Geo3.17

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Find the value of k for which the equation 2x^{2}+2xy-y^{2}+kx+6y-9=0\, represents two straight lines. Find their point of intersection.

From the given a=2,b=-1,c=-9,2h=2,2g=k,2f=6\,

Now abc+2fgh-af^{2}-bg^{2}-ch^{2}=0\,

2(-1)(-9)+2(3){\frac  {k}{2}}(1)-2(9)-(-1){\frac  {k^{2}}{4}}-(-9)(1)=0\,

k^{2}+12k+36=0,k=-6\,

The point of intersection is \left({\frac  {(1)(3)+1(-3)}{-2-1}},{\frac  {(-3)(1)-2(3)}{-2-1}}\right)\,

Therefore,point of intersection is (0,3)\,


Main Page:Geometry:Straight Lines-II