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Find the value of k for which the equation 12x^{2}-10xy+2y^{2}+14x-5y+k=0\, represents two straight lines. Find their point of intersection.

From the given equations, a=12,b=2,c=k,f={\frac  {-5}{2}},g=7,h=-5\,

The condition is that abc+2fgh-af^{2}-bg^{2}-ch^{2}=0\,


Now the equation is S\equiv 12x^{2}-10xy+2y^{2}+14x-5y+2=0\,


Let S\equiv (6x-2y+l)(2x-y+m)=0\,

Equating the quotients 2l+6m=14,-l-2m=-5\,

Solving l=1,m=2\,

Therefore,the equations are 6x-2y+1=0,2x-y+2=0\,

On solving the two equations, we get (6x-2y+1)-2(2x-y+2)=0,6x-4x+1-4=0,2x=3,x={\frac  {3}{2}}\,

Now the value of y is 2({\frac  {3}{2}})-y+2=0,y=5\,

The point of intersection is ({\frac  {3}{2}},5)\,

Main Page:Geometry:Straight Lines-II