57. If a - kẞ=0, a – k'ß = 0, be the equations of two lines,

k

then

will be the anharmonic ratio of the pencil formed by the

four lines a, ß, a – kß, a— k'ß, for (Art. 54)

but this is the anharmonic ratio of the pencil.

The pencil is a harmonic pencil when

k
k

1, for then the

angle AOB is divided internally and externally into parts whose sines are in the same ratio. Hence we have the important theorem, two lines whose equations are a― kẞ=0, a+kß=0, form with a, Bẞ a harmonic pencil.

58. In general the anharmonic ratio of four lines a-kß,

a – IB, a— mß, a – nß is

(n − 1) (mk)

(nm) (l-k)'

For let the pencil be

cut by any parallel to ẞ in the four points K, L, M, N, and the

K L/M

A

β

has the same value for each of these four points, the perpendiculars from these points on a are (by virtue of the equations of the lines) proportional to k, l, m, n; and AK, AL, AM, AN are evidently proportional to these perpendiculars; hence NL is proportional to n-1; MK to m-k; NM to n-m; and LK to l-k.

x

59. The theorems of the last two articles are true of lines represented in the form P-kP', P-IP', &c., where P, P' denote ax+by+c, a'x+by+c', &c. For we can bring P to the form a cos ay sina -p by dividing by a certain factor. The equations therefore P-kP' =0, P– IP' = 0, &c., are equivalent to equations of the form a-kpß =0, a − lpß = 0, &c., where ρ is the ratio of the factors by which P and P' must be divided in order to bring them to the forms a, B. But the expressions

for anharmonic ratio are unaltered when we substitute for k, l, man; kp, lp, mp, p.

It is worthy of remark, that since the expressions for anharmonic ratio only involve the coefficients k, l, m, n, it follows that if we have a system of any number of lines passing through a point, P-kP', P-IP', &c.; and a second system of lines passing through another point, Q—kQ', Q — IQ', &c., the line P-kP' being said to correspond to the line Q-Q', &c.; then the anharmonic ratio of any four lines of the one system is equal to that of the four corresponding lines of the other system. We shall hereafter often have occasion to speak of such systems of lines, which are called homographic systems.

60. Given three lines a, ß, y, forming a triangle ;* the equation of any right line, ax+by+c=0, can be thrown into the form la + mB+ny = 0.

Write at full length for a, B, y the quantities which they represent, and la+mẞ+ ny becomes

(1 cosa+m cosẞ+n cosy) x + (l sina+m sinß + n siny) y

(lp+mp' +np′′) = 0. This will be identical with the equation of the given line, if we have

I cosa+m cosß+n cosy = a, 7 sina+m sinß+n siny=b,

lp+mp' + np" C1

and we can evidently determine l, m, n, so as to satisfy these three equations,

The following examples will illustrate the principle that it is possible to express the equations of all the lines of any figure in terms of any three, a=0, B = 0, y=0.

Ex. 1. To deduce analytically the harmonic properties of a complete quadrilateral. (See figure, next page).

Let the equation of AC be a = 0 0; of AB, ẞ=0; of BD, y=0; of AD la - mß = 0; and of BC, mß — ny = 0. Then we are able to express in terms of these quantities the equations of all the other lines of the figure.

* We say "forming a triangle," for if the lines a, ẞ, y meet in a point, la + mẞ+ny must always denote a line passing through the same point, since any values of the coordinates which make a, ß, y separately = 0, must make la + mß + ny = 0.

I

A

B

D

F

EF joins the point ay to the point (la — mß+ny, ẞ), and its equation will be found to be la + ny = 0. From Art. 57 it appears that the four lines EA, EO, EB, and EF form a harmonic pencil, for their equations have been shown to be

is

Again, the equation of FO, which joins the points (la+ny, ß) and (la-mß, mß-ny)

la - 2mẞ + ny = 0.

Hence (Art. 57) the four lines FE, FC, FO, and FB are a harmonic pencil, for their equations are

la - mẞ + ny = 0, ẞ= 0, and la — mẞ + ny ± mß = 0.

mß

Again, OC, OE, OD, OF are a harmonic pencil, for their equations are

la - mẞ = 0, mẞ — ny = 0, and la – mß ± (mß – ny) = 0.

Ex. 2. To discuss the properties of the system of lines formed by drawing through the angles of a triangle three lines meeting in a point.

Let the equation of AB be y = 0; of AC, ß = 0 ; of BC, a = 0; and let the lines OA, OB, OC, meeting in a point,

be mẞ - ny, ny - la, la - mẞ (see –

Art. 55).

M

Now we can prove that the three points L, M, N are all in one right line, whose equation is

for this line passes through the points (la + mß – ny, y) or N; (la − mß + ny, ß) or M; and (mß + ny − la, a) or L.

The equation of CN is

la + mẞ = 0,

for this is evidently a line through (a, ß) or C, and it also passes through N, since it = (la + mß + ny) ~ ny.

Hence BN is cut harmonically, for the equations of the four lines CN, CA■ CF, CB are

a = 0, ẞ = 0, la — mß = 0, la + mß = 0.

The equations of this example can be applied to many particular cases of frequent occurrence. Thus (see Ex. 3, p. 54) the equation of the line joining the feet of two perpendiculars of a triangle is a cos A + ẞ cos B y cos C = 0; while a cos A + ẞ cos B + y cos C passes through the intersections with the opposite sides of the triangle, of the lines joining the feet of the perpendiculars. In like manner a sin A+ẞ sin B y sin C represents the line joining the middle points of two sides, &c.

Ex. 3. Two triangles are said to be homologous, when the intersections of the corresponding sides lie on the same right line called the axis of homology; prove that the lines joining the corresponding vertices meet in a point [called the centre of homology].

Let the sides of the first triangle be a, ß, y; and let the line on which the corresponding sides meet be la + mß + ny; then the equation of a line through the intersection of this with a must be of the form l'a + mẞ + ny = 0, and similarly those of the other two sides of the second triangle are

la + m'ß + ny = 0, la + mẞ + n'y = 0.

But subtracting successively each of the last three equations from another, we get for the equations of the lines joining corresponding vertices

(1 — 1') a = (m — m') ß, (m — m′) ẞ = (n — n') y, (n − n') y = (1 — l'′) a,

which obviously meet in a point.

61. To find the condition that two lines la+mß + ny, l'a + m'ß + n'y may be mutually perpendicular.

Write the equations at full length as in Art. 60, and apply the criterion of Art. 25, Cor. 2 (AA' + BB' = 0), when we find ll' + mm' + nn' + (mn' + m'n) cos (B − y) + (nl' + n'l) cos (y — a) +(lm' + l'm) cos (a — B) = 0.

Now since and y are the angles made with the axis of x by the perpendiculars on the lines B, Y, B-y is the angle between those perpendiculars, which again is equal or supplemental to the angle between the lines themselves. If we suppose the origin to be within the triangle, and A, B, C to be the angles of the triangle, ß-y is the supplement of A. The condition for perpendicularity therefore is

ll'+mm'+nn'— (mn'+m'n) cosA — (nl'+n'l) cos B—(lm'+l'm) cosC=0. As a particular case of the above, the condition that la+mß +ny may be perpendicular to y is

n = m cos A+ l cos B.

In like manner we find the length of the perpendicular from x'y

on la+mß+ny. Write the equation at full length and apply the formula of Art. 34, when, if we write x' cos a+y' sin a-p=a', &c., the result is

la' + mß' + ny'

√ (l2 + m2 + n2 − 2mn cos A 2nl cos B-2lm cos C')'

Ex. 1. To find the equation of a perpendicular to y through its extremity. The equation is of the form la+ ny = 0. And the condition of this article gives n = l cos B, as in Ex. 6, p. 54.

Ex. 2. To find the equation of a perpendicular toy through its middle point. The middle point being the intersection of y with a sin A – ß sin B, the equation of any line through it is of the form a sin A - ẞ sin B+ny = 0, and the condition of this article gives n = sin (4 --- B).

Ex. 3. The three perpendiculars at middle points of sides meet in a point. For eliminating a, ß, y in turn between

a sin A - ẞ sin B + y sin (A – B) = 0, ß sin B – y sin C + a sin (B − C) = 0, we get for the lines joining to the three vertices the intersection of two perpendículars

α

COS A

=

β

وقع

=

cos B cos C

and the symmetry of the equations proves that the third perpendicular passes through the same point. The equations of the perpendiculars vanish when multiplied by sin2C, sin2A, sin2B, and added together.

Ex. 4. Find, by Art. 25, expressions for the sine, cosine, and tangent of the angle between la+mß + ny, l'a + m'ß + n'y.

Ex. 5. Prove that a cos A + ẞ cos B + y cos C is perpendicular to

a sin A cos A sin (B − C) + ß sin B cos B sin (C − A) + y sin C cos C sin (A – B). Ex. 6. Find the equation of a line through the point a'ß'y' perpendicular to the line y. Ans. a (ß'+ y' cos A) – ß (a′ + y′ cos B) + y (B′ cos B — a′ cos A).

62. We have seen that we can express the equation of any right line in the form la + mẞ+ny = 0, and so solve any problem by a set of equations expressed in terms of a, B, y, without any direct mention of x and y. This suggests a new way of looking at the principle laid down in Art. 60. Instead of regarding a as a mere abbreviation for the quantity x cos a + y sin a−p, we may look upon it as simply denoting the length of the perpendicular from a point on the line a. We may imagine a system of trilinear coordinates in which the position of a point is defined by its distances from three fixed lines, and in which the position of any right line is defined by a homogeneous equation between these distances, of the form

la + mB + ny = 0.

The advantage of trilinear coordinates is, that whereas in