represents the two right lines x-ay = 0, x-by=0, where a and b are the two roots of the quadratic

It is proved, in like manner, that the equation

n-1

-2

(x − a)" − p (x − a)"−1 ( y − b) + q (ix − a)"−2 (y − b)3...+ t (y — b)" = 0 denotes n right lines passing through the point (a, b).

Ex. 1. What locus is represented by the equation xy = 0?

Ans. The two axes; since the equation is satisfied by either of the suppositions x = 0, y = 0,

Ans. The bisectors of the angles between the axes, x+y=0 (see Art. 35).

Ex. 3. What locus is represented by x25xy + 6y2=0? Ans. x-2y=0, x-3y=0. Ex. 4. What locus is represented by x2 - 2xy sec0+ y2 = 0?

Ans. x y tan (45° ± 10).

Ex. 5. What lines are represented by x2

2xy tan 0y2 = 0?

Ex. 6. What lines are represented by x3 - 6x2y + 11xy2 - 6y3 = 0?

73. Let us examine more minutely the three cases of the solution of the equation x-pxy + qy2 = 0, according as its roots are real and unequal, real and equal, or both imaginary.

The first case presents no difficulty: a and b are the tangents of the angles which the lines make with the axis of y (the axes being supposed rectangular), p is therefore the sum of those tangents, and q their product.

In the second case, when ab, it was once usual among geometers to say that the equation represented but one right line (x — ay = 0). We shall find, however, many advantages in making the language of geometry correspond exactly to that of algebra, and as we do not say that the equation above has only one root, but that it has two equal roots, so we shall not say that it represents only one line, but that it represents two coincident right lines.

Thirdly, let the roots be both imaginary. In this case no real coordinates can be found to satisfy the equation, except the coordinates of the origin x = 0, y = 0; hence it was usual to say that in this case the equation did not represent right lines, but was the equation of the origin. Now this language appears to us very objectionable, for we saw (Art. 14) that two equations

are required to determine any point, hence we are unwilling to acknowledge any single equation as the equation of a point. Moreover, we have been hitherto accustomed to find that two different equations always had different geometrical significations, but here we should have innumerable equations, all purporting to be the equation of the same point; for it is obviously immaterial what the values of p and q are, provided only that they give imaginary values for the roots, that is to say, provided that p2 be less than 49. We think it, therefore, much preferable to make our language correspond exactly to the language of algebra; and as we do not say that the equation above has no roots when p2 is less than 49, but that it has two imaginary roots, so we shall not say that, in this case, it represents no right lines, but that it represents two imaginary right lines. In short, the equation x2 - · pxy + qy* = 0 being always reducible to the form (x — ay) (x — by) = 0, we shall always say that it represents two right lines drawn through the origin; but when a and b are real, we shall say that these lines are real; when a and b are equal, that the lines coincide; and when a and b are imaginary, that the lines are imaginary. It may seem to the student a matter of indifference which mode of speaking we adopt; we shall find, however, as we proceed, that we should lose sight of many important analogies by refusing to adopt the language here recommended. Similar remarks apply to the equation

Ax2 + Bxy + Cy2 = 0,

which can be reduced to the form x2 - pxy + qy2 = 0, by dividing by the coefficient of x2. This equation will always represent two right lines through the origin; these lines will be real if B-4AC be positive, as at once appears from solving the equation; they will coincide if B - 4A C=0; and they will be imaginary if B-4AC be negative. So, again, the same language is used if we meet with equal or imaginary roots in the solution of the general homogeneous equation of the nth degree.

74. To find the angle contained by the lines represented by the equation x-pxy + qy2 = 0.

Let this equation be equivalent to (x - ay) (x-by) = 0, then a- - b the tangent of the angle between the lines is (Art. 25) 1+ ab

but the product of the roots of the given equation = q, and their difference = √(p-4q). Hence

COR. The lines will cut at right angles, or tan & will become infinite, if q=-1 in the first case, or if A+ C= 0 in the second.

*If the axes be oblique we find, in like manner,

sin ∞ √(B2-4A C)

Ans. 45° Ans. 0.

tan &=

75. To find the equation which will represent the lines bisecting the angles between the lines represented by the equation

μ

Ax2 + Bxy + Cy' = 0.

Let these lines be x-ay=0, x-by=0; let the equation of the bisector be x-μy = 0, and we seek to determine μ. Now (Art. 18) is the tangent of the angle made by this bisector with the axis of y, and it is plain that this angle is half the sum of the angles made with this axis by the lines themselves. Equating, therefore, tangent of twice this angle to tangent of sum, we get

This gives us a quadratic to determine μ, one of whose roots will be the tangent of the angle made with the axis of y by the internal bisector of the angle between the lines, and the other the tangent of the angle made by the external bisector. We can find the combined equation of both lines by substituting in and we get

the last quadratic for μ its value =

x

y

xy-y=0,*

and the form of this equation shows that the bisectors cut each other at right angles (Art. 74).

The student may also obtain this equation by forming (Art. 35) the equations of the internal and external bisectors of the angle between the lines x-ay=0, x-by=0, and multiplying them together, when he will have

and then clearing of fractions, and substituting for a +b, and ab their values in terms of A, B, C, the equation already found is obtained.

76. We have seen that an equation of the second degree may represent two right lines; but such an equation in general cannot be resolved into the product of two factors of the first degree, unless its coefficients fulfil a certain relation, which can be most easily found as follows. Let the general equation of the second degree be written

or

ax2 + 2hxy + by* + 2gx+2fy + c = 0,†

ax2 + 2 (hy+g) x + by3 + 2fy + c = 0.

It is remarkable that the roots of this last equation will always be real, even the roots of the equation Ax2 + Bxy + Cy2 = 0 be imaginary, which leads to the curious result, that a pair of imaginary lines has a pair of real lines bisecting the angle between them. It is the existence of such relations between real and imaginary lines which makes the consideration of the latter profitable. † It might seem more natural to write this equation

ax2 + bxy + cy2 + dx + ey +ƒ = 0,

but as it is desirable that the equation should be written with the same letters all through the book, I have decided on using, from the first, the form which will hereafter be found most convenient and symmetrical. It will appear hereafter

Solving this equation for x we get

ax=- (hy + g) ± √√ {{h2 — ab) y2 + 2 (hg — aƒ) y + (g2 — ac)}.

In order that this may be capable of being reduced to the form x = my+n, it is necessary that the quantity under the radical should be a perfect square, in which case the equation would denote two right lines according to the different signs we give the radical. But the condition that the radical should be a perfect square is

Expanding, and dividing by a, we obtain the required condition, viz. abc+2fgh — af* — bg* — ch2 = 0.*

1. Verify that the following equation represents right lines, and find the lines: x2 - 5xy +4y2 + x + 2y − 2 = 0.

Ans. Solving for x as in the text, the lines are found to be

xy-10, x- -4y+2=0.

Ex. 2. Verify that the following equation represents right lines: (ax+ By - r2)2 = (a2 + B2 — r2) (x2 + y2 — r2).

Ex. 3. What lines are represented by the equation

x2 - xy + y2 - x-y+1=0?

Ans. The imaginary lines x + Oy + 02 = 0, x + 02y+0=0, where 0 is one of the imaginary cube roots of 1.

Ex. 4. Determine h, so that the following equation may represent right lines:

x2 + 2hxy + y2 — 5x − 7y + 6 = 0.

Ans. Substituting these values of the coefficients in the general condition, we get for h the quadratic 12h2 - 35h + 25 = 0, whose roots are and 4.

*77. The method used in the preceding Article, though the most simple in the case of the equation of the second degree, is not applicable to equations of higher degrees; we therefore give another solution of the same problem. It is required to ascertain

that this equation is intimately connected with the homogeneous equation in three variables, which may be most symmetrically written

ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0.

The form in the text is derived from this by making z = 1. The coefficient 2 is affixed to certain terms, because formulæ connected with the equation, which we shall have occasion to use, thus become simpler and more easy to be remembered.

*If the coefficients f, g, h in the equation had been written without numerical multipliers, this condition would have been