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Coordinate Geometry
- Tuesday September 6, JWR
- 1 Introduction Contents
- 2 Some Fallacies
- 2.1 Every Angle is a Right Angle!?
- 2.2 Every Triangle is Isosceles!?
- 2.3 Every Triangle is Isosceles!? -II
- 3 Affine Geometry
- 3.1 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 Geometry
- 4.1 Orthogonal Matrices
- 4.2 Euclidean Transformations
- 4.3 Congruence
- 4.4 Similarity Transformations
- 4.5 Rotations
- 4.6 Review
- 4.7 Angles
- 4.8 Addition of Angles
- 5 More Euclidean Geometry
- 5.1 Circles
- 5.2 The Circumcircle and the Circumcenter
- 5.3 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
- 6 Projective Geometry
- 6.1 Homogeneous coordinates
- 6.2 Projective Transformations
- 6.3 Desargues and Pappus
- 6.4 Duality
- 6.5 The Projective Line
- 6.6 Cross Ratio
- 6.7 A Geometric Computer
- 7 Inversive Geometry
- 7.1 The complex projective line
- 7.2 Feuerbach’s theorem
- 8 Klein’s view of geometry
- 8.1 The elliptic plane
- 8.2 The hyperbolic plane
- 8.3 Special relativity
- A Matrix Notation
- B Determinants
The reduction of geometry to algebra requires the notion of a transfor- mation 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 is distance preserving transformation carrying one to the other and they are similar iff there is a similarity transformation carrying one to the other. Sec- ondly, in each kind of geometry there are normal form theorems which can be used to simplify coordinate proofs. For example, in affine geometry every tri- angle is equivalent to the triangle whose vertices are A 0 = (0, 0), B 0 = (1, 0), C 0 = (0, 1) (see Theorem 3.13) and in Euclidean geometry every triangle is congruent 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] and many of the exercises and some of the proofs in these notes have been taken from that source.
2 Some Fallacies
Pictures sometimes lead to erroneous reasoning, especially if they are not carefully drawn. The three examples in this chapter illustrate this. I got them from [6]. See if you can find the mistakes. Usually the mistake is a kind of sign error resulting from the fact that some point is drawn on the wrong side of some line.
2.1 Every Angle is a Right Angle!?
R
P
O
E
D C
A B
Q
Figure 1: Every Angle is a Right Angle!?
Let ABCD be a square and E be a point with BC = BE. We will show that ∠ABE is a right angle. Take R to be the midpoint of DE, P to be the midpoint of DC, Q to be the midpoint of AB, and O to be the point 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 is the perpendicular bisector of AB; it follows that AO = BO. The triangles DRO and ERO are congruent since RO is the perpendicular bisector of DE; it follows that DO = EO. Now DA = BE as ABCD is a square and E is a point with BC = BE. Hence the triangles OAD and OBE are congruent because the corresponding sides are equal. It follows that ∠ABE = ∠OBE − ∠ABO = ∠OAD − ∠BAO = ∠BAD.
2.3 Every Triangle is Isosceles!? -II
HH HH HH HH HH HH HH H
D D D D D D
DD B X C
A
Figure 3: AX bisects ∠BAC
In a triangle ABC, let X be the point at which the angle bisector of the angle at A meets the segment BC. By Exercise 2.2 below we have
XB AB
XC
AC
Now ∠AXB = ∠ACX + ∠CAX = ∠C + 12 ∠A since the angles of a triangle sum to 180 degrees. By the Law of Sines (Exercise 2.1 below) applied to triangle AXB we have
XB AB
sin ∠BAX sin ∠AXB
sin 12 ∠A sin(∠C + 12 ∠A)
Similarly ∠AXC = ∠ABX + ∠BAX = ∠B + 12 ∠A so
XC AC
sin 12 ∠A sin(∠B + 12 ∠A)
From (1-3) we get sin(∠C + 12 ∠A) = sin(∠B + 12 ∠A) so ∠C + 12 ∠A = ∠B + 1 2 ∠A^ so^ ∠C^ =^ ∠B^ so^ AB^ =^ AC^ so^ ABC^ is isosceles.
Exercise 2.1. The law of sines asserts that for any triangle ABC we have
sin ∠A BC
sin ∠B CA
sin ∠C AB
Prove this by computing the area of ABC in three ways. Does the argument work 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 to the area of ACX in two different ways.
3 Affine Geometry
3.1 Lines
3.1. Throughout R denotes the set of real numbers and R^2 denotes the set of pairs of real numbers. Thus a point of P ∈ R^2 is an ordered pair P = (x, y) of real numbers.
Definition 3.2. A line in R^2 is a set of form
` = {(x, y) ∈ R^2 : ax + by + c = 0}
where a, b, c ∈ R and either a 6 = 0 or b 6 = 0 (or both). Three or more points are called collinear iff there is a line ` which contains them all. Three or more lines are called concurrent iff they have a common point. Two lines are said to be parallel iff they do not intersect.
3.3. The two most fundamental axioms of plane geometry are
Axiom (1) Two (distinct nonparallel) lines intersect in a (unique) point.
Axiom (2) Two (distinct) points determine a line.
Axiom (1) says that two equations
a 1 x + b 1 y + c 1 = 0, a 2 x + b 2 y + c 2 = 0
for 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 is case if and only if the equations are nonzero multiples of one another). The latter two cases are characterized by the condition a 1 b 2 − a 2 b 1 = 0 and in the first case the intersection point is
x = −
c 1 b 2 − c 2 b 1 a 1 b 2 − a 2 b 1
, y = −
a 1 c 2 − a 2 c 1 a 1 b 2 − a 2 b 1
Axiom (2) says that for any two distinct points P 1 = (x 1 , y 1 ) and P 2 = (x 2 , y 2 ) there is a unique line
` = {(x, y) : ax + by + c = 0}
containing both. Remark 3.5 below gives a formula for this line.
three points Pi are collinear if and only if this matrix equation (viewed as a system of three homogeneous linear equations in three unknowns (a, b, c)) has a nonzero solution. Part (I) thus follows from the following
Key Fact. A homogeneous system of n linear equations in n unknowns has a nonzero solution if and only if the matrix of coefficients has determinant zero.
Part (II) is similar, but there are several cases. The matrix equation
a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3
x 0 y 0 1
says that the point (x 0 , y 0 ) lies on each of the three lines aix + biy + ci = 0. The three lines are parallel (and not vertical) if and only if they have the same slope, i.e. if and only if −a 1 /b 1 = −a 2 /b 2 = −a 3 /b 3. This happens if and only if the matrix equation
a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3
m 0
has a solution m. The lines are vertical (and hence parallel) if and only if b 1 = b 2 = b 3 = 0. This happens if and only if the matrix equation
a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3
holds. This (and the above Key Fact) proves “only if”. For “if” assume that the matrix equation
a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3
u v w
has a nonzero solution (u, v, w). If w 6 = 0, then x 0 = u/w, y 0 = v/w satisfies (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.
(((((((((((((((((((((((((((((((
(
t = − 1 •
P 0
t = 0
t = (^12)
P 1
t = 1 •
t = 2
Figure 4: P = tP 1 + (1 − t)P 0
Remark 3.5. The point P = (x, y) lies on the line joining the distinct points P 1 = (x 1 , y 1 ) and P 2 = (x 2 , y 2 ) if and only if the points P 1 ,, P 2 , P are collinear. Thus Theorem 3.4 implies that an equation for this line is
det
x 1 y 1 1 x 2 y 2 1 x y 1
It has form ax + by + c = 0 where
a = y 1 − y 2 , b = x 2 − x 1 , c = x 1 y 2 − x 2 y 1.
The points P 1 and P 2 satisfy this equation since a determinant vanishes if two of its rows are the same.
Theorem 3.6. The line connecting the two distinct points P 0 = (x 0 , y 0 ) and P 1 = (x 1 , y 1 ) is given by
` = {tP 1 + (1 − t)P 0 : t ∈ R},
i.e. a point P = (x, y) lies on ` if and only if
x = tx 1 + (1 − t)x 0 , y = ty 1 + (1 − t)y 0
for some t ∈ R. (See Figure 4.)
Proof. These are the parametric equations for the line as taught in Math 222. The formula
det
x 0 y 0 1 x 1 y 1 1 x y 1
(^) = t det
x 0 y 0 1 x 1 y 1 1 x 1 y 1 1
(^) + (1 − t) det
x 0 y 0 1 x 1 y 1 1 x 0 y 0 1
3.10. It is convenient to use matrix notation to deal with affine transforma- tions. To facilitate this we will we not distinguish between points and column vectors, i.e. we write both
P = (x, y) and P =
[
x y
]
Then, in matrix notation, an affine transformation takes the form
T (P ) = M P + V,
where
T (P ) =
[
x′ y′
]
, M =
[
a b c d
]
, P =
[
x y
]
, V =
[
p q
]
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 T 1 ◦ T 2 of two affine transformations T 1 and T 2 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 = 1 and b = c = p = q = 0. Suppose T 1 and T 2 are affine, say T 1 (P ) = M 1 P + V 1 and T 2 (P ) = M 2 P + V 2. Then (T 1 ◦ T 2 )(P ) = T 1 (T 2 (P )) = M 1 (M 2 P + V 2 ) + V 1 = M P + V where M = M 1 M 2 and V = M 1 V + V 2. Since det(M 1 M 2 ) = det(M 1 ) det(M 2 ) 6 = 0, this shows that (T 1 ◦ T 2 ) is affine. To compute the inverse transformation we solve the equation P ′^ = T (P ) for P ; we get P ′^ = M P + V ⇐⇒ M P = P ′^ − V ⇐⇒ P = M −^1 P ′^ + M −^1 V. In other words, T −^1 (P ′) = M ′P ′^ + V ′
where
M ′^ =
ad − bc
[
d −b −c a
]
, V ′^ =
ad − bc
[
dp − bq −cp + aq
]
This shows that the inverse T −^1 of the affine transformation T is itself an affine transformation.
Theorem 3.12. An affine transformation maps lines onto lines, line seg- ments onto line segments, and rays onto rays.
Proof. Let be a line and T be an affine transformation; the theorem asserts that the image T () = {T (P ) : P ∈ `}
is again a line. Fix two distinct points P 0 , P 1 ∈ . Let M and V be the matrices which define T , i.e. T (P ) = M P + V. Choose P ∈. Then P = (1 − t)P 0 + tP 1 for some t ∈ R. Hence
T (P ) = M
(1 − t)P 0 + tP 1
+ V
= (1 − t)
M P 0 + V
M P 1 + V
= (1 − t)T (P 0 ) + tT (P 1 )
which shows that T (P ) lies on the line ′^ connecting T (P 0 ) and T (P 1 ). The same argument (reading T −^1 for T ) shows that if P ′^ ∈′^ then T −^1 (P ′) ∈ . Hence T () = `′^ as claimed. Reading 0 ≤ t ≤ 1 for t ∈ R proves the theorem for line segments. Reading t ≥ 0 for t ∈ R proves the theorem for rays.
Theorem 3.13. For any two triangles 4 ABC and 4 A′B′C′^ there is a unique affine transformation T such that T ( 4 ABC) = 4 A′B′C′, i.e. T (A) = A′, T (B) = B′, and T (C) = C′.
Proof. Let A = (a 1 , a 2 ), B = (b 1 , b 2 ), C = (c 1 , c 2 ). Define T 0 by T 0 (x, y) = (x′, y′) where
[ x′ y′
]
[
a 1 − c 1 b 1 − c 1 a 2 − c 2 b 2 − c 2
] [
x y
]
[
c 1 c 2
]
Then T 0 (A 0 ) = A, T 0 (B 0 ) = B, T 0 (C 0 ) = C where A 0 = (1, 0), B 0 = (0, 1), C 0 = (0, 0). As in the proof of Theorem 3.4 we have
det
a 1 a 2 1 b 1 b 1 1 c 1 c 2 1
(^) = det
[
a 1 − c 1 b 1 − c 1 a 2 − c 2 b 2 − c 2
]
and this is nonzero since A, B, C are not collinear. Hence T 0 is an affine transformation. Similarly there is an affine transformation T 1 such that T 1 (A 0 ) = A, T 1 (B 0 ) = B′, T 1 (C 0 ) = C′^ By Theorem 3.11 T := T 1 ◦ T 0 −^1
A′
B′
C′
A
B
C
Z Z
Z Z
Z Z
Z Z
Z Z
Z Z
Z
ZZ
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
HH
Z Z
Z Z
Z Z
Z Z
Z Z
Z Z
Z
ZZ
Z^ •
Y • X
Figure 5: Parallel Pappus’ Theorem
Proof. Choose affine coordinates (x, y) so that the line ABC has equation y = 0 and the line A′B′C′^ has equation y = 1. Then A = (a, 0), B = (b, 0), C = (c, 0), A′^ = (a′, 1), B′^ = (b′, 1), C′^ = (c′, 1). The equation of the line AB′^ is
0 =
a 0 1 b′^1 x y 1
= −x + (b′^ − a)y + a
and similarly the equation of the line A′B is x + (b − a′)y − b = 0. To find the intersection point Z = (z 1 , z 2 ) we solve these two equations. The result is
z 2 =
a − b a − b + a′^ − b′^
, z 1 =
aa′^ − bb′ a − b + a′^ − b′^
Now calculate the coordinates of X = (x 1 , x 2 ) and Y = (y 1 , y 2 ) by cyclically permuting the symbols and then use Theorem 3.4. (Note that the columns of the resulting matrix sum to zero.)
Exercise 3.16. Do the calculations required to complete the proof of The- orem 3.15.
Theorem 3.17 (Parallel Desargues Theorem). Let the corresponding sides of two triangles 4 ABC and 4 A′B′C′^ intersect in
X = BC ∩ B′C′, Y = CA ∩ C′A′, Z = AB ∩ A′B′.
(See Figure 6.) If the lines AA′, BB′, CC′^ are parallel, then the points X,Y , Z are collinear. (This theorem is a special case of Theorem 6.16 below.)
A
• B
• C
A′
• B
′
C′
PP PPP PPP P
A A A A A A
PPP PPP PPP
A A A A A A
PPP PPP
A A A A A A
AA
Z •
• X
• Y
Figure 6: Parallel Desargues Theorem
Proof. Choose coordinates (x, y) so that the lines AA′, BB′, CC′^ are the vertical lines x = a, x = b, x = c. Then A = (a, p), A′^ = (a, p′), B = (b, q), B′^ = (b, q′), C = (c, r), C′^ = (c, r′). The line AB has equation ∣ ∣ ∣ ∣ ∣ ∣ a p 1 b q 1 x y 1
i.e. (p − q)x + (b − a)y + aq − bp = 0. Similarly, the equation of line A′B′^ is (p′^ − q′)x + (b − a)y + aq′^ − bp′^ = 0. The two lines intersect in the solution of the matrix equation [ q − p a − b q′^ − p′^ a − b
] [
x y
]
[
aq − bp aq′^ − bp′
]
so the intersection is Z = (z 1 , z 2 ) where
z 1 =
bp − aq + aq′^ − bp′ p − q − p′^ + q′^
, z 2 =
pq′^ − p′q p − q − p′^ + q′
Similarly X = (x 1 , x 2 ) where
x 1 =
cq − br + br′^ − cq′ q − r − q′^ + r′^
, x 2 =
qr′^ − q′r q − r − q′^ + r′
and Y = (y 1 , y 2 ) where
y 1 =
ar − cp + cp′^ − ar′ r − p − r′^ + p′^
, y 2 =
rp′^ − r′p r − p − r′^ + p′
3.3 Directed Distance
Theorem 3.22. Let P = (1 − t)P 0 + tP 1 and Q = (1 − s)P 0 + sP 1 be two points on the line P 0 P 1. Then the distance^1 |P Q| between P and Q is given by (^) ∣ ∣P Q
∣ (^) = |s − t|
∣P 0 P 1
Proof. P − Q = (s − t)
P 0 − P 1
Definition 3.23. The directed distance (P Q) from P to Q in the direction from P 0 to P 1 is defined by
(P Q) = (s − t)
∣P 0 P 1
3.24. Distances are always nonnegative. However, the directed distance can be negative. For example, this is the case if the points appear on the line in the order P 0 , P 1 , Q, P , i.e. if P 1 is between P 0 and Q and Q is between P 1 and P. (See Definition 3.7.) Interchanging P 0 and P 1 reverses the sign of the directed distance and hence leaves a ratio of directed distances un- changed. Most affine transformations do not preserve distances; those which do preserve distance are called Euclidean transformations and will be stud- ied in Section 4. However, affine transformations preserve ratios of collinear distances. In fact,
Theorem 3.25. Affine transformations preserve ratios of collinear directed distances.
Proof. As in the proof of Theorem 3.12, T (P ) = (1−t)T (P 0 )+tT (P 1 ). Hence T (P ) − T (Q) = (s − t)
T (P 0 ) − T (P 1 )
so (P ′Q′) (P 0 ′P 1 ′)
= s − t =
(P Q)
(P 0 P 1 )
for P 0 ′ = T (P 0 ), P 1 ′ = T (P 1 ), P ′^ = T (P ), Q′^ = 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 the line AB which is equidistant from A and B. It is given by
M = 12 (A + B)
(Read A = P 0 , B = P 1 , M = P and t = 1/2 in the parametric equation P = (1 − t)P 0 + tP for a line.)
(^1) See Definition 4.5 below.
3.4 Points and Vectors
Definition 3.27. The difference W = P 1 − P 0 between two points P 0 and P 1 is called the vector from P 0 to P 1.
Remark 3.28. An affine transformation T (P ) = M P where V = 0 is called a linear transformation.^2 These are studied in the first course in linear algebra. An affine transformation T is a linear transformation if and only if it fixes the origin , i.e. if and only if T (0) = 0. When points undergo an affine transformation, the corresponding vectors undergo a linear transformation. This means the following. If T (P ) = M P + V is an affine transformation, and P 0 ′ = T (P 0 ), P 1 ′ = T (P 1 ) are the images of points P 0 , P 1 under T , then the vectors W = P 1 − P 0 and W ′^ = P 1 ′ − P 0 ′ are related by the formula
W ′^ = M W.
Contrast this with the formula
P ′^ = M P + V
for P ′^ = T (P ). The set of all linear transformations form a group: the fact that the composition of two linear transformations is again linear follows from the associative law for matrix multiplication, i.e. M 2 (M 1 W ) = (M 2 M 1 )W.
Remark 3.29. An affine transformation T (P ) = P + V where M is the identity matrix is called a translation. The set of all translations forms a group: the identity transformation is the translation with V = 0, the composition of two translations T 1 (P ) = P + V 1 and T 2 (P ) = P + V 2 is
T 2 ◦ T 1 (P ) = P + (V 1 + V 2 )
and the inverse transformation of the translation T (P ) = P + V is
T −^1 (P ) = P − V.
3.5 Area
In Math 222 you learn how to compute the area of a parallelogram using determinants (cross products). We’ll take this as the definition of area. The calculations in this section are easy if you are familiar with matrix algebra. The neophyte can skip the proofs in this section.
(^2) The reader is cautioned that some authors use the term linear transformation for what we call affine transformation.
