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Show that the two pairs of lines 12x^{2}+7xy-12y^{2}=0,12x^{2}+7xy-12y^{2}-x+7y-1=0\, form a square.

The equation 12x^{2}+7xy-12y^{2}=0\, represents a perpendicular pair from the origin. equation 1

The pair 12x^{2}+7xy-12y^{2}-x+7y-1=0\, is parallel to equation above. Equation 2.

Again the point of intersection is \left({\frac  {hf-bg}{ab-h^{2}}},{\frac  {gh-af}{ab-h^{2}}}\right)\,

\left({\frac  {1}{25}},{\frac  {7}{25}}\right)\,

Slope of OC is {\frac  {{\frac  {7}{25}}}{{\frac  {-1}{25}}}}=-7=m_{1}\,

Equation of the diagonal AB is equation 1 - equation 2



Therefore,slope of AB={\frac  {1}{7}}=m_{2}\,

Therefore, m_{1}m_{2}=-1,m_{1}m_{2}=(-7)({\frac  {1}{7}})=-1\,

AB is perpendicular to OC.

Therefore OACB is a square.

Main Page:Geometry:Straight Lines-II