Geo3.16

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 $abc+2fgh-af^2-bg^2-ch^2=0\,$

$24k+175-75-98-25k=0,k=2\,$

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

$(6x-2y)(2x-y)+14x-5y+2=0\,$

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 $(\frac{3}{2},5)\,$