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Math Help - locus sum

  1. #1
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    locus sum

    the locus of orthocentre of triangle formed by lines (1+p)x-py+p(1+p)=0,(1+q)x-qy+q(1+q)=0 and y=0 where p not equal to q is


    i know the locus of orthocentre in given question is a straight line by eliminating p and q but the procedure is too long.is there any formula for finding orthocentre
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  2. #2
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    Hi

    Here is a proposal (hope you won't find it too long ^^)
    Let ABC the triangle with A and C lying on the line y=0
    (AB): (1+p)x-py+p(1+p)=0 therefore one direction vector is \vec{u}(p;1+p)
    (BC): (1+q)x-qy+q(1+q)=0 therefore one direction vector is \vec{v}(q;1+q)
    It is easy to show that A(-p;0) and C(-q;0)
    Let H(x;y) be the orthocentre
    \vec{u} \cdot \vec{CH} = 0 gives p(x+q)+(1+p)y=0 [1]
    \vec{v} \cdot \vec{AH} = 0 gives q(x+p)+(1+q)y=0 [2]

    [1]-[2] gives x+y=0, which means that the locus of H is a straight line
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  3. #3
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    what is ch and ah here
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  4. #4
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    These are vectors
    I am sorry if I gave you some information that you cannot use
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