[" 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)" ."]

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)=

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

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