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Show that the pair of lines 6x^{2}+5xy-4y^{2}+7x+13y-3=0\, form a parallelogram with the pair of lines 6x^{2}+5xy-4y^{2}=0\,.Find its area.

From the given


(2x-y)(3x+4y)+7x+13y-3=0\equiv (2x-y+l)(3x+4y+m)=0\,


Solving we get l=3,m=-1.

The two lines are

2x-y+3=0,3x+4y-1=0\, which are parallel to 2x-y=0,3x+4y=0\,

Hence the two pairs form a parallelogram.

The point of intersection of the given pair of lines is (-1,1)\,

Therefore,area of the parallelogram is {\frac  {a\alpha ^{2}+2h\alpha \beta +b\beta ^{2}}{2{\sqrt  {h^{2}-ab}}}}={\frac  {(6)(1)+5(-1)-4(1)}{2{\sqrt  {{\frac  {25}{4}}+24}}}}={\frac  {3}{11}}\, square units.

Main Page:Geometry:Straight Lines-II