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        Prove that the distance from the origin to the orthocentre of the triangle formed by the lines x/m + y/n =1 and ax^2 + 2hxy + by^2 =0 is                (m^2 + n^2)^1/2 {[(a + b) mn] / [am^2 – 2hmn + bn^2]}
2 months ago

							Let H = (p, q) is orthocenter.x/m + y/n =1  => nx + my – mn = 0      --------(1)Opposite vertex to this side is Origin (0, 0).equation to any line perpendicular to eq (1) and passing through origin is        mx-ny = 0  ------(2)Now, Length of one altitude OC which is perpendicular to eq (1) is the perpendicular length from origin to eq (1) i,e.  OC =            =                   as    H lie on this line OC and we have OH = OC – HCHC is the perpendicular distance from point H to eq (1)Which is  HC =            =             as    OH = OC – HC    =          –                                =                ----------(3)Now, let              contains    y – v1x = 0    -----(4) and y – v2x = 0     --------(5)  as its pair of straight lines.This implies that        v1 + v2 = -2h/b      -------(6)     and     v1v2 = a/b      -------(7)Solving eq(6) and (7), we get   v1 =         v2 =  Solving eq (1) and (4) we get the coordinate of another vertex of the triangle. Let us name the vertex and A,then A =           The slope of eq (5) which is the third side of the triangle is    v2 .  the slope of any line perpendicular to (5) is  – 1/v2the equation to a line with point A and slope -1/v2 is                 ---------(8)eq (2) and (8) represent two altitudes of the given triangle.Therefore after substituting v1 and v2    values in eq (8) and solving eq (2) and (8)We get H = (p, q) =          Substituting  p and q values in eq (3) we getOH =      Hence proved.

2 months ago
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