Solve for `x and y` : `q/px+ p/qy = p^2 + q^2; x+y = 2pq`.

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

Solve for x and y by cross-multiplication : (i)px+qy=p-q,qx-py=p+q

Solve the following pair of linear equations : px + qy= p-q qx- py= p+q

The equations of the tangents drawn from the origin to the circle x^2 + y^2 - 2px - 2qy + q^2 = 0 are perpendicular if (A) p^2 = q^2 (B) p^2 = q^2 =1 (C) p = q/2 (D) q= p/2

The equations of the tangents drawn from the origin to the circle x^2 + y^2 - 2px - 2qy + q^2 = 0 are perpendicular if (A) p^2 = q^2 (B) p^2 = q^2 =1 (C) p = q/2 (D) q= p/2

Given that q^2-p r ,0,\ p >0, then the value of |p q p x+q y q r q x+r y p x+q y q x+r y0| is- a. zero b. positive c. negative \ d. q^2+p r