 14 Ax + By + C > 0 Ax + By + C = 0 [A, B] Ax + By + C < 0 Aswecrossthelineinthedirectionof [A, B]thevaluesof Ax + By + Cchangefromnegativetopositive.This iseasytoseeindirectly.if (x, y) = (ta, tb)then Ax + By + C = t(a 2 + B 2 ) + Candfor t 0willdefinitely be positive. Thefunction f(x, y) = Ax + By + Ciscalledan affine function.oneusefulpropertyofaffinefunctionsisthis: If Pand Qaretwopointsintheplane, tarealnumber,and fanaffinefunctionthen f ( (1 t)p + tq ) = (1 t)f(p) + tf(q). Recallthat (1 t)p + tqistheweightedaverageof Pand Q.As tvariesoverallrealnumbersthisexpression producesallpointsonthelinethrough Pand Q.Provingthispropertyisaneasycalculation. Exercise 2.15.Anaffinefunction f(x) = Ax + By + Cisequalto 4at (0, 0)and 7at (1, 2).Whereontheline segmentbetweenthesetwopointsis f(x) = 0? 2.9. Code The file eves-animation.eps is a page turning animation of Eves proof of Pythagoras Theorem. References 1. Euclid, The Elements,translatedbyT.L.Heath. ThisisavailableinacommonlyfoundDoverreprint. The partduetoeuclid,butnotheath sveryvaluablecomments,isalsoavailableonthe Netat djoyce/java/elements/elements.html Thecommentaryusedtobefoundat but at the moment I write this(december, 2003) all links to Heath s comments, other than the initial chapters, are unfortunately dead. 2. Elisha S. Loomis, The Pythagorean proposition: its demonstrations analyzed and classified, and bibliography of sources for data of the four kinds of proofs, National Council of Teachers, Washington, Not wildly exciting. Not as much variety as you might hope for, either. But still curious. 3. H. Eves, In mathematical circles, The idea for the sequence of pictures for Pythagoras Theorem is taken fromp.75.