Geo2.38

The three straight lines $px+qy+r=0,qx+ry+p=0,rx+py+q=0\,$ are concurrent if $p+q+r=0\,$

Let $(\alpha,\beta)\,$ be the point of concurrence of the given three lines

Hence $p\alpha+q\beta+r=0,q\alpha+r\beta+p=0,r\alpha+p\beta+q=0\,$

Let these equations be 1,2,3.

Solving 1 and 2,

$\frac{\alpha}{pq-r^2}=\frac{\beta}{qr-p^2}=\frac{1}{rp-q^2}\,$

Therefore $\alpha=\frac{pq-r^2}{rp-q^2},\beta=\frac{qr-p^2}{rp-q^2}\,$

Since $(\alpha,\beta)\,$ lies on equation 3, we have

$r[\frac{pq-r^2}{rp-q^2}]+p[\frac{qr-p^2}{rp-q^2}]+q=0\,$

$p^3+q^3+r^3-3pqr=0\,$

$(p+q+r)(p^2+q^2+r^2-pq-qr-rp)=0\,$

$p+q+r=0\,$

If $(p^2+q^2+r^2-pq-qr-rp)=0\,$

$\frac{1}{2}[(p-q)^2+(q-r)^2+(r-p)^2]=0\,$ this implies that

$(p-q)^2=0,(q-r)^2=0,(r-p)^2=0\,$ this means $p=q=r\,$ which is absurd.