`px + qy = p-q" "qx - py = p+q`
रेखिक समीकरणो के युग्मो को हल कीजिए :

Solve px+qy=p-q,qx-py=p+q

If px + qy + r = 0 and qx + py + r = 0 (x != y) , then show that the value of x + y " is " (-r)/(p) or (-r)/(q)

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 :

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

Three lines px + qy+r=0, qx + ry+ p = 0 and rx + py + q = 0 are concurrent of

Prove that px^(q - r) + qx^(r - p) + rx^(p - q) gt p + q + r where p,q,r are distinct number and x gt 0, x =! 1 .

Solve for x and y, px + qy = 1 and qx + py = ((p + q)^(2))/(p^2 + q^2)-1 .

If the equations 4x + py = 12 and qx + 3y = 6 are dependent, then the value of p and q are ____ and respectively