Hogg. — On certain Tri polar Relations. 347 



The following particular cases may be noted : — 



Euler's line 2 [(6^ - c") X] = o. 



Brocard diameter 2 [a? {If — c^) X] = o. 



Diameter through in-centre 2 [a (6 — c) X] = o. 



Diameter through symmedian point of medial triangle of triangle 



ABC 2 [cos A cos (B - C) X] = o. 

 Diameter parallel to BC aX — h cos C Y — c cos B Z = o. 

 Diameter through vertex A {h^ — c^) X - ^^Y + c-Z = o. 

 Diameter paiallel to tangent to circle ABC at vertex A a cos (B — C) X 



— h cos B Y — c cos C Z = 0. 



§ 3. If Z + TO -\- n = 0, the equation ZX + toY + »Z = k represents a 

 straight line parallel to ^X + mY + wZ = o. We may now find the 

 equation in tripolar co-ordmates of the line whose trilinear equation is 

 pa+ qfi + ry = o. 



Let the required equation be written 



L' = ZX + wY +w Z + 4 A V = 0, 



where I -\- m -\- n = o and A is the area of the triangle of reference. 



Transforming L' to trilinear co-ordinates and dividing out by 

 aa -f 6/8 + cy, we have 



a Vsin^B sin'^C/ h \sin^C sin^A/ c ysin'^^A sin^B 

 -\r K {aa + bft + Cy) = 0. 



Hence, comparing coefficients, we have 



sin^B sin^C 



sin^C sin'^A 



+ -.-vr7 + '^c' = Xrc, 



sin'^A sin^B 

 which give 



2Z + 4k A sin A cos A = A. sin^A {— pci + qb -\- re) 

 2m + 4kA sin B cos B = X siu^B {pa — qb + re) 

 2n + 4kA sin C cos C = A sin^C {pa + qb — re), 



whence, by addition, we obtain 



AT? 

 K = ^ — {p cos A + g cos B -f r cos C) 

 . ^ A 



21 = Xa {— p + q cos C -f r cos B) 



2w = Xb {p cos G — q + r cos A) 



2n — Xc {p cos B + g cos A — r). 

 Hence the required line is 



L' EZ (— p + q cos C + r cos B) aX + {p cos C — q -\- r cos A) 6Y 

 + {p cos B + (/ cos A — r) cZ = o. 



The tripolar equations of the sides of the triangle of reference are 

 aX — b cos C Y — c cos B Z = abc cos A 



— a cos C X + 6Y — c cos A Z = abc cos B 



— a cos B X — 6 cos A Y -f cZ = abc cos C. 



