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For the external bisector we proceed as follows. Let Q be any point on the line, a, B, y its co-ordinates. Draw QK perpendicular to AC, QL to AB. Then, as before, we have

QK = QL. It will however be observed, that if Q and B lie on the same side of AC, Q and C will lie on opposite sides of AB, and vice versa. Hence, if

QK=B, QL=-9.

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We have therefore

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wn

B+y=0 as the equation of the line AQ, which externally bisects the angle A.

From the form of these equations we see, (1), That the three internal bisectors of the angles of a triangle intersect in a point; (2), That the internal bisector of any one angle, and the external bisectors of the other two, also intersect in a point.

These points may be shewn to be respectively the centres of the inscribed and escribed circles.

We shall hereafter prove that the points, in which the external bisectors of each angle respectively intersect the sides opposite to them, lie in the same straight line; and also that the points in which the external bisector of any one angle, and the internal bisectors of the other two angles, intersect the sides respectively opposite to them, lie in the same straight line.

7. We now proceed to investigate the general equation of a straight line.

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Every straight line may be represented by an equation of the first degree.

Let Q be any point on the straight line AC, R on AB, and P any point on the straight line QR, then we have to

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investigate the relation between the co-ordinates (a, b, y) of the point P.

Fig. 6.

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The property of the straight line, which we shall make the basis of our investigation, is, that it is the locus of a point which moves in such a manner, that the sum of the areas of the triangles PAQ, PAR is constant.

Let AQ=9, AR=r, then the areas of the triangles PAQ, PAŘ will be respectively represented by 19B, try, and the area of QAR by 64 Δ.

2qr Hence

+ roy=

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bc

=

qr
(aa + b3 + cy).

BB

bc This is the equation of the straight line QR, and, since it involves the two arbitrary quantities q, r, it is in the most general form of the equation of the first degree between two variables. Putting

gra qr
= 1,

-q=m, 4-r=n,
bc
the equation may be written

la + + ny=0.

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8. We shall next establish the converse proposition, that every equation of the first degree represents a straight line.

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Let

la + + ny = 0 be the general equation of the first degree, and let f, g, h be the co-ordinates of any fixed point D on the locus of the equation, a, b, u those of any point P.

Draw DE, PM perpendicular to AC, DF, PN perpendicular to AB. Also draw Dm, Dn, perpendicular respectively to PM, PN.

Then Pm=B-9, Pn=y-h.
Also, since f, g, h is a point on the locus,

If + mg + nh=0,
whence la-f) + m (B-9) +n (y-h) = 0,
Again,

ag + b3 + cy=2A,

af + bg +ch=24;
.. a (Q-f) +b (B - g) + (y-h) =0;

a-f B-9

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yーh am - bl

Hence, the ratio of Pm to Pn is constant, whatever point on the locus P may represent. This can only be true when that locus is a straight line.

9. To find the co-ordinates of the point of intersection of two given straight lines. Let the equations of the two straight lines be

la + mp + ny=0,

J'a+ m'8+ m = 0.
Where these intersect, we have

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g mn' - m'n ni' n'l Im' - I'm These equations, combined with

ar + b3 + cy=2A, give the values of a, b, y, at the point of intersection.

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10. To find the equation of the straight line, passing through two given points.

Let f, g, h; f'og', h', be the co-ordinates of the two given points, and

suppose the equation of the required straight line to be

La + MB + Ny=0. We must then have

Lf + Mg + Nh= 0,

Lf' + Mg' + Nh'=0; whence

L
M

N
gh' - g'h hf' h'f fi' f'g'
giving, as the equation of the required line,

(gh' -g'h) a + (hf' h'f) B+(fg' -f'g) y=0.

11. To find the general equation of a straight line, passing through the point of intersection of two given straight lines.

If the equations of the straight lines be

la + + ny = 0,
l'a+m' B+ny=0,

every straight line, passing through their point of intersection, may be represented by an equation of the form

la + + ny=k (l'a + m'ß + n'y), where k is an arbitrary constant. For this equation is satisfied when the equations of the given straight lines are both satisfied, and, being of the first degree, it represents a straight line. It is therefore the equation of a straight line passing through their point of intersection.

12. To find the condition that three points may lie in the same straight line,

Let az, B , %; Qz, B., Y,; as, Bg, Yo, be the co-ordinates of the three given points, then, if these points lie in the same straight line, suppose the equation of that line to be

λα + μβ + γ = 0.

Then , M, v must satisfy the following equations :

aa, +, + 2y = 0,
12, + , + 272 = 0,

λας + μβ, + νY, = 0,
whence, eliminating 2, l, v, by cross multiplication,

QB:7: - 0,337, +aß.7. 2.B.72 + ax B.72 QBxy = 0,

the required condition.

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