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Grade: 12th pass
        
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]}
7 months ago

Answers : (1)

Khaled bin seleem
12 Points
							
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 =         \left | -mn \right | / (m^{2}+n^{2})^{1/2}   =    mn / (m^{2}+n^{2})^{1/2}               as   \left | a \right | = \pm a
 
H lie on this line OC and we have OH = OC – HC
HC is the perpendicular distance from point H to eq (1)
Which is  HC =    \bg_white \left | np+mq-mn \right | / (m^{2}+n^{2})^{1/2}        =        \bg_white (-np-mq+mn) / (m^{2}+n^{2})^{1/2}     as    \left | a \right | = \pm a
OH = OC – HC    =      mn / (m^{2}+n^{2})^{1/2}    –    \bg_white (-np-mq+mn) / (m^{2}+n^{2})^{1/2}
                            =      \bg_white (np+mq) / (m^{2}+n^{2})^{1/2}          ----------(3)
Now, let       ax^{2}+2hxy+by^{2}=0       contains    y – v1x = 0    -----(4) and y – v2x = 0     --------(5)  as its pair of straight lines.
This implies that        v+ v2 = -2h/b      -------(6)     and     v1v2 = a/b      -------(7)
Solving eq(6) and (7), we get   v1 = \frac{(h^{2} - ab)^{1/2} - h}{b}        v2 = \frac{-((h^{2} - ab)^{1/2} + h)}{b}
 
Solving eq (1) and (4) we get the coordinate of another vertex of the triangle. Let us name the vertex and A,
then A =          (\frac{mn}{n +m v_{1}}, \frac{v_{1}mn}{n +m v_{1}})
 
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/v2
the equation to a line with point A and slope -1/v2 is        x + v_{2}y - \frac{v_{1}v_{2}mn + mn}{n + v_{1}m} = 0         ---------(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) =         (\frac{mn^{2} (a + b)}{am^{2} - 2hmn + bn^{2}} , \frac{m^{2}n (a + b)}{am^{2} - 2hmn + bn^{2}} )
 
Substituting  p and q values in eq (3) we get
OH =    (m^{2}+n^{2})^{1/2}(\frac{(a+b)mn}{am^{2}-2hmn+bn^{2}})
 
 
Hence proved.
 
7 months ago
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