*CHAPTER IV.

APPLICATION OF ABRIDGED NOTATION TO THE EQUATION OF THE RIGHT LINE.

53. We have seen (Art. 40) that the line

(x cos a + y sin a − p) − k (x cos ẞ + y sin B −p') = 0 denotes a line passing through the intersection of the lines

x cos a+ y sin a − p = 0, x cos ẞ+ y sin ß − p' = 0. We shall often find it convenient to use abbreviations for these quantities. Let us call

x cos a + y sin a − p, a; x cos ẞ+ y sin ß — p', B.

Then the theorem just stated may be more briefly expressed; the equation a- kß = 0 denotes a line passing through the intersection of the two lines denoted by a=0, B=0. We shall for brevity call these the lines a, B, and their point of intersection the point aß. We shall, too, have occasion often to use abbreviations for the equations of lines in the form Ax + By + C = 0. We shall in these cases make use of Roman letters, reserving the letters of the Greek alphabet to intimate that the equation is in the form

x cos ay sina-p=0.

A

B

P

54. We proceed to examine the meaning of the coefficient k in the equation a-kẞ=0. We saw (Art. 34) that the quantity a (that is, a cosa + y sin a-p) denotes the length of the perpendicular PA let fall from any point xy on the line OA (which we suppose represented by a). Similarly, that ẞ is the O length of the perpendicular PB from the point zy on the line OB, represented by B. Hence the equation a-kẞ=0 asserts that if, from any point of the locus represented by it, perpendiculars be let fall on the lines OA, OB, the ratio of these perpendiculars (that is, PA: PB) will be constant and = k. Hence

the locus represented by a-kẞ=0 is a right line through O, and

a + kB = 0 denotes a right line dividing externally the angle

AOB into parts such that

sin POA
sin POB

= k. It is, of course, assumed

in what we have said that the perpendiculars PA, PB are those which we agree to consider positive; those on the opposite sides of a, ß being regarded as negative.

Ex. 1. To express in this notation the proof that the three bisectors of the angles of a triangle meet in a point.

a - ẞ = 0,

The equations of the three bisectors are obviously (see Arts. 35, 54) a By = 0, ya = 0, which, added together, vanish identically.

Ex. 2. Any two of the external bisectors of the angles of a triangle meet on the third internal bisector.

Attending to the convention about signs, it is easy to see that the equations of two external bisectors are a + ẞ = 0, a + y = 0, and subtracting one from the other we get By = 0, the equation of the third internal bisector.

Ex. 3. The three perpendiculars of a triangle meet in a point.

Let the angles opposite to the sides a, ẞ, y be A, B, C respectively. Then since the perpendicular divides any angle of the triangle into parts, which are the complements of the remaining two angles, therefore (by Art. 54) the equations of the perpendiculars are

a cos A - ẞ cos B = 0, ẞ cos By cos C = 0, y cos C-a cos 4 = 0,

which obviously meet in a point.

Ex. 4. The three bisectors of the sides of a triangle meet in a point. The ratio of the perpendiculars on the sides from the point where the bisector meets the base plainly is sin A: sin B. Hence the equations of the three bisectors are a sin A - ẞ sin B = 0, ẞ sin B − y sin C = 0, y sin C - a sin A 0.

Ex. 5. The lengths of the sides of a quadrilateral are a, b, c, of the line joining middle points of diagonals.

; find the equation

Ans. aa- - bẞ + cy — dồ = 0; for this line evidently passes through the intersection of aa - bẞ, and cy-do; but, by the last example, these are the bisectors of the base of two triangles having one diagonal for their common base. In like manner aa - dò, bß – cy intersect in the middle point of the other diagonal.

-

Ex. 6 To form the equation of a perpendicular to the base of a triangle at its extremity. Ans. a + y cos B = 0.

Ex. 7. If there be two triangles such that the perpendiculars from the vertices of one on the sides of the other meet in a point, then, vice verså, the perpendiculars from the vertices of the second on the sides of the first will meet in a point.

Let the sides be a, ß, y, a', B', y', and let us denote by (aß) the angle between a and B. Then the equation of the perpendicular

The condition that these should meet in a point is found by eliminating ẞ between the first two, aud examining whether the resulting equation coincides with the third. It is

cos (aẞ') cos (By') cos (ya') = cos (a'ẞ) cos (ẞ'y) cos (y'a).

But the symmetry of this equation shews that this is also the condition that the perpendiculars from the vertices of the second triangle on the sides of the first should meet in a point.

55. The lines a-k3=0, and ka-B-0, are plainly such that one makes the same angle with the line a which the other makes with the line 8, and are therefore equally inclined to the bisector a- B.

Ex. If through the vertices of a triangle there be drawn any three lines meeting in a point, the three lines drawn through the same angles, equally inclined to the bisectors of the angles, will also meet in a point.

Let the sides of the triangle be a, ß, y, and let the equations of the first three lines be

la mẞ= 0, mẞ — ny = 0, ny — la = 0,

which, by the principle of Art. 41, are the equations of three lines meeting in a point, and which obviously pass through the points aß, By, and ya. Now, from this Article, the equations of the second three lines will be

56. The reader is probably already acquainted with the following fundamental geometrical theorem :-"If a pencil of four right lines meeting in a point O be intersected by any transverse right line in the four points A, P, P, B, then

the ratio

AP.P'B

AP'.PB

is constant, no matter how

the transverse line be drawn.” This ratio is

called the anharmonic ratio of the pencil. In o

B

P'

P

A

fact, let the perpendicular from O on the transverse line =p; then p.AP OA.OP.sin A OP(both being double the area of the triangle AOP) ; p.PB = OP'.OB sin P OB; p.AP = OA.OP' sin AOP'; p.PB= OP. OB. sin POB; hence

p.AP.PB=0A. OP. OP'. OB. sin AOP.sin P'OB;
p.AP.PBOA. OP. OP. OB. sin AOP'.sin POB;

AP.PB sin A OP. sin P'OB

but the latter is a constant quantity, independent of the position of the transverse line.

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

k

then T

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

=

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, ẞ a harmonic pencil.

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

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

A

β

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―l; MK to m−k; NM to n and LK to l-k.

m;

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, ax+by+c', &c. For we can bring P to the form x 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 p 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, m, n; kp, lp, mp, np.

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−1Q', &c., the line P-kP being said to correspond to the line Q-kQ', &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, B, 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

1 cosa +m cosẞ+n cosy = a, l sina + m sinẞ+n siny=b,

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, 80, 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;

la mẞ = 0; and of BC, mẞny 0.

--

of AB, ẞ=0; of BD, y=0; of AD 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+ng 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.