Geo3.17

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 $(0,3)\,$