3 Analytic Geometry

3.1 The Cartesian Co-ordinate System

Pure Euclidean geometry in the style of Euclid and Hilbert is what we call synthetic: axiomatic, with-out co-ordinates or explicit formulæ for length, area, volume, etc. Nowadays, the practice of ele-mentary geometry is almost entirely analytic: reliant on algebra, co-ordinates, vectors, etc. The majorbreakthrough came courtesy of Rene Descartes (1596–1650) and Pierre de Fermat (1601/16071–1655),whose introduction of an axis, a fixed reference ruler against which objects could be measured usingco-ordinates, allowed them to apply the Islamic invention of algebra to geometry, resulting in moreefficient computations.The new geometry was revolutionary, so much so that Descartes felt the need to justify his argu-ments using synthetic geometry, lest no-one believe his work! This attitude persisted for some time:when Issac Newton published his groundbreaking Principia in 1687, his presentation was largely syn-thetic, even though he had used co-ordinates in his derivations. Synthetic geometry is not without itsbenefits—many results are much cleaner, and analytic geometry presents its own logical difficulties—but, as time has passed, its study has become something of a fringe activity: co-ordinates are simplytoo useful to ignore!

Given that Cartesian geometry is the primary form we learn in grade-school, we merely sketch thefamiliar ideas of co-ordinates and vectors.

• Assume everything necessary about continuity on the real line.

• Perpendicular axes meet at the origin.

• The Cartesian co-ordinates of a point P are measured by project-ing onto the axes: in the picture, P has co-ordinates (1, 2), oftenwritten simply as P = (1, 2).

• Curves are defined using equations. For instance, the Pythagoreandistance function says that the circle of radius 1 centered at theorigin may be described by the equation x2 + y2 = 1.

−2

−1

1

2

3y

−2 −1 1 2 3x

P

In the 1600’s this was not considered a new axiomatic system, but rather a collection of computationaltools built on top of Euclid. We may therefore assume anything from Euclid and mix strategies asappropriate. To see this at work, consider a simple result.

Lemma 3.1. Let O = (0, 0), A = (x, y) and B = (v, w) whereO, A, B are non-collinear. If C = (x + v, y + w), then the quadri-lateral OACB is a parallelogram.

Proof. Calculate distances: |BC| =√

x2 + y2 = |OA|, etc.Now use side-side-side to see that 4OAC ∼= 4CBO. The usualdiscussion of parallel lines/alternate angles from Euclid forcesopposite sides to be parallel.

OA

BC

1There is some argument over Fermat’s birth given that he possibly had a deceased older brother, also named Pierre.

1

Vector Geometry

Vectors come to us courtesy of several mathematicians, most prominently William Rowan Hamilton(1805–1865), Oliver Heaviside (1850–1925) and J. Willard Gibbs (1839–1903). Hamilton had stumbledupon the algebra of quaternions when attempting to extend to three dimensions the contemporary useof complex numbers to describe planar geometry.2 Heaviside and Gibbs independently developedvector calculus. The revolution was rapid; by 1900 vector calculations were dominant in physics.

Definition 3.2. A directed line segment−→AB is a segment together with an orientation.a

The position vector of a point A is the directed line segment−→OA where O is the origin.

A vector is an equivalence class of directed line segments where two segments are equivalent if andonly if they are congruent and oriented in the same direction.

aWe write−→AB for the directed line segment so as to distinguish it from the ray

−→AB of Hilbert’s Euclidean geometry.

A vector has length and direction, but no fixed location. All directed line segments with the samelength and direction represent the same vector. The standard representation of a vector involves placingits tail at the origin: we can then describe the vector by giving the co-ordinates of its head.

Example 3.3. In the picture, the standard representation of a vectorv is shown in blue. The green arrows are other representations of thesame vector. Various common notations include

v = ~v = v =−→OA =

(21

)= 〈2, 1〉 = 2i + j

The usual convenient abuse of terminology is at work: strictly−→OA ∈ v

since v is an equivalence class, but no-one writes this. −1

1

2

3y

−1 1 2 3x

A

O

Addition and Scalar Multiplication are defined algebraically in the familiar manner:

v + w =

(v1v2

)+

(w1w2

)=

(v1 + w1v2 + w2

), λv :=

(λv1λv2

)Vector addition can be visualized by placing representative segments nose-to-tail. In view of Lemma 3.1, the commutativity of addition v + w =w + v is often known as the parallelogram law. Indeed the Lemma maybe rephrased in this language: the parallelogram OACB is spanned by

−→OA =

(xy

)and

−→OB =

(vw

)v + w

v

v

w

w

O

The scalar multiple λv may be viewed as stretching or shrinking v, and reversing its direction whenλ < 0.

2Quaternions are objects of the form a + bi + cj + dk where a, b, c, d ∈ R, i2 = j2 = k2 = −1 and i, j, k multiply as ifusing the cross-product (ij = k = −ji, etc.). Hamilton couldn’t make the three-dimensional part (a = 0) into a suitablealgebra, but a fourth dimension fixed things. Hamilton eventually realized that if he dropped his requirement of having awell-defined multiplication, he could apply his vector approach to the study of geometry in any dimension.

2

We finish with a famous result showing how easy it can be to work in analytic geometry.

Theorem 3.4. The medians of a triangle meet at a point 1/3 of the way along each median.

a

b

13(a + b)

O

A

B

M

G

Proof. Given4OAB, let a =−→OA and b =

−→OB.

If M is the midpoint of AB, then−−→OM = 1

2 (a + b).

Let G lie 23 of the distance along

−−→OM: that is

−→OG =

23· 1

2(a + b) =

13(a + b)

The point 13 of the way along the median through A has position

vector

12−→OB +

13

(−→OA− 1

2−→OB)=

13(a + b)

Similarly, the point 13 of the way along the median through B has

position vector

12−→OA +

13

(−→OB− 1

2−→OA

)=

13(a + b)

All three points have the same position vector, and are therefore the same point G!

Compare this to the argument/exercise using Ceva’s Theorem at the end of our discussion of Eu-clidean Geometry!The proof shows another important aspect of analytic geometry. The standard approach is to placethe origin and orient a figure in a manner which makes calculations simple. This is essentially Eu-clid’s (sketchy) superposition principle, or Hilbert’s congruence, but can be more rigorously groundedin a full discussion of isometries.

Exercises. 3.1.1. Given a quadrilateral ABCD, let W, X, Y, Z be the midpoints of the sides AB, BC,CD and DA respectively. Use vectors to prove that WXYZ is a parallelogram.

3.1.2. (a) Let A = (a1, a2) and B = (b1, b2) be given. Show that any point on the line segmentjoining A and B has co-ordinates ((1− t)a1 + tb1, (1− t)a2 + tb2) where 0 ≤ t ≤ 1.

(b) Show that the midpoint of A and B has co-ordinates( 1

2 (a1 + b1), 12 (a2 + b2)

).

(You’ll probably find it easiest to think in terms of vectors for both parts)

3.1.3. (a) Perform a pure co-ordinate proof (no vectors) of Theorem 3.4.(b) Descartes and Fermat did not, in fact, have a fixed perpendicular second axis! Their

approach was equivalent to choosing a a second axis whose angle to the first was chosento make a problem as easy as possible. Given4OAB, where B = (b, 0), choose the secondaxis to point along

−→OA so that A has co-ordinates (0, a). Now give an even simpler proof

of the centroid theorem.

3

3.2 Angles in Planar Analytic Geometry

We define angle measure a little differently: this time we use radians and extend to any angle.

Definition 3.5. Suppose A, B, C are distinct and draw any circle centered at A. The radian-measure]BAC is the ratio of the arc-length to the radius measured counter-clockwise from

−→AB to

−→AC.

Being ratios, radians are naturally unitless.In analytic geometry it is common to label angles using theirradian measure. In part this is since. . .Angles are congruent if and only if their radian measures areequal or negative each other modulo 2π.In the picture, θ = ]BAC and ]CAB = 2π − θ. This last isoften taken instead to be −θ.

θ

−θ

2π − θ

A

BC

Definition 3.6. Let P = (x, y) lie on a circle of radius r such that thesegment

−→OP makes angle θ radians measured counter-clockwise from

the positive x-axis. The cosine and sine of θ are defined by

x = r cos θ, y = r sin θ

The AAA theorem for similar triangles says we can view these as well-defined functions of radian-measure, not merely angle!

x

yr

θ

P

O

The definitions work for any angles: if, say, π2 < θ < 3π

2 , simply take x < 0. As co-ordinates, it is noproblem for x or y to be negative.Basic relationships should also be obvious from the picture: e.g.

cos(π2 − θ) = sin θ = sin(π − θ)

(= y

r

)Moreover, by drawing an isosceles right-triangle and an equilateral triangle, the most well-knownvalues of the sine and cosine are easily recovered:

θφ

1

cos θ

sin θcos φ

sin φ

π4

π4

1

1

√2

π3

π6

1

√32

cos(π2 − θ) = sin θ sin π

4 = cos π4 = 1√

2sin π

3 = cos π6 =

√3

2

sin(π2 − θ) = cos θ sin π

6 = cos π3 = 1

2

4

Solving Triangles

In analytic geometry, a triangle is described by six data values: three side lengths and three anglemeasures. Euclid’s triangle congruence theorems (SAS, ASA, SSS, SAA) say that knowing three ofthese in a suitable combination should be enough to recover the remainder. The way this is done isusing the cosine and sine rules.

Theorem 3.7 (Cosine Rule). In a trianglea 4ABC, we have c2 = a2 + b2 − 2ab cos γ

aWe follow the convention that a is the length of the side opposite vertex A, which has (interior) angle measure α.

x > 0 z > 0

h

B

γ

AC D

a c

b

Proof. Let the base be b and drop a perpendicular from B to AC.Call the height h and split b = x + z, where x = a cos γ. ApplyingPythagoras’ twice, we have:

a2 = h2 + x2, c2 = h2 + z2

Eliminating the h terms yields

c2 = a2 − x2 + z2 = a2 + (z + x)(z− x)

= a2 + b(x + z− 2x) = a2 + b(b− 2x)

= a2 + b2 − 2bx = a2 + b2 − 2ab cos γ

The argument works for all possible arrangements: just be careful with the signs of x and z!

x < 0z > 0

B

h

γ

ACD

ac

bz < 0

x > 0

h

B

γ

AC D

ac

b

Example 3.8. The SSS congruence corresponds to solving a triangle usingthe cosine rule. For instance, the given triangle has angles satisfying

cos α =62 + 72 − 32

2 · 6 · 7 =1921≈ 25°

cos β =32 + 72 − 62

2 · 3 · 7 =1121≈ 58°

cos γ =32 + 62 − 72

2 · 3 · 6 =−19≈ 96°

We use degree measure for clarity. In modern times computing inversecosines is easy, but historically this required large data tables.

7

3

6α

β

γ

5

Theorem 3.9 (Sine Rule). If d is the diameter of the circumcircle of4ABC, then

sin α

a=

sin β

b=

sin γ

c=

1d

Proof. Draw the circumcircle and construct4BCD with diameter BD.This is right-angled at C by Thales’ Theorem. There are two cases:

1. If A lies on the major arc

)

BC, then A and D share the same arc,whence ]BDC = α and so

sin α = sin]BDC =ad

2. If A lies on the minor arc

)

BC, then the quadrilateral ABDC lieson a circle whence opposite angles A, B are supplementary. Thus

sin α = sin(π − α) = sin]BDC =ad

The two other angle-side combinations are similar, or can be seen sim-ply by permutation.

ad

α

α

A

B

CD

ad

α π − αA

B

C

D

The sine and cosine rules, together with the fact that angles sum to a straight edge, allow one to solveany triangle. For instance, given an angle-side-angle combination (α, c, β):

1. Compute the remaining angle γ = π − α− β;

2. Use the sine rule twice to compute

a =sin α

sin γc, b =

sin β

sin γc c

ab

αβ

γ

Multiple-angle formulæ

A quick appeal to the dot product and vector notation recovers the usual multiple-angle formulæ.

Definition 3.10. The dot product of vectors u = ( u1u2 ) and v = ( v1

v2 ) is u · v = u1v1 + u2v2.

Theorem 3.11. If θ is the angle (either one!) between u and v, then u · v = |u| |v| cos θ.

The proof is an easy exercise following the cosine rule.

Corollary 3.12. If α and β are any angles, then

cos(α± β) = cos α cos β∓ sin α sin β

sin(α± β) = sin α cos β± cos α sin β

6

ab

αβ

Proof. With respect to a circle of radius 1, the vectors making anglesα, β with the positive x-axis are

a =

(cos αsin α

)b =

(cos βsin β

)Simply take dot products and apply Theorem 3.11:

cos α cos β + sin α sin β = a · b = cos(α− β)

since the vectors have length 1 and the angle between them is α− β.While the picture has been drawn as if α ≥ β ≥ 0, the calculation works regardless or α, β, sinceanalytic geometry (and cosine!) is comfortable with negative angles. The remaining rules are anexercise: use the even/oddness of cosine/sine and apply the identity sin α = cos(π

2 − α), etc.

Exercises. 3.2.1. (a) As we did on page 6, describe the process for solving triangles give for theremaining combinations SAS and SAA.

(b) A triangle has an angle of 2π3 radians between sides of lengths

√2 and

√2−√

3. Find thelength of the remaining side, and the angles.(Hint: (

√3− 1)2 = 4− 2

√3)

3.2.2. Prove Theorem 3.11 by applying the cosine rule to the triangle formed by u, v and v − u.Explain why the comment ‘either one!’ is appropriate.

3.2.3. Use a multiple angle formula to find an exact value for cos 7π12 .

3.2.4. Complete the proof of Corollary 3.12 by supplying an argument for each of the remainingthree cases.

3.2.5. Consider the given picture. Find x, the radian measure α and theexact value of cos α.(Hint: you have similar isosceles triangles. . . ) 1

x

x x

1 − xα

3.3 The Complex Plane

Italian mathematicians of the 16th century (Tartaglia, Cardano, Bombelli, etc.) began experimentingwith square-roots of negative numbers, mostly out of curiosity. The application of complex num-bers to geometry starts with Leonhard Euler (1707–1783), who extended the real line of Descartes toplaces where equations such as x2 = −1 had solutions. This produced a two-dimensional picturewith both real and imaginary axes. The set of complex numbers C allows a description of 2D ge-ometry where basic transformations (rotations and reflections) may be expressed algebraically. Thisapproach indeed predated vectors and, in part, provided the inspiration for their development.

We briefly review complex numbers, since you should have encountered them elsewhere.

7

Definition 3.13. Let i be an abstract symbol satisfying the propertyi2 = −1, and let x, y be real numbers. The complex number z = x + iy isthe point with co-ordinates (x, y). In this context, the Cartesian plane isknown as the Argand diagram.a

Addition, scalar multiplication (by real numbers) and complex multiplica-tion follow the natural commutative and distributive laws for the realnumbers, while using i2 = −1 to simplfy.The complex conjugate of z is z = x− iy.The modulus of z is |z| =

√zz =

√x2 + y2.

aIn the language of linear algebra, C is a vector space over R with basis {1, i}.

1 2 3 4

z = 2− 3i

z = 2 + 3i

|z| =√

13i

2i

3i

−i

−2i

−3i

Example 3.14. We compute a simple complex multiplication: make sure this is clear to you!

(2 + 3i)(4 + 5i) = 2 · 4 + 2 · 5i + 3i · 4 + 3i · 5i = 8 + 10i + 12i− 15 = −7 + 22i

The complex numbers are of interest to us because the rules scream geometry! In particular:

• Addition by z translates all points by z.

• Scalar multiplication scales points.

• Conjugation is reflection across the x-axis: (x, y) 7→ (x,−y).

• The modulus is the distance from the origin; or the length of a vector.

The ability of complex multiplication to describe rotations is particularly useful. For example,

iz = i(x + iy) = −y + ix

Is the result of rotating z counter-clockwise π2 radians about the origin. To obtain all rotations and

reflections, we need an alternative description of a complex number.

Lemma 3.15. 1. (Euler’s Formula) For any θ ∈ R, eiθ = cos θ + i sin θ.

2. (Exponential laws) eiθeiφ = ei(θ+φ) and (eiθ)n = einθ for any n ∈ Z.

Evaluating at θ = π yields the famous Euler identity eiπ = −1.

Proof. 1. This requires either power series or elementary differential equations, topics best de-scribed elsewhere.

2. Apply the multiple-angle formulæ (Corollary 3.12):

ei(θ+φ) = cos(θ + φ) + i sin(θ + φ)

= cos θ cos φ− sin θ sin φ + i sin θ cos φ + i cos θ sin φ

= (cos θ + i sin θ)(cos φ + i sin φ)

= eiθeiφ

Performing an induction with θ = φ and combining with eiθe−iθ = 1 gives the final result.

8

Definition 3.16. Let z = x + iy be a non-zero complex number.Writing x = r cos θ and y = r sin θ, we obtain the polar form

z = reiθ = r(cos θ + i sin θ)

Here r = |z| =√

x2 + y2 is the modulus and θ = arg(z) the argu-ment of z; the angle measured counter-clockwise from the positivereal axis to the vector ( x

y ).

1 2

z = 1 +√

3i = 2eiπ3

|z| = 2

arg(z) = π3

i

2i

Interpreting Lemma 3.15 in the language of the polar form, we find:

Corollary 3.17. 1. For any non-zero complex numbers z, w,

|zw| = |z| |w| , arg(zw) ≡ arg(z) + arg(w) (mod 2π)

2. The complex number eiθz is obtained by rotating z counter-clockwise about the origin throughan angle θ radians.

Proof. Simply let w = reiθ and z = seiφ: the first result is immediate from

reiθseiψ = rsei(θ+ψ)

For the second part, take r = 1 and observe that eiθz has the same modulus as z, while its argumenthas increased by θ (modulo 2π).

Example 3.18. Rotate z = 1+√

3i counter-clockwise by 3π4

radians around the origin.

We simply multiply by

e3πi/4 = cos3π

4+ i sin

3π

4=

1√2(−1 + i)

to obtain

e3πi/4z =1√2(−1 + i)(1 +

√3i)

=1√2

(−1−

√3 + i(1−

√3))

Alternatively, we can keep the calculation in polar form:

z = 2eπi/3 =⇒ e3πi/4z = 2e3πi

4 + πi3 = 2e13πi/12

−2 −1 1

z

e3πi

4 z

3π4

i

2i

−i

9

Complex conjugation takes care of reflections. Since z 7→ z reflectsacross the real axis, we can use this together with rotations and trans-lations to describe all reflections.To reflect across the line making angle θ with the positive real axis,we do three things:

1. Rotate by −θ.

2. Reflect across the real axis.

3. Rotate back by θ.

Otherwise said:

Theorem 3.19. To reflect z ∈ C across the line making angle θ withthe positive real axis, we compute

z 7→ eiθ(e−iθz) = e2iθz

θ

1

2

3

Example 3.20. Reflect z = −2 + 3i across the line through the origin and w =√

3 + i.First compute θ = arg(w) = tan−1 1√

3= π

6 . The desired point is then

eiπ/3(−2− 3i) =

(12+

√3

2i

)(−2− 3i) =

(3√

32− 1

)−(√

3 +32

)i

Corollary 3.21. 1. To rotate z by θ about a point w, compute z 7→ w + eiθ(z− w).

2. To reflect z across the line with slope θ through a point w, compute z 7→ w + e2iθ(z− w).

Mathematicians later tried to replicate this algebraic encoding of simple geometric transformations inthree and higher dimensions. The attempt led (via Hamilton’s quaternions) first to vectors and thento linear algebra/matrix calculations and a unified theory of basic transformations in any dimension.

Exercises. 3.3.1. (a) Express each of the following fractions as complex numbers by rationalizingthe denominator (multiplying through by the complex conjugate. . . )

12i

,1 + i1− i

,1

2 + 4i

(b) Prove that C is closed under multiplicative inverses: i.e., ∀z ∈ C \ {0}, prove that 1z ∈ C.

3.3.2. Use complex numbers to compute the result of the following transformations: you can answerin either standard or polar form.

(a) Rotate 3− 5i counter-clockwise around the origin by 3π4 radians.

(b) Reflect 2− i across the joining 1 + i√

3 and the origin.(c) Reflect 1 + i across the line through the origin making angle π

5 radians with the positivereal axis.

(d) Reflect −2 + 3i across the line joining the origin and the point (√

2 +√

2,√

2−√

2).

10

3.3.3. By letting n = 3 in Lemma 3.15, prove that

cos 3θ = 4 cos3θ − 3 cos θ

Find a corresponding trigonometric identity for sin 3θ.

3.4 Birkhoff’s Axiomatic System for Analytic Geometry (non-examinable)

Analytic geometry was originally built as an addition to Euclidean geometry. In 1932, courtesy ofGeorge David Birkhoff, it was axiomatized in its own right.

Background Assume the usual properties/axioms of the real numbers as a complete ordered field.Birkhoff’s system is typical of modern axiomatic systems in that it is built on top of pre-existingsystems (set theory, complete ordered fields, etc.).

Undefined terms Point, line, distance and angle measure. Let the set of points be denoted S , then,

Distance is a function d : S × S → R+0

Angle measure is a function ] : S × S × S → [0, 2π)

Axioms

Euclidean Given two distinct points, there exists a unique line containing them.

Ruler Points on a line ` are in bijective correspondence with the real numbers in such a way that iftA, tB correspond to A, B ∈ `, then |tA − tB| = d(A, B).

Protactor The rays emanating from a point O are in bijective correspondence with the set [0, 2π) sothat if a, b correspond to rays

−→OA,−→OB, then ]AOB ≡ a− b (mod 2π). This correspondence is

continuous in A, B.

SAS similarity 3 If triangles have a pair of angles with equal measure, and the sides adjacent to saidangles are in the same ratio, then the remaining angles have equal measure and the final sidesare in the same ratio.

Definitions As with Hilbert, some of these are required before later axioms make sense. In partic-ular, the definition of ray is required before the protractor axiom.

Betweenness B lies between A and C if d(A, B) + d(B, C) = d(A, C)

Segment AB consists of the points A, B and all those between

Ray−→AB consists of the segment AB and all points C such that B lies between A and C.

Basic shapes Triangles, circles, etc.

3As with Hilbert, Birkhoff makes SAS an axiom: Birkhoff’s version is stronger, for it also applies to similar triangles

11

Analytic geometry as a model

The axioms should feel familiar. Being shorter than Hilbert’s list, and being built on familiar notionssuch as the real line, it is somewhat easier for us to understand what the axioms are saying and tovisualize them. There is something to prove however; indeed the major point of Birkhoff’s system!

Theorem 3.22. Cartesian Analytic Geometry is a model of Birkhoff’s axioms.

Recall what this requires: we must provide a definition of each of the undefined terms and prove thatthese satisfy each of Birkhoff’s axioms. Here are suitable definitions for Cartesian analytic geometry:

Point An ordered pair (x, y) of real numbers.

Distance d(A, B) =√(Ax − Bx)2 + (Ay − By)2

Line all points satisfying a linear equation ax + by + c = 0.

Angle Define vectors as ordered pairs and consider the matrix J =(

0 −11 0

)whose action on vector is

to rotate counter-clockwise by π2 . Now define angle via

cos θ =v ·w|v| |w| where θ ∈

{[0, π] ⇐⇒ v · Jw ≥ 0(π, 2π) ⇐⇒ v · Jw < 0

Cosine may be defined using power series, so no pre-existing geometric meaning is required.

Proof. (Euclidean axiom) If (x1, y1) and (x2, y2) satisfy ax + by + c = 0 then

a(x1 − x2) + b(y1 − y2) = 0,

whence a = y1 − y2, b = x2 − x1 up to scaling. It follows that the line has equation

(y1 − y2)x + (x2 − x1)y + x1y2 − x2y1 = 0

unique up to multiplication of all three of a, b, c by a non-zero constant.The remaining axioms are exercises.

Exercises. 3.4.1. Prove that the ruler axiom is satsisfied:

(a) First show that if P 6= Q lie on `, then any point A on the line has the form

A = P +tA

d(P, Q)(Q− P) where tA ∈ R

(b) Use this formula to compute d(A, B)2 = (tA − tB)2.

3.4.2. Let I = (1, 0) and i =(

10

). Given v =

−→OA, define a = cos−1 v·i

|v| as above. Clearly a is acontinuous function of v and thus A. Now use the multiple-angle formula (Corollary 3.12) toprove that the protractor axiom is satisfied.

3.4.3. Use the cosine rule (Theorem 3.7) to prove that the SAS similarity axiom is satisfied.

12