Geo3.21

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If ax^{2}+2hxy+by^{2}+2gx+2fy+c=0\,represents a pair of lines then show that the square of the distance from the origin to their point of intersection is {\frac  {c(a+b)-f^{2}-g^{2}}{ab-h^{2}}}\,

Let the equations be l_{1}x+m_{1}y+n_{1}=0,l_{2}x+m_{2}y+n_{2}=0\, let the equations be 1 and 2.

(l_{1}x+m_{1}y+n_{1})(l_{2}x+m_{2}y+n_{2})\equiv ax^{2}+2hxy+by^{2}+2gx+2fy+c\,

l_{1}l_{2}=a,m_{1}m_{2}=b,n_{1}n_{2}=c,l_{1}m_{2}+l_{2}m_{1}=2h,l_{1}n_{2}+l_{2}n_{1}=2g,m_{1}n_{2}+m_{2}n_{1}=2f\,

Solving the equations 1 and 2,

{\frac  {x}{m_{1}n_{2}-m_{2}n_{1}}}={\frac  {y}{l_{2}n_{1}-l_{1}n_{2}}}={\frac  {1}{l_{1}m_{2}-l_{2}m_{1}}}\,

The point of intersection is P[{\frac  {m_{1}n_{2}-m_{2}n_{1}}{l_{1}m_{2}-l_{2}m_{1}}},{\frac  {l_{2}n_{1}-l_{1}n_{2}}{l_{1}m_{2}-l_{2}m_{1}}}]\,

OP^{2}={\frac  {(m_{1}n_{2}-m_{2}n_{1})^{2}+(l_{2}n_{1}-l_{1}n_{2})^{2}}{(l_{1}m_{2}-l_{2}m_{1})^{2}}}\,

Simplifying the erxpression,we have

{\frac  {4f^{2}-4bc+4g^{2}-4ca}{4h^{2}-4ab}}={\frac  {c(a+b)-f^{2}-g^{2}}{ab-h^{2}}}\,


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