AREA OF THE TRIANGLE OF REFERENCE. 3 negative, while B, y are positive. Hence, twice the area PBC will be represented by aa, and we shall therefore have as before aa+bB+cy=2A. Thirdly, let P lie between AB, AC, produced backwards (fig. 3), so that ẞ, y are negative while a is positive. Twice Fig. 3. B the areas of PBC, PCA, PAB, are now represented by aa, - bß, - cy respectively, so that we still have aa+bB+cy=2A. In all cases, therefore, aa+bB+cy=2A. The importance of the above proposition arises from its enabling us to express any equation in a form homogeneous with respect to the trilinear co-ordinates of any point to which it relates. Any locus may be represented, as in the ordinary system, by means of a relation between two coordinates, B and y for example, and this may be made homoax+bB+cy geneous in a, B, y by multiplying each term by 24 raised to a suitable power. Thus, the equation B2+hy+ k2 = 0 is equivalent to the homogeneous equation 4A3ß2+2Ahy (ax+bB+cy) + k2 (ax+bB+cy)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, A A 2. The co-ordinates of the centre of the circumscribed circle are R cos-A, R cos B, R cos C, where 3. The co-ordinates of the centre of the inscribed circle are 2A a+b+c each equal to What are the co-ordinates of the centres of the escribed circles? 4. The co-ordinates of the centre of gravity are 5. Prove that a sin A + ẞ 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 a1, B12.1; 2, B2, Y2, be the co-ordinates of two given points, r the distance between them. Then, will be a rational integral function of a, -a,, B1- B2, Y12, 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 Similar expressions may be found for (B,- B1), (Y1-2)2. Hence, 2 will be of the form 1 dicular to AB, PN, QN' to AC. Draw Qm perpendicular to PM, Qn to PN, and join mn. Then •. mn2= (ẞ1 − ẞ2)2 + (Y1 − Y2)2 + 2(B1 - B2) (Y1 - Y2) cos A, whence r2= 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 This is one form of the expression for 2. It may also be proved in a similar manner that 4. We next proceed to investigate the equation of a straight line; and first, we shall consider the cases of certain straight lines bearing important relations to the triangle of reference. 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. EQUATIONS OF THE BISECTORS. 7 In AD take any point P, and let a, B, y be its co-ordiFig. 5. nates. From D, P draw DE, PG perpendicular to AC, DF PH perpendicular to AB. Then by similar triangles for each is equal to the area of the triangle ABC. or Hence PG.AC-PH. 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 bB = cy, cy = aa, aa=BB, and, therefore, all pass through the point for which aa= =bB=cy. |