(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

a = 0, y = 0, and la±ny = 0.

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

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

la − mẞ + ny = 0, B = 0, and la − mẞ + ny ± mẞ = 0.

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,

M

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

la + mẞ + ny=" 0,

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

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 By 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

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

which obviously meet in a point.

61. To find the condition that two lines la + mB + ny, l'a+m'B+ 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.

made with the axis of x by B-y is the angle between

Now since B and y are the angles the perpendiculars on the lines B, y, 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, ẞy is the supplement of A. The condition for perpendicularity therefore is

ll'+mm'+nn'— (mn'+m'n) cos▲ — (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+ cos B.

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

on la+m3 + ny. formula of Art. 34, &c., the result is

Write the equation at full length and apply the when, if we write x' cos a+y' sin a − p = a',

la' + mB' + 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 = cos B, as in Ex. 6, p. 54.

Ex. 2. To find the equation of a perpendicular to y 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 (A – 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 perpendiculars

α

β

=

cos B COS

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 + B 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 (AB). Ex. 6. Find the equation of a line through the point a'ß'y' perpendicular to the line y. Ans. a (B'y' cos A) − ẞ (a' + y' cos B) + y (B' cos B- a' cos.1).

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, ẞ, v, 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 cosa + 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 position of a point is defined

of trilinear coordinates in which the 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 + mẞ + ny = 0.

The advantage of trilinear coordinates is, that whereas in

Cartesian (or x and y) coordinates the utmost simplification we can introduce is by choosing two of the most remarkable lines in the figure for axes of coordinates, we can in trilinear coordinates obtain still more simple expressions by choosing three of the most remarkable lines for the lines of reference a, B, y. The reader will compare the brevity of the expressions in Art. 54 with those corresponding in Chap. II.

63. The perpendiculars from any point O on a, B, y are connected by the relation aa+b+cy=M, where a, b, c, are the sides, and M double the area, of the triangle of reference. For evidently aa, bß, cy are respectively double the areas of the triangles OBC, OCA, OAB. The reader may suppose that this is only true if the point O be taken within the triangle; but he is to remember that if the point O were on the other side of any of the lines of reference (a), we must give a negative sign to that perpendicular, and the quantity aa+bß +cy would then be double OCA + OAB- OBC, that is, still double the area of the triangle. Since sin A is proportional to a, it is plain that a sin A + B sin B + y sin C is also constant, a theorem which may otherwise be proved by writing a, ß, y at full length, as in Art. 60, multiplying by sin(3 − y), sin (y − a), sin (a—ß), respectively, and adding, when the coefficients of x and y vanish, and the sum is therefore constant.

=

The theorem of this article enables us always to use homogeneous equations in a, B, y, for if we are given such an equation as a = 3, we can throw it into the homogeneous form

64. To express in trilinear coordinates the equation of the parallel to a given line la+ mß +ny.

In Cartesian coordinates two lines Ax + By + C, Ax+ By + C', are parallel if their equations differ only by a constant. follows then that

la + mẞ + ny + k (a sin A +ẞ sin B+ y sin C) = 0

It

denotes a line parallel to la+ mẞ+ny, since the two equations differ only by a quantity which has been just proved to be

constant.

In the same case Ax+ By + C + (Ax + By + C') denotes a line also parallel to the two given lines and half-way between them; hence if two equations P=0, P'=0 are so connected that P- P' constant, then P+ P' denotes a parallel to P and P' half-way between them.

=

Ex. 1. To find the equation of a parallel to the base of a triangle drawn through the vertex. Ans. a sin A+ ẞ sin B = 0.

For this, obviously, is a line through aß; and writing the equation in the form sin C (a sin A + ẞ sin B + γ

sin C') = 0,

it appears that it differs only by a constant from y = = 0.

We see, also, that the parallel a sin A+ ẞ sin B, and the bisector of the base a sin A - ẞ sin B, form a harmonic pencil with a, ß, (Art. 57).

Ex. 2. The line joining the middle points of sides of a triangle is parallel to the base. Its equation (see Ex. 2, p. 58) is

a sin A+ẞ sin B- y sin C = 0, or 2y sin C a sin A + ẞ sin B + y sin C.

Ex. 3. The line aa - bẞ+cy - do (see Ex. 5, Art. 54) passes through the middle point of the line joining ay, ẞd. For (aa+cy) + (bß + dô) is constant, being twice the area of the quadrilateral; hence aa + cy, bß + dò are parallel, and (aa+cy) − (bß + dô) is also parallel and half-way between them. It therefore bisects the line joining (ay), which is a point on the first line, to (B8) which is a point on the second.

65. To write in the form la+mß + ny= 0 the equation of the line joining two given points x'y', x"y".

Let a', as before, denote the quantity x' cos a + y' sin a — p. Then the condition that the coordinates x'y' shall satisfy the equation la+mẞ + ny=0 may be written

in the given form, we obtain for the equation of the line joining

the two points

a (B'y" — y'ß') + B (y'a" — y'a') + y (a′B" — a′′B′) = 0.

It is to be observed that the equations in trilinear coordinates being homogeneous, we are not concerned with the actual lengths of the perpendiculars from any point on the lines of reference, but only with their mutual ratios. Thus the preceding equation is not altered if we write pa', pß', py', for a', B', y'. Accordingly, if a point be given as the intersection of the lines

we may take l, m, n as the trilinear coordinates