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 lines px +qy+r=0, qx + ry + p =0,rx + py+q=0, are concurrant then

The lines px +qy+r=0, qx + ry + p =0,rx + py+q=0, are concurrant then

The lines px +qy+r=0, qx + ry + p =0,rx + py+q=0, are concurrant then

Let the homogeneous system of linear equations px+y+z=0,x+qy+z=0, and x+y+rz=0 ,where p,q,r!=1, have anon-zero solution,then the value of (1)/(1-p)+(1)/(1-q)+(1)/(1-r) is

If the equations px^(2)+qx+r=0 and rx^(2)+qx+p=0 have a negative common root then the value of (p-q+r)=

If the roots of the equation q^2 x^2 + p^2 x + r^2 =0 are the squares of the roots of the equation qx^2 + px + r=0 , then p,q,r are in ………………….. .