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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\,




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.

Main Page:Geometry:Straight Lines-I