An Elementary Treatise on Trilinear Co-ordinates: The Method of Reciprocal Polars, and the Theory of Projections |
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Page 3
... Hence , twice the area PBC will be represented by az , and we shall therefore have as before - aa + bB + cy = 2A . Thirdly , let P lie between AB , AC , produced backwards ( fig . 3 ) , so that B , y are negative while a is positive ...
... Hence , twice the area PBC will be represented by az , and we shall therefore have as before - aa + bB + cy = 2A . Thirdly , let P lie between AB , AC , produced backwards ( fig . 3 ) , so that B , y are negative while a is positive ...
Page 5
... Hence , r2 will be of the form - - 1 ( B1 − B2 ) ( Y1 − Y1⁄2 ) + m ( Y1 — Y1⁄2 ) ( α , — α ) + n ( ï ̧ − x1⁄2 ) ( B1 — B2 ) , - - dicular to AB , PN , QN ' to AC . Draw Qm perpendicular to PM , Qn to PN , and join mn . Then r = PQ ...
... Hence , r2 will be of the form - - 1 ( B1 − B2 ) ( Y1 − Y1⁄2 ) + m ( Y1 — Y1⁄2 ) ( α , — α ) + n ( ï ̧ − x1⁄2 ) ( B1 — B2 ) , - - dicular to AB , PN , QN ' to AC . Draw Qm perpendicular to PM , Qn to PN , and join mn . Then r = PQ ...
Page 6
... Hence Similarly 2A 2A a2 = -l ; a2bc 4A2 ab2c m = - 4A2 , abc2 n = - 4A ** abc Hence r2 4A2 - { a ( B1 — B2 ) ( Y1 — Y2 ) + 6 ( Y1 − Y2 ) ( α2 — α2 ) + c ( α , -α ) ( B1- B2 ) } . -- - This is one form of the expression for . It may ...
... Hence Similarly 2A 2A a2 = -l ; a2bc 4A2 ab2c m = - 4A2 , abc2 n = - 4A ** abc Hence r2 4A2 - { a ( B1 — B2 ) ( Y1 — Y2 ) + 6 ( Y1 − Y2 ) ( α2 — α2 ) + c ( α , -α ) ( B1- B2 ) } . -- - This is one form of the expression for . It may ...
Page 7
... Hence PG.AC - PH.AB , or 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 ...
... Hence PG.AC - PH.AB , or 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 ...
Page 8
... Hence PG Cos C PH = or cos B .. PG cos BPH cos C , B cos By 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 ...
... Hence PG Cos C PH = or cos B .. PG cos BPH cos C , B cos By 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 ...
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angular points asymptotes ax+bB+cy b₁ b₂ C₁ Cambridge centre Chap co-ordinates coefficients College common tangents conic section Crown 8vo determinant directrix Edition equation Fcap find the equation fixed point follows four points given conic given point given straight line Hence imaginary investigated Let the equation line at infinity line joining locus meets the conic nine-point circle numerous Illustrations obtain opposite sides Owens College pair parabola Pascal's Theorem perpendicular point f points of intersection pole Professor prove radical axis ratios rectangular hyperbola represented respect right angles second degree shewn similar and similarly sin POS tangents tangents drawn term a,b,c theorem three straight lines tion touches the line triangle of reference Ua² ux² V'ca v'f+u'g+wh va² values Vb² vß² W. K. CLIFFORD W'ab whence wy² λα