# locus sum

• Mar 5th 2011, 10:33 PM
prasum
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
• Mar 6th 2011, 01:12 AM
running-gag
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 $\displaystyle \vec{u}$(p;1+p)
(BC): (1+q)x-qy+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
• Mar 6th 2011, 10:26 AM
prasum
what is ch and ah here
• Mar 9th 2011, 10:21 AM
running-gag
These are vectors
I am sorry if I gave you some information that you cannot use(Doh)