1 1

= ¦’ n ’1 PR n ’ PR ¦= ¦’ n ’1 ( PR n + n PR ) ¦=

2 2 n

This expression is the modulus of the projection of the vector PR upon the perpendicular

direction. Both formulas of the distance are equivalent because the normal and direction

vectors anticommute:

nv=’vn ” ”

nv+vn=0 n·v=0

Written with components:

n1 v1 + n2 v2 = 0

From each vector the other one is easily obtained by exchanging components and altering

a sign:

n = ( n1 , n2 ) = ( v2 , ’v1 )

As an application, let us calculate the equation of the line r passing through the

points (2, 3) and (7, 6). A direction vector is the segment having as extremes both points:

v = (7, 6) ’ (2, 3) = (5, 3)

and an equation for this line is:

x ’2 y ’3

=

5 3

From the direction vector we calculate the perpendicular vector n and the general equation

of the line:

n = (3, ’5) 3 (x ’ 2) ’ 5 (y ’ 3) = 0 ” 3x’5y+9=0

Now, let us calculate the distance from the point R = (’2, 1) to the line r. We choose the

point (2, 3) as the point P on the line:

PR = R ’ P = (’2, 1) ’ (2, 3) = (’4, ’2)

’ 4 … 3 + ( ’2) … ( ’5)

PR § v PR · n 2

=

d( (’2, 1), r ) = = =

v n 17

34

For another point R = (’3, 0 ), we will find a null distance indicating that this point lies on

r.

RAMON GONZALEZ CALVET

40

Slope and intercept equations of a line

If the ordinate is written as a function of the abscissa, the equation so obtained is

the slope-intercept equation:

y=mx+b

Note that this expression cannot describe the vertical lines whose equation is x = constant.

The coefficient m is called the slope because it is a measure of the inclination of the line.

For any two points P and Q on the line we have:

yP = m xP + b

yQ = m xQ + b

Subtracting both equations:

yQ ’ yP = m (xQ ’ xP)

we see that the slope is the quotient of

ordinate increment divided by the abscissa

increment (figure 5.9):

Figure 5.9

yQ ’ y P

m=

xQ ’ x P

This quotient is also the trigonometric tangent of the oriented angle between the line and

the positive abscissa semiaxis. For angles larger than π/2 the slope becomes negative.

b is the ordinate intercepted at the origin, the y-intercept:

’

x=0 y=b

If we know the slope and a point on a line, the point-slope equation may be used:

y ’ yP = m (x ’ xP)

Also, one may write for the intercept equation of a line:

xy

+ =1

ab

where a and b are the x-intercept and y-intercept of the line with each coordinate axis.

Polar equation of a line

In this equation, the distance from a fixed point F to the generic point P on the line

is a function of the angle ± with regard to the perpendicular direction (figure 5.10). If d is

the distance from F to the line, then:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 41

d

FP =

cos ± Figure 5.10

This equation allows to relate easily the

straight line with the circle and other conic

sections.

Intersection of two lines and pencil of

lines

The calculation of the intersection

of two lines is a very usual problem. Let us

suppose that the first line is given by the point P and the direction vector u and the second

one by the point Q and the direction vector v. Denoting by R the intersection point, which

belongs to both lines, we have:

” k u ’ l v = Q ’ P = PQ

R=P+ku=Q+lv

In order to find the coefficients k and l, the vector PQ must be resolved into a linear

combination of u and v, what results in:

k = PQ § v (u § v) ’1 l = ’ u § PQ (u § v) ’1

The intersection point is also obtained by directly solving the system of the

general equations of both lines:

± n1 x + n 2 y + c = 0

n1' x + n 2' y + c' = 0

The pencil of lines passing through this point is the set of lines whose equations

are linear combinations of the equations of both given lines:

(1 ’ p) [ n1 x + n2 y + c ] + p [ n'1 x + n'2 y + c' ] = 0

where p is a real parameter and

’ ∞ ¤ p ¤ ∞ . Each line of the pencil

determines a unique value of p,

independently of the fact that the

general equation for a line is not

unique

If the equation system has a

unique point as solution, all the lines

of the pencil are concurrent at this

point. But in the case of an

incompatible system, all the lines of Figure 5.11

the pencil are parallel, that is, they

RAMON GONZALEZ CALVET

42

intersect in a point at infinity, which gives generality to the concept of pencil of lines.

If the common point R is known or given, then the equation of the pencil is written

as:

[ ( 1 ’ p ) n1 + p n'1 ] ( x ’ xR ) + [ ( 1 ’ p ) n2 + p n'2 ] ( y ’ yR ) = 0

For 0<p<1, the normal and direction vectors of the p-line are respectively comprised

between the normal and direction vectors of both given lines4. In the other case the vectors

are out of this region (figure 5.11)

For example, calculate the equation of the line passing through the point (3, 4) and

the intersection of the lines 3 x + 2 y + 4 = 0 and 2 x ’ y + 3 = 0. The equation of the

pencil of lines is:

(1’p)(3x+2y+4)+p(2x’y+3)=0

The line of this pencil to which the point (3, 4) belongs must fulfil:

(1’p)(3…3+2…4+4)+p(2…3’4+3)=0

yielding p = 21/16 and the line:

27 x ’ 31 y + 43 = 0

Which is the meaning of the coefficient of linear combination p? Let us write the

pencil of lines P determined by the lines R and S as:

P=(1’p)R+pS

which implies the same relation for the normal vectors:

nP = ( 1 ’ p) nR + p nS

Taking outer products we obtain:

nP § nR nP § nR

p

= p=

’

nP § ( nR ’ nS )

nP § nS p ’ 1

These equalities are also valid for direction vectors whenever they have the same modulus

than the normal vectors. When the modulus of the normal vectors of the lines R and S are

equal we can simplify:

sin (RP )

p

=

1 ’ p sin (PS )

In this case, the value p = 1/2 corresponds to the bisector line of R and S. We see from this

expression and the foregoing ones that each value of p corresponds to a unique line.

4

This statement requires that the angle from the direction vector to the normal vector be a positive

right angle.

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 43

Dual coordinates

The duality principle states that any theorem relating incidence of lines and points

implies a dual counterpart where points and lines have been exchanged. The phrase «two

points determine a unique line passing through these points» has the dual statement «two

lines determine a unique point, intersection of these lines». All the geometric facts have an

algebraic counterpart, and the duality is not an exception. With the barycentric coordinates

every point R on the plane can be written as linear combination of three non aligned points

O, P and Q:

R = ( 1 ’ p ’ q ) O + p P + q Q = (p, q)

where p and q are the coordinates and O the origin.

These three points determine three non parallel lines A, B, and C in the following

way:

A = PQ B = QO C = OP

Then the direction vectors vA = PQ, vB = OP, vC = QO are related by:

v A + v B + vC = 0

Note that vB and ’vC are the base of the vectorial plane.

Any line D on the plane can be written as linear combination of these lines:

D=(1’b’c)A+bB+cC

This means that the general equation (with point coordinates) of D is a linear combination

of the general equations of the lines A, B and C. I call b and c the dual coordinates of the

line D. In order to distinguish them from point coordinates, I shall write D = [b, c]. The

choice of the equation for each line must be unambiguous and therefore the normal vector

in the implicit equations for A, B and C will be obtained by turning the direction vector

over π/2 counterclockwise.

Let us see some special cases. If the dual coordinate b is zero we have:

D=(1’c)A+cC

which is the equation of the pencil of lines passing through the intersection of the lines A

and C, that is, the point P. Then P = [0, c] (for every c). Analogously, c = 0 determines the

pencil of lines passing through the intersection of the lines A and B, which is the point Q,

and then Q = [b, 0] (for every b). The origin of coordinates is the intersection of the lines

B and C and then O = [b, 1 ’ b] (for every b). Compare the dual coordinates of these

points:

O = [b, 1’b] ∀ b, c

P = [0, c] Q = [b, 0]

with the point coordinates of the lines A, B and C:

RAMON GONZALEZ CALVET

44

A = (p, 1’ p) ∀ p, q

B = (0, q) C = (p, 0)

Let us see an example: calculate the dual Cartesian coordinates of the line 2x + 3y

+ 4 = 0. The points of the Cartesian base are O = (0, 0), P = (1, 0) and Q = (0, 1). Then the

lines of the Cartesian base are A: ’x ’ y +1= 0, B: x = 0, C: y = 0. We must solve the

identity:

∀ x, y

2 x + 3 y + 4 ≡ a' ( ’ x ’ y + 1) + b' x + c' y

x ( 2 + a' ’ b' ) + y ( 3 + a' ’ c' ) + 4 ’ a' ≡ 0

whose solution is:

a' = 4 b' = 6 c' = 7

Dividing by the sum of the coefficients we obtain:

2x +3y +4 4 6 7

≡ ( ’ x ’ y +1)+ x+ y

17 17 17 17

from where the dual coordinates of this line are obtained as [b, c] = [6/17, 7/17]. Let us

see their meaning. The linear combination of both coordinates axes is a line of the pencil

of lines passing through the origin:

6 7

x+ y=0 6x+7y=0

or

13 13

This line intersects the third base line ’x ’y + 1 = 0 at the point (7, ’6), whose pencil of

lines is described by:

«6 7

a ( ’ x ’ y + 1) + ( 1 ’ a ) ¬ x + y· = 0

13 13

Then 2 x + 3 y + 4 = 0 is the line of this pencil determined by a = 4/17.

On the other hand, how may we know whether three lines are concurrent and

belong to the same pencil or not? The answer is that the determinant of the dual

coordinates must be zero:

D=(1’b’c)A+bB+cC

E = ( 1 ’ b' ’ c' ) A + b' B + c' C

F = ( 1 ’ b'' ’ c'' ) A + b'' B + c'' C

1’ b ’ c b c

1 ’ b' ’ c' c' = 0

b'

”

D, E and F concurrent

1 ’ b'' ’ c'' b'' c''

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 45

When it happens, we will say that the lines are linearly dependent, and we can write

anyone of them as linear combination of the others:

F=(1’k)D+kE

We can express a point with an equation for dual coordinates in the same manner

as we express a line with an equation for point coordinates. For example, the point (5, 3)

is the intersection of the lines x = 5, y = 3 whose dual coordinates we calculate now:

x ’ 5 ≡ a' ( 1 ’ x ’ y ) + b' x + c' y ’ a' = ’5 b' = ’4 c' = ’5

From where a = 5/14, b = 4/14 and c = 5/14. The dual coordinates of x = 5 are [4/14,

5/14]. Analogously:

y ’ 3 ≡ a' ( 1 ’ x ’ y ) + b' x + c' y ’ a' = ’3 b' = ’3 c' = ’2

From where a = b = 3/8, c = 2/8. The dual coordinates of y = 3 are [3/8, 2/8]. The linear

combinations of both lines are the pencil of the point (5, 3), which is described by the

parametric equation:

[b, c] = ( 1 ’ k ) ® 4 ,

5 ®3 2

+k ,

°14 14

°8 8 »

»

By removing the parameter k, the general equation is obtained:

b ’ 4 / 14 c ’ 5 / 14

= 12 b + 10 c ’ 7 = 0

”

’6

5

That is, the dual direction vector of the point (5, 3) is [5, ’6] and the dual normal vector is

[6, 5]. In the dual plane, an algebra of dual vectors can be defined. A dual direction vector

for a point may be obtained as the difference between the dual coordinates of two lines

whose intersection be the point. Then, there exist dual translations of lines:

dual + : L — W ’ L L ={plane lines} W = {dual vectors}

(A, w) ’ B = A + w

That is, we add a fixed dual vector w to the line A in order to obtain another line B.

Let us prove the following theorem: all the points whose dual direction vectors are

proportional are aligned with the centroid of the coordinate system .

The proof begins from the dual continuous equation for a point P:

b ’ b0 c ’ c 0

=

v1 v2

which can be written in a parametric form:

RAMON GONZALEZ CALVET

46

[b, c] = (1 ’ k )[b0 , c0 ] + k [b0 +`v1 , c 0 + v 2 ]

P:

This equation means that P is the intersection point of the lines with dual coordinates [b0,

c0] and [b0 +v1, c0 +v2]. Then P must be obtained by solving the system of equations of

each line for the point coordinates p, q (x, y if Cartesian):

(1 ’ b0 ’ c 0 ) A + b0 B + c 0 C = 0

±

(1 ’ b0 ’ c 0 ’ v1 ’ v 2 ) A + (b0 + v1 ) B + (c 0 + v 2 ) C = 0

By subtraction of both equations, we find an equivalent system:

±(1 ’ b0 ’ c 0 ) A + b0 B + c 0 C = 0

’ (v1 + v 2 ) A + v1 B + v 2 C = 0

Now, if we consider a set of points with the same (or proportional) dual direction vector,

the first equation changes but the second equation remains constant (or proportional). That

is, the first line changes but the second line remains constant. Therefore all points will be

aligned and lying on the second line:

’ (v1 + v 2 ) A + v1 B + v 2 C = 0

However in which matter do two points differ whether having or not a proportional dual

direction vector? We cannot say that two points are aligned, because we need three points

at least. Let us search the third point X, rewriting the foregoing equality:

v1 v

(B ’ A) + 2 (C ’ A) = 0

v1 + v 2 v1 + v 2

Now the variation of the components of the dual vector generates the pencil of the lines B

’ A and C ’ A. That is, the intersection of both lines is the point X searched:

±B ’ A = 0

X:

C ’ A = 0

Then, two points have a proportional dual direction vector if they are aligned with the

point X, intersection of the lines B ’ A and C ’ A. However, note that the addition of

coefficients of each line is zero instead of one, so that one dual coordinate of each line is

infinite:

B ’ A = [∞, c ] C ’ A = [b, ∞]

I call X the point at the dual infinity or simply the dual infinity point. The dual infinity

point has finite coordinates and a very well defined position. Then the points with

proportional dual vector are always aligned with the dual infinity point, and I shall say

that they are parallel points in a dual sense, of course. In order to precise which point is X

let us take in mind that the lines B ’ A and C ’ A are the medians of the triangle OPQ (the

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 47

point base). Hence the dual infinity point is the centroid of the base triangle OPQ. With

Cartesian coordinates O = (0, 0), P = (1, 0), Q = (0, 1) and X = (1/3, 1/3). This ends the

proof.

Summarising, we can say that two parallel points are aligned with the dual infinity

point, which is the dual statement of the fact that two parallel lines meet at the infinity,

that is, there is a line at the infinity, in the usual sense. This is the novelty of the projective

geometry in comparison with the Euclidean geometry. In fact, the problem arises because

the point coordinates take infinite values, but the line L at the infinity is a well defined line

with finite dual coordinates [1/3, 1/3]:

®1 1 A + B + C

L= , =

° 3 3» 3

This means that the line L at the infinity belongs to the pencil of the lines (A + B ) /2 and C

. But both lines are parallel because their direction vectors are proportional:

v A + v B = ’v C

That is, the lines (A + B ) /2 and C meet at a point located at an infinite distance from the

origin of coordinates. Since any other parallel line meets them also at the infinity, this

argument does not suffice. However we can understand that the line L also belongs to the

pencil of the lines A and (B + C ) /2, which are also parallel with another direction, that is,

they meet at another point of the infinity. Any other pencil we take has always its point of

intersection located at the infinity. Then L only has points located at the infinity and

because of this it is called the line at the infinity. Summarising we can say that the line at

the infinity is the centroid of the three base lines in spite of the incompatibility of its

equation:

(’ x ’ y + 1) + x + y = 0

A+ B +C 1

L= =0

’ ’

3 3 3

Moreover, we may interpret the parallel lines as those lines aligned (in a dual sense) with

the line at the infinity:

E = [bE, cE] F = [bF, cF]

bF ’ b E b F ’ 1 / 3

=

”

E || F

cF ’ cE cF ’ 1 / 3

This is a useful equality because it allows us to know whether two lines are parallel from

their dual coordinates.

Many of the incidence theorems (and also their dual theorems) can be solved by

means of line equations or dual coordinates. A proper example is the proof of the

Desargues theorem.

The Desargues theorem

RAMON GONZALEZ CALVET

48

Given two triangles ABC and A'B'C', let P be the intersection of the prolongation

of the side AB with the side A'B', Q the intersection of BC with B'C', and R the intersection

of CA with C'A'. The points P, Q and R are aligned if and only if the lines AA', BB' and

CC' meet at the same point.

Proof ’ The hypothesis states that the lines AA', BB' and CC' intersect at the same

point O (figure 5.12):

O = a A + ( 1 ’ a ) A'

O = b B + ( 1 ’ b ) B'

O = c C + ( 1 ’ c ) C'

with a, b and c being real. Equating

the first and second equations we

obtain:

Figure 5 12

a A + ( 1 ’ a ) A' = b B + ( 1 ’ b ) B'

which can be rearranged as:

a A ’ b B = ’ ( 1 ’ a ) A' + ( 1 ’ b ) B'

Dividing by a ’ b the sum of the coefficients becomes the unity, and then the equation

represents the intersection of the lines AB and A'B', which is the point P:

a ’1 b ’1

a b

P= A’ B= A' ’ B'

a’b a’b a’b a’b

By equating the second and third equations for the point O, and the third and first ones,

analogous equations for Q and R are obtained:

b ’1 c ’1

b c

Q= B’ C= B' ’ C'

b’c b’c b’c b’c

c ’1 a ’1

c a

R= C’ A= C' ’ A'

c’a c’a c’a c’a

Now we must prove that P, Q and R are aligned, that is, fulfil the equation:

R=dP+(1’d)Q with d real

With the substitution of the former equations into the last one we have:

®a ®b

c a b c

B + ( 1 ’ d )

C’ A=d A’ B’ C

c’a c’a °a ’ b b’a » °b ’ c b’c »

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 49

Arranging all the terms at the left hand side, the expression obtained must be identical to

zero because A, B and C are non aligned points:

b (1 ’ d ) c (1 ’ d )

® da a ® db ®c

’ ’ A+ ’ B+ + C ≡0

a ’ b c ’ a b’c b’c

a’b c’a

° » ° » ° »

This implies that the three coefficients must be null simultaneously yielding a unique

value for d,

a’b

d=

a’c

fact which proves the alignment of P, Q and R, and gives the relation for the distances

between points:

QR QP ’1 = ( A'O A'A ’1 ’ B'O B'B ’1 ) ( A'O A'A-1 ’ C'O C'C ’1 ) ’1

Proof ⇐ We will prove the Desargues theorem in the other direction following the

same algebraic way but applying the duality, that is, I shall only change the words. O, A,

B, C, P, Q and R will be now lines (figure 5.13).

The hypothesis states that the

points AA', BB' and CC' belong to the

same line O, that is, O belongs to the

pencil of the lines A and A', but also

to the pencil of the lines BB' and CC':

O = a A + ( 1 ’ a ) A'

O = b B + ( 1 ’ b ) B'

Figure 5.13

O = c C + ( 1 ’ c ) C'

with a, b and c being real

coefficients. Equating the first and

second equations we obtain:

a A + ( 1 ’ a ) A' = b B + ( 1 ’ b ) B'

which can be rearranged as:

a A ’ b B = ’ ( 1 ’ a ) A' + ( 1 ’ b ) B'

Dividing by a ’ b the sum of the coefficients becomes the unity, and then the equation

represents the line passing through the points AB and A'B' (belonging to both pencils),

which is the line P:

a ’1 b ’1

a b

P= A’ = A' ’ B'

a’b a’b a’b a’b

RAMON GONZALEZ CALVET

50

By equating the second and third equations for the line O, and the third and first ones,

analogous equations for Q and R are obtained:

b ’1 c ’1

b c

Q= B’ C= B' ’ C'

b’c b’c b’c b’c

c ’1 a ’1

c a

R= C’ A= C' ’ A'

c’a c’a c’a c’a

Now we must prove that the lines P, Q and R belong to the same pencil, that is, fulfil the

equation:

R=dP+(1’d)Q with d real

With the substitution of the former equations into the last one we have:

®a ®b

c a b c

B + ( 1 ’ d )

C’ A=d A’ B’ C

c’a c’a °a ’ b b’a » °b ’ c b’c »

Arranging all the terms at the left hand side, the expression obtained must be identical to

zero because A, B and C are independent lines (not belonging to the same pencil):

b (1 ’ d ) c (1 ’ d )

® da a ® db ®c

’ ’ A+ ’ B+ + C ≡0

a ’ b c ’ a b’c b’c

°a ’ b °c ’ a

° » » »

This implies that the three coefficients must be null simultaneously yielding a unique

value for d,

a’b

d=

a’c

fact which proves that P, Q and R are lines of the same pencil.

Exercises

Let A = (2, 4), B = (4, ’3) and C = (2, ’5) be three consecutive vertices of a

5.1

parallelogram. Calculate the fourth vertex D and the area of the parallelogram.

Prove the Euler™s theorem: for any four points A, B, C and D the product AD BC +

5.2

BD CA + CD AB vanishes if and only if A, B and C are aligned.

Consider a coordinate system with vectors { e1, e2 } where ¦e1¦= 1, ¦e2¦= 1 and

5.3

the angle formed by both vectors is π/3.

a) Calculate the area of the triangle ABC being A = (2, 2), B = (4, 4), C = (4, 2).

b) Calculate the distance between A and B, B and C, C and A.

Construct a trapezoid whose sides ¦AB¦, ¦BC¦, ¦CD¦ and ¦DA¦ are known and

5.4

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 51

AB is parallel to CD. Study for which values of the sides does the trapezoid exist.

Given any coordinate system with points {O, P, Q} and any point R with

5.5

coordinates (p, q) in this system, show that:

Area OPR

Area RPQ Area ORQ

1’ p ’ q = p= q=

Area OPQ Area OPQ Area OPQ

Let the points A = (2, 3), B = (5, 4), C = (1, 6) be given. Calculate the distance

5.6

from C to the line AB and the angle between the lines AB and AC.

Given any barycentric coordinate system defined by the points {O, P, Q} and the

5.7

non aligned transformed points {O', P', Q'}, an affinity (or an affine

transformation) is defined as that geometric transformation which maps each point

D to D' in the following way:

D = ( 1 ’ x ’ y ) O + x P + y Q’ D' = ( 1 ’ x ’ y ) O' + x P' + y Q'

Prove that:

a) An affinity maps lines onto lines.

b) An affinity is equivalent to a linear mapping of the coordinates, that is, the

coordinates of any transformed point are linear functions of the coordinates of the

original point.

c) An affinity preserves the coordinates expressed in any other set of independent

points different of the given base:

D = ( 1 ’ b ’ c ) A + b B + c C’ D' = ( 1 ’ b ’ c ) A' + b B' + c C'

d) An affinity maps a parallelogram to another parallelogram, and hence, parallel

lines to parallel lines.

e) An affinity preserves the ratio DE DF ’1 for any three aligned points D, E and

F.

If {A, B, C} is a base of lines and {A', B', C'} their transformed lines -the lines of

5.8

each set being independent-, consider the geometric transformation which maps

each line D to D' in the following way:

D = ( 1 ’ b ’ c ) A + b B + c C’ D' = ( 1 ’ b ’ c ) A' + b B' + c C'

a) Prove that every pencil of lines is mapped to another pencil of lines.

b) The dual coordinates of D' are linear functions of the dual coordinates of D.

c) This transformation preserves the coefficients which express a line as a linear

combination of any three non concurrent lines, that is, in the foregoing mapping

{A, B, C} have not to be necessarily the dual coordinate base and can be any other

set of independent lines.

d) Parallel points are mapped to parallel points.

e) For any three concurrent lines P, Q and R, the single ratio of the dual vectors

PQ PR’1 is preserved.

f) Using the formula of the cross ratio of a pencil of any four lines (ABCD) as a

RAMON GONZALEZ CALVET

52

function of their direction vectors5:

v A § vC v B § v D

( ABCD ) =

v A § v D v B § vC

show that it is preserved.

Calculate the dual coordinates of the lines x “ y + 1 = 0 and x “ y + 3 = 0 . See

5.9

that they are aligned in the dual plane with the line at the infinity, whose dual

coordinates are [1/3, 1/3], and therefore are parallel.

5.10 Calculate the dual equations of the points (2, 1) and (“3, “1) and their direction

vectors. See that they are parallel points. Hence prove that they are aligned with

the dual infinity point, the centroid of the coordinate system (1/3, 1/3).

5

This formula is deduced in the chapter devoted to the cross ratio.

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 53

6. ANGLES AND ELEMENTAL TRIGONOMETRY

Here, the basic identities of the elemental trigonometry are deduced in close

connection with basic geometric facts, a very useful point of view for our pupils.

Sum of the angles of a polygon

Firstly let us see the special case

of a triangle. For any triangle ABC the

following identity holds:

CA AB ’1 AB BC ’1 BC CA ’1 = 1

Let ±, β and γ be the exterior angles

between the sides CA and AB, AB and

Figure 6.1

BC, BC and CA respectively (figure 6.1).

Applying the definition of geometric

quotient, the modulus of all the sides are

simplified and only the exponentials of the arguments remain:

exp(± e12 ) exp(β e12 ) exp(γ e12 ) = exp[ (± + β + γ ) e12 ] = 1

The three angles have the same orientation, which we suppose positive, and are

lesser than π. Hence, since the exponential is equal to the unity, the addition of the three

angles must be equal to 2π:

± + β + γ = 2π

The interior angles, those formed by AB and AC, BC and CA, CA and CB are

supplementary of ±, β, γ (figure 6.1). Therefore the sum of the angles of a triangle is equal

to π:

(π ’ ± ) + (π ’ β ) + (π ’ γ ) = π

This result is generalised to any polygon from the following identity:

AB BC ’1 BC ... YZ ’1 YZ ZA’1 ZA AB ’1 = 1

Let ±, β, γ ... ω be the exterior angles formed by the sides ZA and AB, AB and BC,

BC and CD, ..., YZ and ZA, respectively. After the simplification of the modulus of all

vectors, we have:

exp[ (± + β + γ + ... + ω ) e12 ] = 1

Let us suppose that the orientation defined by the vertices A, B, C ... Z is

counterclockwise, although the exterior angles be not necessarily all positive1. Translating

1

That is, it is not needed that the polygon be convex.

RAMON GONZALEZ CALVET

54

them to a common vertex, each angle is placed close each other following the order of the

perimeter and summing one turn:

± + β + γ + ... + ω = 2π

The interior angles formed by the sides AB and ZA, BC and BA, CD and CB,... ZA and ZY

are supplementary of ±, β, γ, ... ω. Therefore the sum of the angles of a polygon is:

( π ’ ± ) + ( π ’ β ) + ( π ’ γ ) + ... + ( π ’ ω ) = n π ’ 2π = ( n ’ 2 )π

n being the number of sides of the polygon. The deduction for the clockwise orientation of

the polygon is analogous with the only difference that the result is negative.

Definition of trigonometric functions and fundamental identities

Let us consider a circle with radius r (figure 6.2). The extreme of the radius is a

point on the circumference with coordinates (x, y). The arc between the positive X

semiaxis and this point (x, y) has an oriented length s, positive if counterclockwise and

otherwise negative. Also, the X-axis,

the arc of circumference and the

radius delimit a sector with an

Figure 6.2

oriented area A. An oriented angle

± is defined as the quotient of the arc

length divided by the radius2:

s

±=

r

Since the area of the sector is

proportional to the arc length and the

area of the circle is 2πr, it follows

that:

± r2 2A

±=

A= ”

r2

2

The trigonometric functions3, sine, cosine and tangent of the angle ±, are respectively

defined as the ratios:

y x y

sin ± = cos ± = tg ± =

r r x

and the cosecant, secant and cotangent as their inverse fractions:

2

With this definition it is said that the angle is given in radians, although an angle is a quotient of

lengths and therefore a number without dimensions.

3

Also called circular functions due to obvious reasons, to be distinguished from the hyperbolic

functions.

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 55

r 1 r 1 x 1

csc ± = sec ± = cot ± =

≡ ≡ ≡

x cos ±

y sin ± y tg ±

being r related with x and y by the Pythagorean theorem:

r2 = x2 + y2

Then the radius vector v is:

v = x e1 + y e2 = r ( cos± e1 + sin± e2 )

From these definitions the fundamental identities follow:

sin ± 1

tg ± ≡ sin 2± + cos 2 ± ≡ 1 1 + tg 2 ± ≡ ≡ sec 2 ±

cos ± cos ±

2

If we take the opposite angle “± instead of ±, the sign of y is changed while x and r are

preserved, so we obtain the parity relations:

sin(’ ± ) ≡ ’sin ± cos(’ ± ) ≡ cos ± tg(’ ± ) ≡ ’ tg ±

The sine and tangent are odd functions while the cosine is an even function.

Look at the figure 6.2: β = π/2’± is the complementary angle of ± . If ± is higher

than π/2, the angle β becomes negative. On the other hand, if the angle ± is negative then

β is higher than π/2. The trigonometric ratios for β give the complementary angle

identities:

x y x 1

sin β = ≡ cos ± cos β = ≡ sin ± tg β = ≡

y tg ±

r

r

Angle inscribed in a circle and double angle identities

Let us draw any diameter PQ and any

Figure 6.3

radius OA (figure 6.3) in a circle with centre O.

Since OP is also a radius, the triangle POA is

isosceles and the angles OPA and PAO, which

will be denoted as ±, are equal. Because the

addition of the three angles is equal to π, the

angle AOP is π ’ 2± . The angle QOA is

supplementary of AOP, and therefore is equal to

2±, the double of the angle APQ. Let us draw

from A a segment perpendicular to the diameter

and touching it at the point N. By the definition of

sine we have:

RAMON GONZALEZ CALVET

56

NA NA PA

sin 2± = =

OA PA OA

The first quotient is sin± for the triangle PNA. If M is the midpoint of the segment

PA, then ¦PA¦ = 2 ¦MA¦ and the second quotient is equal to 2 cos± for the triangle

MOA:

MA

sin 2± = 2 sin ± ≡ 2 sin ± cos ±

OA

Through an analogous way we obtain cos 2± :

( )

PN ’ PO

ON PA

cos 2± = ≡ 2 cos 2 ± ’ 1 ≡ cos 2 ± ’ sin 2±

=

OA PA OA

Also this result may be obtained from the second fundamental identity. In order to obtain

the tangent of the double angle we make use of the first fundamental identity:

2 sin ± cos ±

sin 2± 2 tg ±

tg 2± ≡ ≡ ≡

cos2± cos 2 ± ’ sin 2± 1 ’ tg 2 ±

Finally, let us draw any other segment PB (figure 6.3). The angle BPQ will be

denoted as β. By the same arguments as above the angle BOQ is 2β. While the angle BPA

is ± + β, the angle BOA is 2± +2β : an angle inscribed in a circumference is equal to the

half of the central angle (angle whose vertex is the centre of the circumference) which

intercepts the same arc of circle. Consequently, all the angles inscribed in the same circle

and intercepting the same arc are equal independently of the position of the vertex on the

circle.

Addition of vectors and sum of trigonometric functions

Let us consider two unitary vectors forming the angles ± and β with the e1

direction (figure 6.4):

u = e1 cos ± + e 2 sin± v = e1 cos β + e 2 sin β

u + v = e1 (cos ± + cos β ) + e 2 (sin ± + sin β )

Figure 6.4

The addition of both vectors, u + v , is the

diagonal of the rhombus which they form, whence

it follows that the long diagonal is the bisector of

the angle ± ’ β between both vectors. The short

diagonal cuts the long diagonal perpendicularly

forming four right triangles. Then the modulus of

u + v is equal to the double of the cosine of the

half of this angle:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 57

±’β

u + v = 2 cos

2

Moreover, the addition vector forms an angle (± + β) /2 with the e1 direction:

±’β « ±+β ±+β

u + v = 2 cos + e 2 sin

¬ e1 cos ·

2 2 2

By identifying this expression for u + v with that obtained above, we arrive at two

identities, one for each component:

±+β ±’β

cos ± + cos β ≡ 2 cos cos

2 2

±+β ±’β

sin ± + sinβ ≡ 2 sin cos

2 2

In a similar manner, but using a subtraction of unitary vectors, the other pair of

identities are obtained:

±+β ±’β

cos ± ’ cos β ≡ ’2 sin sin

2 2

±+β ±’β

sin ± ’ sin β ≡ 2 cos sin

2 2

The addition and subtraction of tangents are obtained through the common

denominator:

sin ± sinβ sin ± cos β ± cos ± sin β sin (± ± β )

tg ± ± tg β ≡ ± ≡ ≡

cos ± cosβ cos ± cos β cos± cosβ

Product of vectors and addition identities

Let us see at the figure 6.4, but now we calculate the product of both vectors:

v u = ( e1 cos β + e 2 sin β ) ( e1 cos ± + e 2 sin ± ) =

= cos ± cos β + sin ± sin β + e12 (sin ± cos β ’ cos ± sin β )

Since u and v are unitary vectors, their product is a complex number with unitary

modulus and argument equal to the oriented angle between them:

v u = cos(± ’ β ) + e12 sin(± ’ β )

The identification of both equations gives the trigonometric functions of the angles

RAMON GONZALEZ CALVET

58

difference:

sin(± ’ β ) ≡ sin± cos β ’ cos ± sin β

cos(± ’ β ) ≡ cos ± cos β + sin ± sin β

Taking into account that the sine and the cosine are odd and even functions respectively,

one obtains the identities for the angles addition:

sin(± + β ) ≡ sin± cos β + cos ± sin β

cos(± + β ) ≡ cos ± cos β ’ sin ± sin β

Rotations and De Moivre™s identity

If v' is the result of turning the vector v over an angle ± then:

v' = v (cos ± + e12 sin ± )

To repeat a rotation of angle ± by n times is the same thing as to turn over an angle n± :

v'' = v ( cos ± + e12 sin± ) = v ( cos n± + e12 sin n± )

n

cos n± + e12 sin n± ≡ ( cos ± + e12 sin± )

n

’

This is the De Moivre™s identity4, which allows to calculate the trigonometric

functions of multiplies of a certain angle through the binomial theorem. For example, for n

= 3 we have:

cos 3± + e12 sin 3± ≡ (cos ± + e12 sin ± )

3

≡ cos 3 ± ’ 3 cos ± sin 2± + e12 ( 3 cos 2 ± sin ± ’ sin 3± )

Splitting the real and imaginary parts one obtains:

cos 3± ≡ cos 3 ± ’ 3 cos ± sin 2± sin3± ≡ 3 cos 2± sin± ’ sin 3±

And dividing both identities one arrives at:

3 tg ± ’ tg 3 ±

tg 3± ≡

1 ’ 3 tg 2 ±

4

With the Euler™s identity, the De Moivre™s identity is:

exp( n± e12 ) ≡ [exp(± e12 )]

n

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 59

Inverse trigonometric functions

The arcsine, arccosine and arctangent are defined as the inverse functions of the

sine, cosine and tangent. They are multivalued functions and the principal values are taken

in the following intervals:

π π

y = arcsin x ” x = sin y and ’ ¤ y¤

2 2

x = cos y and 0 ¤ y ¤ π

y = arccos x ”

π π

y = arctg x ” x = tg y and ’ < y<

2 2

y = arccot x ” x = cot y and 0 < y < π

From the definitions the parity of the inverse functions follow immediately:

arcsin x ≡ ’arcsin (’ x ) arccos x ≡ arccos(’ x )

arctg x ≡ ’ arctg (’ x )

Through the fundamental identities for the circular functions one obtains the

following identities5 for the inverse functions:

[ ] x

arcsin x ≡ arccos 1 ’ x 2 ≡ arctg

1’ x2

[ ] ®

1’ x2

arccos x ≡ arcsin 1 ’ x ≡ arctg

2

x

° »

® 1

x

arctg x ≡ arcsin ≡ arccos

1+ x2 ° 1+ x2 »

where the brackets indicate that the identity only holds for positive values.

From the complementary angles identities we have:

π π

arcsin x ≡ ’ arccos x arctg x ≡ ’ arc cot x

2 2

5

The only inverse circular function predefined in the language Basic is the arctangent ATN(X), so

these identities allows us to program the arcsine and arccosine.

RAMON GONZALEZ CALVET

60

Exercises

6.1 Prove the law of sines, cosines and tangents: if a, b and c are the sides of a triangle

respectively opposite to the angles ±, β and γ, then:

a b c

= = (law of sines)

sin ± sin β sin γ

c2 = a2 + b2 ’ 2 ¦a¦ ¦b¦ cos γ (law of cosines)

±+β

tg

a+b 2

= (law of tangents)

±’β

a’b

tg

2

Hint: use the inner and outer products of sides.

6.2 Prove the following trigonometric identities:

±’β β ’γ γ ’±

cos(± ’ β ) + cos(β ’ γ ) + cos(γ ’ a ) ≡ 4 cos ’1

cos cos

2 2 2

±’β β ’γ γ ’±

sin(± ’ β ) + sin(β ’ γ ) + sin(γ ’ ± ) ≡ ’4 sin sin sin

2 2 2

6.3 Express the trigonometric functions of 4± as a polynomial of the trigonometric

functions of ± .

6.4 Let P be a point on a circle arc whose extremes are the points A and B. Prove that the

sum of the chords AP and PB is maximal when P is the midpoint of the arc AB.

6.5 Prove the Mollweide™s formulas for a triangle:

±’β ±’β

cos sin

a+b a’b

2 2

= =

γ γ

c c

sin cos

2 2

6.6 Deduce also the projection formulas:

a = b cos γ + c cos β b = c cos ± + a cos γ c = a cos β + b cos ±

6.7 Prove the half angle identities:

± 1 ’ cos ± 1 ’ cos ± sin±

± 1 ’ cos ± ± 1 + cos ±

≡± ≡± tg ≡ ± ≡ ≡

sin cos

1 + cos ± sin± 1 + cos ±

2 2 2 2 2

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 61

7. SIMILARITIES AND SINGLE RATIO

Two geometric figures are similar if they have the same shape. If the orientation of both

figures is the same, they are said to be directly similar. For example, the dials of clocks are

directly similar. On the other hand, two figures can have the same shape but different

orientation. Then they are said to be oppositely similar. For example our hands are oppositely

similar. However these intuitive concepts are insufficient and the similarity must be defined

with more precision.

Direct similarity (similitude)

If two vectors u and v on the plane

form the same angle as the angle between the

Figure 7.1

vectors w and t, and they have proportional

lengths (figure 7.1), then they are

geometrically proportional:

± (u, v ) = ± (w, t )

u w u v ’1 = w t ’1

’

=

v t

that is, the geometric quotient of u and v is equal to the quotient of w and t, which is a complex

number. This definition of a geometric quotient is also valid for vectors in the space provided

that the four vectors lie on the same plane. In this case, the geometric quotient is a quaternion, as

Hamilton showed.

The geometric proportionality for vectors allows to define the similarity of triangles.

Two triangles ABC and A'B'C' are said to be directly similar and their vertices and sides denoted

with the same letters are homologous if:

AB BC ’1 = A'B' B'C' ’1

that is, if the geometric quotient of two sides of the first triangle is equal to the quotient of the

homologous sides of the second triangle. Arranging the vectors of this equation one obtains the

equality of the quotients of the homologous sides:

AB’1 A'B' = BC ’1 B'C'

One can prove easily that the third quotient of homologous sides also coincides with the

other quotients:

AB’1 A'B' = BC ’1 B'C' = CA’1 C'A' = r

The similarity ratio r is defined as the quotient of every pair of homologous sides, which

is a complex constant. The modulus of the similarity ratio is the size ratio and the argument is

the angle of rotation of the triangle A'B'C' with respect to the triangle ABC.

RAMON GONZALEZ CALVET

62

A' B'

exp[± ( AB, A' B' ) e12 ] Figure 7.2

r=

AB

The definition of similarity is

generalised to any pair of polygons in the

following way. Let the polygons ABC...Z and

A'B'C'...Z' be. They are said to be directly

similar with similarity ratio r and the sides

denoted with the same letters to be homologous

if:

r = AB ’1 A'B' = BC ’1 B'C' = CD ’1 C'D' = ... = YZ ’1 Y'Z' = ZA ’1 Z'A'

One of these equalities depends on the others and we do not need to know whether it is

fulfilled. Here also, the modulus of r is the size ratio of both polygons and the argument is the

angle of rotation. The fact that the homologous exterior and interior angles are equal for directly

similar polygons (figure 7.2) is trivial because:

B'A' B'C' ’1 = BA BC ’1 ’ angle A'B'C' = angle ABC

C'B' C'D' ’1 = CB CD ’1 ’ angle B'C'D' = angle BCD etc.

The direct similarity is an equivalence relation since it has the reflexive, symmetric and

transitive properties. This means that there are classes of equivalence with directly similar

figures.

A similitude with ¦r¦=1 is called a displacement, since both polygons have the same

size and orientation.

Opposite similarity

Two triangles ABC and A'B'C' are

oppositely similar and the sides denoted with

the same letters are homologous if:

AB BC ’1 = ( A'B' B'C' ’1 )* = B'C' ’1 A'B'

The former equality cannot be arranged in a

quotient of a pair of homologous sides as we

have made before. Because of this, the

similarity ratio cannot be defined for the Figure 7.3

opposite similarity but only the size ratio,

which is the quotient of the lengths of any two

homologous sides. An opposite similarity is always the composition of a reflection in any line

and a direct similarity.

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 63

AB BC ’1 = v ’1 A'B' B'C' ’1 v BC ’1 v ’1 B'C' = AB ’1 v ’1 A'B' ”

”

BC ’1 ( v ’1 B'C' v ) = AB ’1 ( v ’1 A'B' ’1 v ) = r

where r is the ratio of a direct similarity whose argument is not defined but depends on the

direction vector v of the reflection axis. Notwithstanding, this expression allows to define the

opposite similarity of two polygons. So two polygons ABC...Z and A'B'C'...Z' are oppositely

similar and the sides denoted with the same letters are homologous if for any vector v the

following equalities are fulfilled:

AB ’1 ( v ’1 A'B' ’1

v ) = BC ’1 ( v ’1 B'C' v ) = .... = ZA ’1 ( v ’1 Z'A' v )

that is, if after a reflection one polygon is directly similar to the other. The opposite similarity is

not reflexive nor transitive: if a figure is oppositely similar to another, and this is oppositely

similar to a third figure, then the first and third figures are directly similar. Then there are not

classes of oppositely similar figures.

An opposite similarity with ¦r¦=1 is called a reversal, since both polygons have the

same size both opposite orientations.

The theorem of Menelaus

For every triangle ABC (figure 7.4), Figure 7.4

three points D, E and F lying respectively on

the sides BC, CA and AB or their prolongations

are aligned if and only if:

AF FB ’1 BD DC ’1 CE EA ’1 = ’1

Proof ’ Let us suppose that D, E and F are

aligned on a crossing straight line. Let us

denote by p, q and r the vectors with origin at

the vertices A, B and C and going

perpendicularly to the crossing line. Then every pair of right angle triangles having the

hypotenuses on a common side of the triangle ABC are similar so that we have:

BF ’1 AF = q ’1 p CD ’1 BD = r ’1 q AE ’1 CE = p ’1 r

Multiplying the three equalities one obtains:

AE ’1 CE CD ’1 BD BF ’1 AF = 1

Taking the complex conjugate expression and changing the sign of BF, CD and AE, the theorem

is proved in this direction:

AF FB ’1 BD DC ’1 CE EA ’1 = ’1