If `p+1/p=q,` then for `p > 0`

The locus of the orthocentre of the triangle formed by the lines (1+p) x-py + p(1 + p) = 0, (1 + q)x-qy + q(1 +q) = 0 and y = 0, where p!=q , is

[[ 0 , p-q , p-r ],[q-p , 0 , q-r],[ r-p , r-q , 0 ]] =

The locus of the orthocentre of the triangle formed by the lines : (1 + p)x - py + p(1+ p) = 0, (1 + q)x- qy + q(1+ q) = 0, and y = 0 , where p ne q , is :

(p^^~q) is logically equal to 1. p -> q 2. ~ p->q 3. p -> ~ q 4. ~(p ->q)

If p != 0, q != 0 and the roots of x^(2) + px +q = 0 are p and q, then (p, q) =

If p != 0, q != 0 and the roots of x^(2) + px +q = 0 are p and q, then (p, q) =

If p, q, are real and p ne q then the roots of the equation (p-q) x^(2)+5(p+q) x-2(p-q)=0 are