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raised to a suitable power. Thus, the equation B2 +hy+k+=0) is equivalent to the homogeneous equation
44°22 + 2Ahy (aa + bB+ cy) + k* (aa + b3 + cry)2 = 0.
The following examples may familiarize the reader with this system of co-ordinates.
1. Prove that the co-ordinates of the middle point of the line BC are 0,
2. The co-ordinates of the centre of the circumscribed circle are R cos A, R cos B, R cos C, where
a cos A + b cos B + c cos ("*
3. The co-ordinates of the centre of the inscribe circle are
2A each equal to
a + b + c What are the co-ordinates of the centres of the escribed circles ?
4. The co-ordinates of the centre of gravity are
2A 2A 2A
Prove that a sin A + B sin B + y sin C is equal to where R is the radius of the circumscribing circle.
3. To find the distance between two given points, in terms of their trilinear co-ordinates.
Let 04, B., %; Qg, B2, yz, be the co-ordinates of two given points, r the distance between them.
Then, gut will be a rational integral function of a, - og, B.-By, - Yg, of the second degree*.
* This, if not self-evident, may be proved as follows: Let P, Q be the two given points. Join PQ, and draw PM, QM' perpen.
aa, + bß, + cy, = 24,
aa, + bB, + cy, =24;
6 - Q.)' = 7. ) ) -a Similar expressions may be found for (B.-B.), (Y. - Y).
Hence, pa will be of the form 2 (B-B2) (7.-Y) + m (Y. - Y) (Qz - Q) +n (az -22) (B-B2),
dicular to AB, PN, QN' to AC. Draw Qm perpendicular to PM, Qn to PN, and join mn. Then
.. mn=(B1-B2)' +(71-72) +2(B1-B2)(71-72) cos A,
(B1-B2)+(71-72)2 + 2(B1 – B2)(71-72) cos A whence ge
sin? A a rational integral function of the second degree.
where l, m, n are certain functions of a, b, c, which we proceed to determine.
Since the values of l, m, n are independent of the positions of the points, the distance of which we wish to find, suppose these points to be B and C. Then
2A Q=0, B.=
2A Ag = 0, B,= 0, Y,= Hence
24 2A a*=-7
+c(Qz - Q) (B.-B.)}.
+c cos C (y - y)).
To find the equation of the straight line drawn through one of the angular points of the triangle of reference, so as to bisect the opposite side.
Let D be the middle point of the side BC, we have then to investigate the equation of the straight line AD.
In AD take any point P, and let a, B, y be its co-ordi
nates. From D, P draw DE, PG perpendicular to AC, DF, PH perpendicular to AB. Then by similar triangles
PG : DE :: PH : DF.
DE, AC=DF. AB,
bB=cy. This is a relation between the co-ordinates of any point on the line AD, it therefore is the equation of that line.
COR. It hence may be proved that the three straight lines, drawn through the angular points of a triangle to bisect the opposite sides, intersect in a point. For these straight lines will be represented by the equations
αα = bß, and, therefore, all pass through the point for which
In the next three propositions the reader will easily be able to draw a figure for himself, by comparison with fig. 5.
5. To find the equation of the straight line drawn through one of the angular points of the triangle of reference, perpendicular to the opposite side.
Making a construction similar to that in the last proposition, it will be seen that we have here
each of these fractions being equal to the area of the triangle
B cos B=y cos C.
This will be the equation of the straight line, drawn through A, at right angles to BC.
COR. It may hence be shewn that the three straight lines drawn through the angular points of a triangle, perpendicular to the opposite sides, intersect in the point determined by the equations
a cos A = ß cos B=y cos C.
6. To find the equations of the internal and external bisectors of an angle of the triangle of reference.
For the internal bisector of the angle A, we shall have, making the same construction as before,
The straight line will be therefore represented by the equation B=y.