### Transcription

Coordinate GeometryJWRTuesday September 6, 2005Contents1 Introduction32 Some Fallacies2.1 Every Angle is a Right Angle!? . . . . . . . . . . . . . . . . .2.2 Every Triangle is Isosceles!? . . . . . . . . . . . . . . . . . . .2.3 Every Triangle is Isosceles!? -II . . . . . . . . . . . . . . . . .45673 Affine Geometry3.1 Lines . . . . . . . . . . . . . . .3.2 Affine Transformations . . . . .3.3 Directed Distance . . . . . . . .3.4 Points and Vectors . . . . . . .3.5 Area . . . . . . . . . . . . . . .3.6 Parallelograms . . . . . . . . . .3.7 Menelaus and Ceva . . . . . . .3.8 The Medians and the Centroid .4 Euclidean Geometry4.1 Orthogonal Matrices . . .4.2 Euclidean Transformations4.3 Congruence . . . . . . . .4.4 Similarity Transformations4.5 Rotations . . . . . . . . .4.6 Review . . . . . . . . . . .4.7 Angles . . . . . . . . . . .1.8812192020232426.3030313233343637

4.8Addition of Angles . . . . . . . . . . . . . . . . . . . . . . . . 395 More Euclidean Geometry5.1 Circles . . . . . . . . . . . . . . . . . .5.2 The Circumcircle and the Circumcenter5.3 The Altitudes and the Orthocenter . .5.4 Angle Bisectors . . . . . . . . . . . . .5.5 The Incircle and the Incenter . . . . .5.6 The Euler Line . . . . . . . . . . . . .5.7 The Nine Point Circle . . . . . . . . .5.8 A Coordinate Proof . . . . . . . . . . .5.9 Simson’s Theorem . . . . . . . . . . . .5.10 The Butterfly . . . . . . . . . . . . . .5.11 Morley’s Theorem . . . . . . . . . . . .5.12 Bramagupta and Heron . . . . . . . . .5.13 Napoleon’s Theorem . . . . . . . . . .5.14 The Fermat Point . . . . . . . . . . . .4343444446464747495054545454546 Projective Geometry6.1 Homogeneous coordinates6.2 Projective Transformations6.3 Desargues and Pappus . .6.4 Duality . . . . . . . . . . .6.5 The Projective Line . . . .6.6 Cross Ratio . . . . . . . .6.7 A Geometric Computer . .5555576063646667.7 Inversive Geometry697.1 The complex projective line . . . . . . . . . . . . . . . . . . . 697.2 Feuerbach’s theorem . . . . . . . . . . . . . . . . . . . . . . . 698 Klein’s view of geometry708.1 The elliptic plane . . . . . . . . . . . . . . . . . . . . . . . . . 708.2 The hyperbolic plane . . . . . . . . . . . . . . . . . . . . . . . 708.3 Special relativity . . . . . . . . . . . . . . . . . . . . . . . . . 70A Matrix Notation71B Determinants732

C Sets and Transformations175IntroductionThese are notes to Math 461, a course in plane geometry I sometimes teachat the University of Wisconsin. Students who take this course have completed the calculus sequence and have thus seen a certain amount of analyticgeometry. Many have taken (or take concurrently) the first course in linearalgebra. To make the course accessible to those not familiar with linear algebra, there are three appendices explaining matrix notation, determinants,and the language of sets and transformations.My object is to explain that classical plane geometry is really a subsetof algebra, i.e. every theorem in plane geometry can be formulated as atheorem which says that the solutions of one system of polynomial equationssatisfy another system of polynomial equations. The upside of this is thatthe criteria for the correctness of proofs become clearer and less reliant onpictures.The downside is evident: algebra, especially complicated but elementaryalgebra, is not nearly so beautiful and compelling as geometry. Even theweakest students can appreciate geometric arguments and prove beautifultheorems on their own. For this reason the course also includes syntheticarguments as well. I have not reproduced these here but instead refer tothe excellent texts of Isaacs [4] and Coxeter & Greitzer [3] as needed. It ismy hope that the course as a whole conveys the fact that the foundationsof geometry can be based on algebra, but that it is not always desirable toreplace traditional (synthetic) forms of argument by algebraic arguments.The following quote of a quote which I got from page 31 of [3] should serveas a warning.The following anecdote was related by E.T. Bell [1] page 48.Young Princess Elisabeth had successfully attacked a problem inelementary geometry using coordinates. As Bell states it, “Theproblem is a fine specimen of the sort that are not adapted tothe crude brute force of elementary Cartesian geometry.” Herteacher René Descartes (who invented the coordinate method)said that “he would not undertake to carry out her solution . . .in a month.”3

The reduction of geometry to algebra requires the notion of a transformation group. The transformation group supplies two essential ingredients.First it is used to define the notion of equivalence in the geometry in question.For example, in Euclidean geometry, two triangles are congruent iff there isdistance preserving transformation carrying one to the other and they aresimilar iff there is a similarity transformation carrying one to the other. Secondly, in each kind of geometry there are normal form theorems which can beused to simplify coordinate proofs. For example, in affine geometry every triangle is equivalent to the triangle whose vertices are A0 (0, 0), B0 (1, 0),C0 (0, 1) (see Theorem 3.13) and in Euclidean geometry every triangle iscongruent to the triangle whose vertices are of form A (a, 0), B (b, 0),C (0, c) (see Corollary 4.14).This semester the official text is [3]. In past semesters I have used [4] andmany of the exercises and some of the proofs in these notes have been takenfrom that source.2Some FallaciesPictures sometimes lead to erroneous reasoning, especially if they are notcarefully drawn. The three examples in this chapter illustrate this. I gotthem from [6]. See if you can find the mistakes. Usually the mistake is akind of sign error resulting from the fact that some point is drawn on thewrong side of some line.4

2.1Every Angle is a Right Angle!?DPCREAQBOFigure 1: Every Angle is a Right Angle!?Let ABCD be a square and E be a point with BC BE. We willshow that ABE is a right angle. Take R to be the midpoint of DE, Pto be the midpoint of DC, Q to be the midpoint of AB, and O to be thepoint where the lines P Q and the perpendicular bisector of DE intersect.(See Figure 2.1.) The triangles AQO and BQO are congruent since OQ isthe perpendicular bisector of AB; it follows that AO BO. The trianglesDRO and ERO are congruent since RO is the perpendicular bisector ofDE; it follows that DO EO. Now DA BE as ABCD is a squareand E is a point with BC BE. Hence the triangles OAD and OBEare congruent because the corresponding sides are equal. It follows that ABE OBE ABO OAD BAO BAD.5

2.2Every Triangle is Isosceles!?ARQOBDCFigure 2: An Isosceles Triangle!?Let ABC be a triangle; we will prove that AB AC. Let O be the pointwhere the perpendicular bisector of BC and the angle bisector at A intersect,D be the midpoint of BC, and R and Q be the feet of the perpendicularsfrom O to AB and AC respectively (see Figure 2.2.) The right trianglesODB and ODC are congruent since OD OD and DB DC. HenceOB OC. Also the right triangles AOR and AOQ are congruent since RAO QAO (AO is the angle bisector) and AOR AOQ (theangles of a triangle sum to 180 degrees) and AO is a common side. HenceOR OQ. The right triangles BOR and COQ are congruent since we haveproved OB OC and OR OQ. Hence RB QC. Now AR AQ(as AOR and AOQ are congruent) and RB QC (as BOR and COQ arecongruent) so AB AR RB AQ QC AC as claimed.6

2.3Every Triangle is Isosceles!? -IIABDHHHHDHHDHDHHDHHHHDHDDHXFigure 3: AX bisects BACCIn a triangle ABC, let X be the point at which the angle bisector of theangle at A meets the segment BC. By Exercise 2.2 below we haveXBXC .ABAC(1)Now AXB ACX CAX C 12 A since the angles of a trianglesum to 180 degrees. By the Law of Sines (Exercise 2.1 below) applied totriangle AXB we havesin 12 AXBsin BAX ABsin AXBsin( C 12 A)(2)Similarly AXC ABX BAX B 12 A sosin 21 AXC .ACsin( B 12 A)(3)From (1-3) we get sin( C 12 A) sin( B 21 A) so C 12 A B 1 A so C B so AB AC so ABC is isosceles.2Exercise 2.1. The law of sines asserts that for any triangle ABC we havesin Bsin Csin A BCCAABProve this by computing the area of ABC in three ways. Does the argumentwork for an obtuse triangle? What is the sign of the sine?Exercise 2.2. Prove (1). Hint: Compute the ratio of the area of ABX tothe area of ACX in two different ways.7

33.1Affine GeometryLines3.1. Throughout R denotes the set of real numbers and R2 denotes the set ofpairs of real numbers. Thus a point of P R2 is an ordered pair P (x, y)of real numbers.Definition 3.2. A line in R2 is a set of form {(x, y) R2 : ax by c 0}where a, b, c R and either a 6 0 or b 6 0 (or both). Three or more pointsare called collinear iff there is a line which contains them all. Three ormore lines are called concurrent iff they have a common point. Two linesare said to be parallel iff they do not intersect.3.3. The two most fundamental axioms of plane geometry areAxiom (1) Two (distinct nonparallel) lines intersect in a (unique) point.Axiom (2) Two (distinct) points determine a line.Axiom (1) says that two equationsa1 x b1 y c1 0,a 2 x b2 y c2 0for lines have a unique common solution (the usual case), no common solution(this means that the lines are parallel), or else define the same line (which iscase if and only if the equations are nonzero multiples of one another). Thelatter two cases are characterized by the condition a1 b2 a2 b1 0 and in thefirst case the intersection point isx c1 b2 c2 b1,a1 b2 a2 b1y a 1 c2 a 2 c1.a1 b2 a2 b1Axiom (2) says that for any two distinct points P1 (x1 , y1 ) and P2 (x2 , y2 )there is a unique line {(x, y) : ax by c 0}containing both. Remark 3.5 below gives a formula for this line.8

Theorem 3.4. (I) Three points Pi x1 det x2x3 (xi , yi ) are collinear if and only if y1 1y2 1 0.y3 1(II) Three distinct lines i {(x, y) : ai x bi y ci 0} are concurrent orparallel if and only if a1 b1 c1det a2 b2 c2 0.a3 b3 c3Proof. A determinantHence x1 det x2x3is unchanged if one row is subtracted from another. y1 1x1 x3 y1 y3 0y2 1 det x2 x3 y2 y3 0 .y3 1x3y31Evaluating the determinant on the right gives x1 y 1 1det x2 y2 1 (x1 x3 )(y2 y3 ) (x2 x3 )(y1 y3 ).x3 y 3 1Dividing by (x1 x3 )(x2 x3 ) shows that the determinant vanishes if andonly ify1 y3y2 y3 .x1 x3x2 x3This last equation asserts that the slope of the line P1 P3 equals the slope ofthe line P2 P3 . Since P3 lies on both lines, this occurs if and only if the linesare the same, i.e. if and only if the points P1 , P2 , P3 are collinear.The above proof assumes that x1 , x2 6 x3 ; a special argument is requiredin the contrary case. We give another proof which handles both cases at thesame time. The matrix equation x1 y1 1a0 x2 y2 1 b 0 x3 y3 1c0says that the points Pi lie on the line ax by c 0. Any nonzero solution(a, b, c) of this equation must have either a 6 0 or b 6 0 or both. Hence the9

three points Pi are collinear if and only if this matrix equation (viewed asa system of three homogeneous linear equations in three unknowns (a, b, c))has a nonzero solution. Part (I) thus follows from the followingKey Fact. A homogeneous system of n linear equations in nunknowns has a nonzero solution if and only if the matrix ofcoefficients has determinant zero.Part (II) is similar, but a1 a2a3there are several cases. The matrix equation b1 c1x00 b2 c2y0 0 (1)b3 c310says that the point (x0 , y0 ) lies on each of the three lines ai x bi y ci 0.The three lines are parallel (and not vertical) if and only if they have thesame slope, i.e. if and only if a1 /b1 a2 /b2 a3 /b3 . This happens ifand only if the matrix equation a1 b1 c110 a2 b2 c2 m 0 (2)a3 b3 c300has a solution m. The lines are vertical (and hence parallel) if and only ifb1 b2 b3 0. This happens if and only if the matrix equation 00a1 b1 c1 a2 b2 c2 1 0 .(3)00a3 b3 c3holds. This (and the above Key Fact) proves “only if”. For “if” assume thatthe matrix equation ua1 b1 c10 a2 b2 c2 v 0 a3 b3 c3w0has a nonzero solution (u, v, w). If w 6 0, then x0 u/w, y0 v/wsatisfies (1). If w 0 and u 6 0, then m v/u satisfies (2). If w u 0,then v 6 0 so (1) holds.10

t 2t 0t 1(((((( t 12( ((((((( ((((( (t 1(((((((((((( P1P0Figure 4: P tP1 (1 t)P0Remark 3.5. The point P (x, y) lies on the line joining the distinctpoints P1 (x1 , y1 ) and P2 (x2 , y2 ) if and only if the points P1 ,, P2 , P arecollinear. Thus Theorem 3.4 implies that an equation for this line is x1 y1 1det x2 y2 1 0.x y 1It has form ax by c 0 wherea y1 y2 ,b x2 x1 ,c x1 y2 x2 y1 .The points P1 and P2 satisfy this equation since a determinant vanishes iftwo of its rows are the same.Theorem 3.6. The line connecting the two distinct points P0 (x0 , y0 ) andP1 (x1 , y1 ) is given by {tP1 (1 t)P0 : t R},i.e. a point P (x, y) lies on if and only ifx tx1 (1 t)x0 ,y ty1 (1 t)y0for some t R. (See Figure 4.)Proof. These are the parametric equations for the line as taught in Math 222.The formula x0 y 0 1x0 y 0 1x0 y0 1det x1 y1 1 t det x1 y1 1 (1 t) det x1 y1 1 0.x y 1x1 y 1 1x0 y0 111

shows that any point P of form P tP1 (1 t)P0 lies on the line. Converselyin P lies on the line, choose t to satisfy one of the two parametric equationsand then the equation for the line in the form ax by c 0 shows that theother parametric equation holds as well.Definition 3.7. The line segment connecting points P0 and P1 is the set[P0 , P1 ] {tP1 (1 t)P0 : 0 t 1}.We say that a point P on the line joining P0 and P1 lies between P0 and P1iff it lies in the segment [P0 , P1 ]. We call P0 and P1 the end points of theline segment. The ray emanating from P0 in the direction of P1 is the set{tP1 (1 t)P0 : t 0}. We call P0 the initial point of the ray.Definition 3.8. (Some general terminology.) An ordered sequence(P1 , P2 , . . . , Pn )of n distinct points no three of which are collinear is called a polygon or ann-gon. The points are called the vertices of the polygon. Two consecutivevertices in the list are said to be adjacent and also the vertices Pn and P1 areadjacent. The lines and line segments joining adjacent vertices are called thesides of the polygon and the other lines and line segments joining verticesare called diagonals. The term extended side is employed if we want toemphasize that the line and not the line segment is intended. for emphasis.but the adjective is often omitted. A 3-gon is also called a triangle, a 4-gonis also called a quadrangle, a 5-gon is also called a pentagon, a 6-gon isalso called a hexagon, etc. For triangles the two notations (A, B, C) and4ABC are synonymous; the latter is more common.3.2Affine TransformationsDefinition 3.9. A transformation T : R2 R2 is called affine iff it has theformx0 ax by p(x0 , y 0 ) T (x, y) 0.y cx dy qwhere ad bc 6 0.12

3.10. It is convenient to use matrix notation to deal with affine transformations. To facilitate this we will we not distinguish between points and columnvectors, i.e. we write both xP (x, y)andP .yThen, in matrix notation, an affine transformation takes the formT (P ) M P V,where T (P ) x0y0 ,M a bc d ,P xy ,V pq ,and det(M ) 6 0.Theorem 3.11. The set of all affine transformations is a group, i.e.(1) the identity transformation I(P ) P is affine,(2) the composition T1 T2 of two affine transformations T1 and T2 is affine,and(3) the inverse T 1 of an affine transformation T is affine.Proof. The identity transformation I has the requisite form with a d 1and b c p q 0. Suppose T1 and T2 are affine, say T1 (P ) M1 P V1 and T2 (P ) M2 P V2 . Then (T1 T2 )(P ) T1 (T2 (P )) M1 (M2 P V2 ) V1 M P V where M M1 M2 and V M1 V V2 . Sincedet(M1 M2 ) det(M1 ) det(M2 ) 6 0, this shows that (T1 T2 ) is affine. Tocompute the inverse transformation we solve the equation P 0 T (P ) for P ;we get P 0 M P V M P P 0 V P M 1 P 0 M 1 V . Inother words,T 1 (P 0 ) M 0 P 0 V 0where1M ad bc0 d b ca 1V ad bc0, dp bq cp aq .This shows that the inverse T 1 of the affine transformation T is itself anaffine transformation.13

Theorem 3.12. An affine transformation maps lines onto lines, line segments onto line segments, and rays onto rays.Proof. Let be a line and T be an affine transformation; the theorem assertsthat the imageT ( ) {T (P ) : P }is again a line. Fix two distinct points P0 , P1 . Let M and V be thematrices which define T , i.e. T (P ) M P V . Choose P . ThenP (1 t)P0 tP1 for some t R. Hence T (P ) M (1 t)P0 tP1 V (1 t) M P0 V t M P1 V (1 t)T (P0 ) tT (P1 )which shows that T (P ) lies on the line 0 connecting T (P0 ) and T (P1 ). Thesame argument (reading T 1 for T ) shows that if P 0 0 then T 1 (P 0 ) .Hence T ( ) 0 as claimed. Reading 0 t 1 for t R proves the theoremfor line segments. Reading t 0 for t R proves the theorem for rays.Theorem 3.13. For any two triangles 4ABC and 4A0 B 0 C 0 there is aunique affine transformation T such that T (4ABC) 4A0 B 0 C 0 , i.e. T (A) A0 , T (B) B 0 , and T (C) C 0 .Proof. Let A (a1 , a2 ), B (b1 , b2 ), C (c1 , c2 ). Define T0 by T0 (x, y) (x0 , y 0 ) where 0 c1xa1 c1 b1 c1x . c2y0a2 c2 b2 c2yThen T0 (A0 ) A, T0 (B0 ) B, T0 (C0 ) C where A0 (1, 0), B0 (0, 1),C0 (0, 0). As in the proof of Theorem 3.4 we have a1 a2 1a cb c1111det b1 b1 1 deta2 c2 b2 c2c1 c2 1and this is nonzero since A, B, C are not collinear. Hence T0 is an affinetransformation. Similarly there is an affine transformation T1 such thatT1 (A0 ) A, T1 (B0 ) B 0 , T1 (C0 ) C 0 By Theorem 3.11 T : T1 T0 114

is an affine transformation. It satisfies the conclusion of the theorem. Forexample, T (A) T2 (T1 1 (A)) T2 (A0 ) A0 .To prove uniqueness let T 0 be another affine transformation such thatT 0 (A) A0 , T 0 (B) B 0 , and T 0 (C) C 0 . Let I T2 1 T 0 T1 . ByTheorem 3.11 I is affine so I(P ) M P V for some 2 2 matrix M and2 1 matrix V . Also I(C0 ) C0 , I(A0 ) A0 , and I(B0 ) B0 . FromV M C0 V I(C0 ) C0 0 it follows that V is the zero matrix andfrom M A0 A0 and M B0 B0 it follows that M is the identity matrix.Hence I is the identity transformation. ThusT T2 T1 1 T2 I T1 1 T2 T2 1 T 0 T1 T1 1 T 0as required.Remark 3.14. Often it is possible to find a coordinate proof of a theoremwhich is both straight forward and uncomplicated by using affine transformations. For example, imagine a theorem involving five points A, B, C, D,and three lines a, b, c, and suppose the hypothesis includes the condition thatA, B, C are the vertices of a triangle, i.e. they are not collinear. We can theprove the theorem by arguing as follows: Choose an affine transformation Twith T (A0 ) A, T (B0 ) B, T (C0 ) C. Let D0 T 1 (D), a0 T 1 (a),b0 T 1 (b), c0 T 1 (c). Prove the theorem for A0 , B0 , C0 , D0 , a0 , b0 ,c0 . Check that the hypotheses and conclusion are preserved by affine transformations. This will be true if the hypotheses and conclusion involve onlyassertions about lines (see Theorem 3.12, ratios of collinear distances (seeTheorem 3.25), and ratios of areas (see Theorem 3.32). For example, if thelines a and b are parallel, then so are a0 and b0 . We can now conclude thatthe theorem holds for A, B, C, D, a, b, c. We will often signal this kind ofproof by saying Choose affine coordinates (x, y) so that . . . The followingtheorem illustrates this technique.Theorem 3.15 (Parallel Pappus’ Theorem). Assume that the three pointsA, B, C are collinear and that the three points A0 , B,0 C 0 are collinear. Letthe lines joining them in pairs intersect as follows:X BC 0 B 0 C,Y CA0 C 0 A,Z AB 0 A0 B.(See Figure 5.) If the lines ABC and A0 B 0 C 0 are parallel, then the pointsX,Y , Z are collinear. (This theorem is a special case of Theorem 6.19 below.)15

A0 HZHB0 ZC0 ZZH ZHZ H Z H Z Z HZ Z HHZ Z Z ZHH Z ZHH ZZ Z HH Z YHHZ X ZZZ HH ZZH ZZH H ZZ ABCFigure 5: Parallel Pappus’ TheoremProof. Choose affine coordinates (x, y) so that the line ABC has equationy 0 and the line A0 B 0 C 0 has equation y 1. Then A (a, 0), B (b, 0),C (c, 0), A0 (a0 , 1), B 0 (b0 , 1), C 0 (c0 , 1). The equation of the lineAB 0 isa 0 10 b0 1 1 x (b0 a)y ax y 1and similarly the equation of the line A0 B is x (b a0 )y b 0. To findthe intersection point Z (z1 , z2 ) we solve these two equations. The resultisaa0 bb0a b,z .z2 1a b a0 b0a b a0 b0Now calculate the coordinates of X (x1 , x2 ) and Y (y1 , y2 ) by cyclicallypermuting the symbols and then use Theorem 3.4. (Note that the columnsof the resulting matrix sum to zero.)Exercise 3.16. Do the calculations required to complete the proof of Theorem 3.15.Theorem 3.17 (Parallel Desargues Theorem). Let the corresponding sidesof two triangles 4ABC and 4A0 B 0 C 0 intersect inX BC B 0 C 0 ,Y CA C 0 A0 ,Z AB A0 B 0 .(See Figure 6.) If the lines AA0 , BB 0 , CC 0 are parallel, then the points X,Y ,Z are collinear. (This theorem is a special case of Theorem 6.16 below.)16

B0 A A A AA0 A C0Y AA AA X C Z PP PP PP PP A PP P PPPP BFigure 6: Parallel Desargues TheoremProof. Choose coordinates (x, y) so that the lines AA0 , BB 0 , CC 0 are thevertical lines x a, x b, x c. Then A (a, p), A0 (a, p0 ), B (b, q),B 0 (b, q 0 ), C (c, r), C 0 (c, r0 ). The line AB has equationa p 1b q 1x y 1 0i.e. (p q)x (b a)y aq bp 0. Similarly, the equation of line A0 B 0 is(p0 q 0 )x (b a)y aq 0 bp0 0. The two lines intersect in the solutionof the matrix equation xaq bpq p a b yaq 0 bp0q 0 p0 a bso the intersection is Z (z1 , z2 ) wherez1 bp aq aq 0 bp0,p q p0 q 0z2 pq 0 p0 qp q p0 q 0x2 qr0 q 0 rq r q 0 r0y2 rp0 r0 pr p r 0 p0Similarly X (x1 , x2 ) wherex1 cq br br0 cq 0,q r q 0 r0and Y (y1 , y2 ) wherey1 ar cp cp0 ar0,r p r 0 p017

At this point we could complete the proof using Theorem 3.4 (see Remark 3.18 below), but here is a trick which finishes the proof more easily.The proof uses the following two assertions:(I) An affine transformation of formT (x, y) (x0 x, y0 wx y)transforms each vertical line x k to another vertical line.(II) Given any line {(x, y) : y mx y0 } there is an affine transformation T as in part (I) such that T ( ) is the x-axis.Using these two facts we may suppose w.l.o.g. that the line XY is the xaxes, i.e. that x2 y2 0. From the above formulas it follows that qr0 q 0 rand rp0 r0 p. Multiplying these two equations and dividing by rr0 givesqp0 q 0 p, i.e. z2 0. Hence Z lies on the x-axis as well, i.e. the points X,Y , Z are collinear as required.Remark 3.18. To show that X, Y , Z in Theorem 3.17 are collinear we couldshow that the determinantx1 x2 1cq br br0 cq 0qr0 q 0 rq r q 0 r01y1 y2 1 ar cp cp0 ar0rp0 r0 pr p r0 p0m0000z1 z2 1bp aq aq bppq p qp q p0 q 0vanishes. Here m (q r q 0 r0 )(r p r0 p0 )(p q p0 q 0 ). Athree by three determinant has six terms each of which has three factors:one from the first column, one from the second, and one from the third. Inthe case at hand the first and third factors have four terms each and thesecond has two. Thus the fully expanded determinant has 6 4 2 4 192terms. I evaluated it using a computer program (Maple) which does symboliccalculation and all the terms cancel leaving zero. If you believe in computers,this is an alternative proof.Exercise 3.19. Complete the proof of Theorem 3.17 by proving (I) and (II)in the proof.Exercise 3.20. The proof of Theorem 3.17 assumes that m 6 0, i.e. thatthe three numbers p p0 , q q 0 , r r0 are distinct. What if this is false?Exercise 3.21. The last step in the proof of Theorem 3.17 assumes thatrr0 6 0. What if this is false? Hint: If rr0 0 then either r 0 or r0 0.There are three cases: r r0 0, r 6 r0 0, r0 6 r 0, and the last twoare treated the same way.18

3.3Directed DistanceTheorem 3.22. Let P (1 t)P0 tP1 and Q (1 s)P0 sP1 be twopoints on the line P0 P1 . Then the distance1 P Q between P and Q is givenbyP Q s t P0 P1 Proof. P Q (s t) P0 P1 .Definition 3.23. The directed distance (P Q) from P to Q in the directionfrom P0 to P1 is defined by(P Q) (s t) P0 P1 .3.24. Distances are always nonnegative. However, the directed distance canbe negative. For example, this is the case if the points appear on the linein the order P0 , P1 , Q, P , i.e. if P1 is between P0 and Q and Q is betweenP1 and P . (See Definition 3.7.) Interchanging P0 and P1 reverses the signof the directed distance and hence leaves a ratio of directed distances unchanged. Most affine transformations do not preserve distances; those whichdo preserve distance are called Euclidean transformations and will be studied in Section 4. However, affine transformations preserve ratios of collineardistances. In fact,Theorem 3.25. Affine transformations preserve ratios of collinear directeddistances.Proof. As in the proof of Theorem 3.12, T (P ) (1 t)T (P0 ) tT (P1 ). HenceT (P ) T (Q) (s t) T (P0 ) T (P1 ) so(P 0 Q0 )(P Q) s t 0 0(P0 P1 )(P0 P1 )for P00 T (P0 ), P10 T (P1 ), P 0 T (P ), Q0 T (Q).Corollary 3.26. Affine transformations preserve midpoints of segments.Proof. The midpoint of the segment [A, B] is the unique point M on theline AB which is equidistant from A and B. It is given byM 12 (A B)(Read A P0 , B P1 , M P and t 1/2 in the parametric equationP (1 t)P0 tP for a line.)1See Definition 4.5 below.19

3.4Points and VectorsDefinition 3.27. The difference W P1 P0 between two points P0 andP1 is called the vector from P0 to P1 .Remark 3.28. An affine transformation T (P ) M P where V 0 is calleda linear transformation.2 These are studied in the first course in linearalgebra. An affine transformation T is a linear transformation if and only ifit fixes the origin , i.e. if and only if T (0) 0. When points undergo an affinetransformation, the corresponding vectors undergo a linear transformation.This means the following. If T (P ) M P V is an affine transformation,and P00 T (P0 ), P10 T (P1 ) are the images of points P0 , P1 under T , thenthe vectors W P1 P0 and W 0 P10 P00 are related by the formulaW 0 M W.Contrast this with the formulaP 0 MP Vfor P 0 T (P ). The set of all linear transformations form a group: the factthat the composition of two linear transformations is again linear follows fromthe associative law for matrix multiplication, i.e. M2 (M1 W ) (M2 M1 )W .Remark 3.29. An affine transformation T (P ) P V where M is theidentity matrix is called a translation. The set of all translations formsa group: the identity transformation is the translation with V 0, thecomposition of two translations T1 (P ) P V1 and T2 (P ) P V2 isT2 T1 (P ) P (V1 V2 )and the inverse transformation of the translation T (P ) P V isT 1 (P ) P V.3.5AreaIn Math 222 you learn how to compute the area of a parallelogram usingdeterminants (cross products). We’ll take this as the definition of area. Thecalculations in this section are easy if you are familiar with matrix algebra.The neophyte can skip the proofs in this section.2The reader is cautioned that some authors use the term linear transformation for whatwe call affine transformation.20

Definition 3.30. The oriented area of a triangle 4ABC is half the determinant(ABC) : 21 det[B A C A]of the 2 2 matrix whose columns are the edge vectors B A and C Afrom A. Thus if A (a1 , a2 ), B (b1 , b2 ), C (c1 , c2 ), then b1 a1 c1 a11(ABC) 2 detb2 a2 c2 a2The area is the absolute value of the oriented area.Remark 3.31. The oriented area of 4ABC is positive (and hence equal tothe area) when the points A, B, C occur in counter clockwise order.Theorem 3.32. An affine transformation preserves ratios of oriented areas,i.e. it changes all oriented areas by the same factor.Proof. Suppose that T (P ) M P V . Then T (B) T (A) M (B A) andT (C) T (A) M (C A). Hence we have an equality of 2 2 matrices T (B) T (A) T (C) T (A) M B A C A .Taking the determinant of both sides and using the fact that the determinantof a product is the product of the determinants shows that the oriented areaof T (4ABC) is the determinant of M times the oriented area of 4ABC.Theorem 3.33. The oriented areas of the four triangles obtained by deletinga vertex of a quadrangle satisfy(ABD) (DBC) (ABC) (CDA).Proof. (See Figure 7.) The determinant is linear in its columns, reverses signif the columns are interchanged, and vanishes if two of the columns are thesame. From this we see that(ABC) det[A

8 Klein's view of geometry 70 . 1 Introduction These are notes to Math 461, a course in plane geometry I sometimes teach at the University of Wisconsin. Students who take this course have com-pleted the calculus sequence and have thus seen a certain amount of analytic geometry. Many have taken (or take concurrently) the rst course in linear