given eq of line are
(1 + p)x – py + p(1+p) = 0 ...(i)
(1 + q)x – qy + q(1+q) = 0 ...(ii)
y = 0 ...(iii)
solving eq (i) and (ii) we get
C(pq, (1+p)(1+q))
and A(-p,0) and B(-q,0)
equation of altitude CM passing through C and perpendicular to AB is
x = p q ..(iv)
slope of line (ii) is (1+q)/q
slope of altitude BN (N is on line AC) is -q/(1+q)
equation of BN is
....(v)
let H be orthocenter (h,k) intersection of line (iv) and (v)
solve this
x = pq and y = -pq
so, h=pq and y=-pq
x+y=0
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