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but the product of the roots of the given equation =q, and their

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If the equation had been given in the form
Ax2 + Bxy + Cy2 = 0,

it would have been found that

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COR. The lines will cut at right angles, or tan & will become

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1 in the first case, or if A+ C=0 in the second.

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

Ans. 45°

Ans. 0.

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

му

Ax2 + Вxy + Cy2 = 0.

=

Let these lines be x-ay=0, x-by=0; let the equation of the bisector be x 0, and we seek to determine μο Now (Art. 18) μ 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

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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 =

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X

y

xy-y2=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

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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 + by2+2gx + 2fy + c = 0,†

ax2 + 2 (hy + g) x + by2+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 — af) y + (g3 − ac)}.

In order that this may be capable of being reduced to the form xmy+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

(h2 — ab) (g2 — ac) = (hg — af)3.

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

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Ex. 2. Verify that the following equation represents right lines:

(ax + By — p2)2 = (a2 + ß2 — p2) (x2 + y2 — p2).

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 3 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

4abc+fgh-af2 — bg2 — ch2 = 0.

whether the given equation of the second degree can be identical with the product of the equations of two right lines

(ax+ By − 1) (a'x + B'y − 1) = 0.

Multiply out this product, and equate the coefficient of each term to the corresponding coefficient in the general equation of the second degree, having previously divided the latter by c, so as to make the absolute term in each equation = 1. We thus obtain five equations, viz.

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from which eliminating the four unknown quantities a, a', B, B', we obtain the required condition. The first four of the equations at once give us two quadratics for determining a, a'; ß, ß'; which indeed might have been also obtained from the consideration that these quantities are the reciprocals of the intercepts made by the lines on the axes; and that the intercepts made by the locus on the axes are found (by making alternately x = 0, y=0, in the general equation) from the equations

ax2+2gx+c=0, by2+2fy+c=0.

We can now complete the elimination by solving the quadratics, substituting in the fifth equation and clearing of radicals; or we may proceed more simply as follows: Since nothing shews whether the root a of the first quadratic is to be combined with

the root B or B' of the second, it is plain that

either of the values aß' + a'ß or aß + a'ß'. geometrically, since if the locus meet the

2h

с

may have

This is also evident

axes in the points

L, I'; M, M'; it is plain that if it represent right lines at all, these must be either the pair LM, L'M', or else LM', L'M, whose equations are

(ax+ By−1) (a'x + B'y−1) = 0, or (ax + B'y − 1) (a'x+By − 1) = 0. The sum then of the two quantities aß' + a'ß, aß + a'ß'

and their product

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a

= aa′ (B2 + ß'2) + ßß′ (a" + a") = a (4f" — 2bc) + b (4ga — 2ac)

BB'

c2

2

c2

L

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which, cleared of fractions, is the condition already obtained.

Ex. To determine h so that x2 + 2hxy + y2 — 5x − 7y+6=0 may represent right lines (see Ex. 4, p. 72).

The intercepts on the axes are given by the equations

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whose roots are x = 2, x = 3; y = 1, y = 6. Forming, then, the equation of the lines joining the points so found, we see that if the equation represent right lines, it must be of one or other of the forms

(x + 2y - 2) (2x + y − 6) = 0, (x + 3y − 3) (3x + y − 6) = 0,

whence, multiplying out, h is determined.

*78. To find how many conditions must be satisfied in order that the general equation of the nth degree may represent right lines. We proceed as in the last Article; we compare the general equation, having first by division made the absolute term = 1, with the product of the n right lines

= 0.

(ax + By − 1) (a'x + B'y − 1) (a′′x + B'y − 1) &c. Let the number of terms in the general equation be N; then from a comparison of coefficients we obtain N-1 equations (the absolute term being already the same in both); 2n of these equations are employed in determining the 2n unknown quantities a, a', &c., whose values being substituted in the remaining equations afford N-1-2n conditions. Now if we write the general equation

A

+ Bx + Cy

+ Dx2 + Exy + Fy*

+ Gx3 + Hx2y + Kxy2 + Ly3

+ &c. = 0,

it is plain that the number of terms is the sum of the arithmetic series

N=1+2+3+... (n + 1) (n + 1) (n + 2);

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=

1.2

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