M 460 Supplement on nalytic Geometry Donu rapura In the 1600 s Descartes introduced cartesian coordinates which changed the way we now do geometry. This also paved for subsequent developments such as calculus. Here we revisit some parts of Euclidean geometry from this perspective. Notation: To avoid confusion, we label theorems by letters here. Numbered theorems refer to theorems in version 5 of McClure s notes. Given a pair of points and B, B may refer the line through these points, the line segment B, or the length of this line segment depending on the context which will be made clear. 1 Cartesian Coordinates We choose a point on the plane O called the origin, and draw two perpendicular lines call the x-axis and the y-axis. t the risk of sounding pedantic, we should also choose directions along these axes which tells us whether we are to the left or right of O along the x-axis, or above or below O along the y-axis. Given a point on the plane, we can construct a line through parallel to the y-axis. This line will meet the x-axis at a point, say P. We can measure the distance OP. The x-coordinate a 1 of is +OP if P is to the right of O, and OP if P is the left of O. We also construct a line Q parallel to the x-axis with Q on y-axis. The y-coordinate a 2 of is defined similarly as +OQ if Q is above O and OP otherwise. The quadrilateral P OQ is automatically a rectangle by Theorem. parallelogram with one right angle is a rectangle. Proof. You should check this yourself! (It s too easy to be a homework problem.) 1
y axis Q a 2 O a 1 P x axis The point is determined by the pair (a 1, a 2 ). Conversely, any pair of real numbers corresponds to a point in the plane by reversing this process. To summarize: Theorem B. There is a one to one correspondence between points on the plane and the set of pairs of real numbers denoted by R 2. We will simply regard a point as equal to the pair (a 1, a 2 ) of its coordinates. Note that O = (0, 0). The set of points where a 1 0 and a 2 0 is called the first quadrant. The remaining quadrants are where some or all these numbers are negative. Theorem C. Given two points = (a 1, a 2 ) and B = (b 1, b 2 ). 1. If a 1 = b 1, the distance from to B is b 2 a 2. 2. If a 2 = b 2, the distance from to B is b 1 a 1. Proof. We treat case 2. The line l through perpendicular to the y-axis contains B. By swicthing and B if necessary, we can assume that b 1 a 1, which means that B is to the right of on l. There are 3 cases: (a) and B are to the right of O, that is 0 a 1 b 1. (b) O is in between and B, that is a 1 0 b 1. (c) and B are to the left of O, that is a 1 b 1 0. In case (a), we have b 1 = BO = O + B = a 1 + B. Therefore B = b 1 a 1 = b 1 a 1 In case (b), B = O + BO = a 1 + b 1 = b 1 a 1 2
Finally, in case (c), b 1 = BO = O + B = a 1 + B. Therefore B = a 1 b 1 = b 1 a 1 a 1 b a 1 1 O B a 1 b 1 O B a b 1 1 b 1 B O Theorem D. The distance between = (a 1, a 2 ) and B = (b 1, b 2 ) is (b 1 a 1 ) 2 + (b 2 a 2 ) 2. Proof. Draw the right triangle BC, where C = (b 1, a 2 ). B By the previous theorem, C = b 1 a 1 and BC = b 2 a 2. Then by Pythagoreas theorem (theorem 9), so that B 2 = C 2 + BC 2 = (b 1 a 1 ) 2 + (b 2 a 2 ) 2 B = (b 1 a 1 ) 2 + (b 2 a 2 ) 2 C 2 Vector Sum We construct a geometric operation on points called the vector sum. (Incidentally, notions involving vectors evolved in the 19th century long after Descartes.) Given distinct points and B, we define + B as the vertex C of the parallelogram OCB. It is clear that this does not depend on the order, so that + B = B +. 3
C=+B B O Theorem E. If = (a 1, a 2 ) and B = (b 1, b 2 ) are distinct, then +B = (a 1 +b 1, a 2 +b 2 ). In other words, the coordinates of + B are the sum of the coordinates of and B. Proof. We treat the case where and B both lie in the first quadrant. Other cases will be considered in the homework. Let C = + B = (c 1, c 2 ), we have to show that c 1 = a 1 + b 1. By interchanging and B if necessary, we can assume that a 1 b 1. Consider the picture below. 4
H C B G O D E F x axis We assume that D, BE and CF are parallel to the y-axis. Therefore we have a 1 = OD, b 1 = OE and c 1 = OF. We also let G be the point on CF such that BG is parallel to the x-axis. Therefore BEF G is a parallelogram because the sides are parallel. It also follows that BGC = 90. Since D and the x-axis are perpendicular, we have that ( ) BGC = OD Since CBO is a parallelogram, theorem 11 implies that O = BC By BF5, we see that OD = OHF and also that OHF = BCG. Therefore ( ) OD = BCG 5
Since the sum of the angles of OD and BCG are both 180, we deduce from this together with (*) and (**) that OD = CBG. Therefore we may conclude from S (BF3) that OD and BCG are congruent. Thus BG = OD = a 1. Since BEF G is a parallelogram (in fact a rectangle), we deduce that EF = BG = a 1. Consequently The proof that c 2 = a 2 + b 2 is similar. c 1 = OF = OE + EF = b 1 + a 1 We can extend the sum operation to allow = B, by defining + = (2a 1, 2a 2 ) We also write this as 2. More generally, for any real number let t = (ta 1, ta 2 ) Theorem F. The distance from O to t is t O. If t 0, then t lies on the ray O. If t < 0, t lies on the line O on the opposite side from (i.e. not on the ray O). Proof. This will be left as homework. We define subtraction by B = + ( 1)B = (a 1 b 1, a 2 b 2 ) The geometric interpretation is explained by the picture below. B B B 6
3 Midpoints Theorem G. If = (a 1, a 2 ) and B = (b 1, b 2 ) are two distinct points, then M = 1 2 + 1 ( 2 B = a1 + b 1, a ) 2 + b 2 2 2 is the midpoint of B. Proof. We treat the case a 1 b 1 and a 2 b 2. Let M = (m 1, m 2 ) denote the midpoint of B and let C = (b 1, a 2 ), D = (m 1, a 2 ) and E = (b 1, m 2 ). B M E D Since MD and BC are both perpendicular to the x-axis, the x-axis forms a transversal such that the corresponding angles are the same. Therefore M D and BC are parallel. Therefore by theorem 17 of McClure s notes, D is the midpoint of C. Therefore m 1 a 1 = D = 1 2 (b 1 a 1 ) C and so m 1 = 1 2 (b 1 + a 1 ) Similarly, E is a midpoint of BC. Therefore m 2 a 2 = CE = 1 2 (b 2 a 2 ) and so m 2 = 1 2 (b 2 + a 2 ) Theorem H. If = (a 1, a 2 ), B = (b 1, b 2 ) and C = (c 1, c 2 ) are three distinct points, then C is on the line segment B if and only if C = (1 t) + tb, for some 0 t 1. Proof. We have to show that c 1 = (1 t)a 1 + tb 1 and c 2 = (1 t)a 2 + tb 2 for some 0 t 1. ssume for simplicity that a 1 b 1 and a 2 b 2. Let D = (b 1, a 2 ) and let E = (c 1, a 2 ) 7
B C E D Let t = E D 0 Since E is a part of D, E D so that t 1. By theorem D, E = c 1 a 1, CE = c 2 a 2 D = b 1 a 1, BD = b 2 a 2 The triangles BD and CE share the angle. We have CE = BD = 90. Therefore CE = 180 CE = 180 BD = BD It follows that BD and CE are similar, since they have equal angles. Therefore by BF4. By substitution, we obtain CE BD = E D = t c 1 a 1 b 1 a 1 = t By algebra, c 2 a 2 b 2 a 2 = t c 1 = a 1 + t(b 1 a 1 ) = (1 t)a 1 + tb 1 c 2 = a 2 + t(b 2 a 2 ) = (1 t)a 2 + tb 2 8
4 Centroid Recall that a median of triangle is a line joining a midpoint to the opposite vertex. Theorem I. If = (a 1, a 2 ), B = (b 1, b 2 ) and C = (c 1, c 2 ) are three distinct points, then G = 1 3 + 1 3 B + 1 3 C lies on each of the medians of BC. In particular, the medians are concurrent Note that this gives a new proof of theorem 29. The formula shows that the centroid G coincides with the center of mass that you have learned about in other courses. Proof. Let M, N and P be the midpoints of BC, C and B respectively. By theorem G, M = 1 (B + C) 2 By algebra, G = 1 3 + 2 3 M By theorem H, G lies on M. By a similar argument, we can see that G lies on BN and CP. Theorem J. If = (a 1, a 2 ), B = (b 1, b 2 ) and C = (c 1, c 2 ) be the vertices of a triangle, then D lies inside BC if and only if D = r + sb + tc where 0 < r < 1, 0 < s < 1 and 0 < t < 1 and r + s + t = 1. Proof. Homework. Given a quadrilateral BCD, we can define its centroid as 1 4 + 1 4 B + 1 4 C + 1 4 D We will leave it for homework to discover what this means geometrically. 5 Dot Products We start with a converse to Pythagoreas theorem. Theorem K. Given a triangle with sides labeled a, b, c. If c 2 = a 2 + b 2, then the triangle is a right triangle with c its hypotenuse. 9
Proof. Let 1 be the angle opposite c. There are three cases 1 < 90, 1 > 90 and 1 = 90. It suffices to prove that the first two cases are impossible if c 2 = a 2 + b 2. We treat only the first in detail. Draw in the altitude from the angle opposite b. Let h be the height. The side b is divided into two nonzero parts b = d + e as pictured. a h c 1 By Pythagoreas theorem d e a 2 = d 2 + h 2 c 2 = e 2 + h 2 Since 1 < 90, we must have d 0. Therefore d 2 > 0. dding h 2 to both sides and using the first equation implies a 2 > h 2 Since b > e, we have dding these two inequalities yields which contradicts c 2 = a 2 + b 2. b 2 > e 2 a 2 + b 2 > c 2 We define the dot product of two points by (a 1, a 2 ) (b 1, b 2 ) = a 1 b 1 + a 2 b 2 Theorem L. Let, B, C, D be points, and t be a number. Then 1. B = B. 2. t( B) = (t) B = (tb) 3. ( + B) (C + D) = C + D + B C + B D. 4. ( B) ( B) equals B 2 (the length of B squared). 10
Proof. The proof of the first three formulas is straight forward algebra. For the last item, we have ( B) ( B) = (a 1 b 1, a 2 b 2 ) (a 1 b 1, a 2 b 2 ) = (a 1 b 1 ) 2 + (a 2 b 2 ) 2 This is B 2 by theorem D. Theorem M. B = 0 if and only if O and BO are perpendicular. Proof. Consider the triangle OB. By the previous theorem, O 2 =, BO 2 = B B and B 2 = ( B) ( B) = B B + B B Therefore ( ) B 2 = O 2 + BO 2 2 B If OB = 90, then Pythagoreas theorem implies B 2 = O 2 + BO 2 When combined with ( ) this forces B = 0. Conversely, if B = 0. Then ( ) implies Therefore, by theorem K, OB = 90. By an similar argument, we get B 2 = O 2 + BO 2 Theorem N. ( C) (B C) = 0 if and only if CB = 90. 6 Circumcenter We can now give a proof of theorem 24 of McClure s notes using analytic geometry. In order to reduce the complexity of the proof, we start with a preliminary and somewhat technical result called a lemma. Lemma 1. Let D, E and X be three distinct points, and let Q be the midpoint of DE. Then XQ is a perpendicular bisector of DE if and only if 2X (E D) = E E D D Proof. By theorem N, XQ is perpendicular to DE if and only if or equivalently if and only if (X Q) (E Q) = 0 ( ) X (E Q) = Q (E Q) 11
Since Q is a midpoint, Q = 1 2 D + 1 2 E by theorem G. Substituting this into the left and right side of (*) gives X (E Q) = 1 X (E D) 2 and Q (E Q) = 1 4 (E + D) (E D) = 1 (E E D D) 4 Substituting these into (*) yields or 1 2 X (E D) = 1 (E E D D) 4 2X (E D) = E E D D Theorem O. The perpendicular bisectors of the sides of a triangle BC are concurrent. Proof. Let M, N, P be the midpoints of B, C and BC respectively. Let X be the intersection of the perpendicular bisectors of B and C. Then by lemma 1, and Subtracting these equations yields which simplifies to 2X (B ) = B B 2X (C ) = C C 2X (C ) 2X (B ) = C C B B + 2X (C B) = C C B B Lemma 1 will then imply that X lies on the perpendicular bisector of BC as required. 12