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1. In the systems of co-ordinates with which we have hitherto been concerned, we have considered a point as determined, directly or indirectly, by means of its distances from three given straight lines; and we have regarded a curve as the aggregation of all points, the co-ordinates of which satisfy a certain equation. It is equally possible, however, to consider a straight line as determined by means of its distances from three points, which distances may be termed its coordinates; and to regard à curve as the envelope of all straight lines, the co-ordinates of which satisfy a certain equation.
This system is closely connected with the theory of reciprocal polars. In fact, it may be looked upon as a means of so interpreting equations as at once to obtain the results which the method of reciprocal polars would deduce from the ordinary method of interpretation.
We may then define the co-ordinates of a straight line to be the perpendiculars let fall upon it from three given points A, B, C. The lengths of these perpendiculars we will denote by the letters p, q, r, respectively, the lengths BC, CA, AB being represented as before by the letters a, b, c, and the angles of the triangle of reference ABC being denoted by A, B, C, and its area by A.
2. Any two co-ordinates, q and r for example, will be considered to have contrary signs if the line of which they
are the co-ordinates cuts the line BC in a point lying between B and C, otherwise to have the same sign. Thus, the internal bisector of the angle A has its co-ordinates of contrary signs, the external bisector of the same sign. The sign of p relatively to q and r will be determined in the same
If D be any point on the line BC, q, r, the co-ordinates of any line passing through it, and BD =a, CD=2,, distances measured along the line BC from B to being considered positive, and from C to B negative, it will readily be seen that
a a, Since this is a relation between the co-ordinates of any line passing through the point D, it may be considered as the equation of the point D.
If D be the middle point of BC, a, =-2g, hence it appears that the middle points of the sides of the triangle of reference are represented by the equations,
q+r=0, r+p=0, p+q=0. It may also be proved that the points where the internal bisectors of the angles meet the opposite sides, are represented by
bq + cr=0, cr+ ap= 0, ap+bq = 0. The points where the external bisectors of the angles meet the opposite sides, by
bq - cr=0, cr- ap=0, ap – bq=0. The feet of the perpendiculars from the angular points on the opposite sides, by q tan B+r tan C= 0, gtan C+p tan A = 0,
ptan A +q tan B=0. The points of contact of the inscribed circle, by a +2=0, L+
3. We shall next prove the following proposition; that if O be any given point within the triangle ABC, then the co-ordinates p, q, r°(their signs being taken in the manner already explained) of any straight line passing through it, will be connected by the following equation,
Let the angle AOP=0; then, considering p as negative, and q and r as positive,
2+ BO cos A OB.sin 0 – BO sin A OB.cos 0 =0,
Hence, eliminating sin 0 and cos 0, -p.BO.CO sin BOC-9.00.A O sin COA
-r.A0.BO sin A OB=0, whence
ABOC.p+ACOA.q+AAOB.r=0. This equation may be regarded as the equation of the point o.
A similar equation may be proved to hold for any point without the triangle, BOC being considered negative, if A and O be on opposite sides of BC.
The following are the equations of some important points connected with the triangle of reference: Centre of gravity,
p +9+r=0. Centre of circumscribing circle, psin 2 A+qsin 2B+rsin 2C=0. Centre of inscribed circle,
ap+bq + cr=0. Centres of escribed circles,
ap – bq+cr=0,
ap+bq - cr=0. Intersection of perpendiculars,
ptan A +q tan B+r tanC= 0.
4. We proceed to investigate the identical relation which holds between the co-ordinates of any straight line.
Let any straight line cut the sides AB, AC of the triangle of reference in Ď, E. From A, B, C let fall AP, BQ, CR, perpendiculars on the line, then BQ=4, CR=r, AP=-p.
Through A draw Q'R' parallel to DE, and produce BQ, CR, to meet it in Q, R'.
BQ=q-p, CR=r-p, and sin Q'AB=95%, sin R’AC="/?, and BAC= A.
Hence cos Q'AB cos R'AC- sin Q AB sin R'AC=-cos A,
(4-p) (r-p) cos Q AB cos R'AC=
- cos A.......(1). bc
cos Q AB= cos (BAC +R'AC)
= cos A.cos R'AC-"72sin A ....(2)
cos R'AC=cos A.cos Q'AB-L-P sin A .....(3).
Hence, by (2) and (3),