
locus sum
the locus of orthocentre of triangle formed by lines (1+p)xpy+p(1+p)=0,(1+q)xqy+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

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)xpy+p(1+p)=0 therefore one direction vector is $\displaystyle \vec{u}$(p;1+p)
(BC): (1+q)xqy+q(1+q)=0 therefore one direction vector is $\displaystyle \vec{v}$(q;1+q)
It is easy to show that A(p;0) and C(q;0)
Let H(x;y) be the orthocentre
$\displaystyle \vec{u} \cdot \vec{CH} = 0$ gives p(x+q)+(1+p)y=0 [1]
$\displaystyle \vec{v} \cdot \vec{AH} = 0$ 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


These are vectors
I am sorry if I gave you some information that you cannot use(Doh)