Question

For what value of $\lambda$ does the equation $12x^2-10xy+2y^2+11x-5y+\lambda =0$ represent a pair of straight lines? Find their equations.

Answer 1

Here $a=12,b=2,c=\lambda ,f=-\frac{5}{2},g=\frac{11}{2},h=-5$.
Putting these values in the condition
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$
we get $24\lambda +\frac{275}{2}-\frac{121}{2}-25\lambda =0$
$\therefore \lambda =2$
$\therefore 12x^2-10xy+2y^2+11x-5y+2=0$
represents a pair of straight lines and proceeding as in part (a), their equations are
$4x-2y+1=0$ and $3x-y+2=0$.
Alt. The factors of $2$nd degree terms are $(2x-y)(6x-2y)$. Hence the two lines are $(2x-y+p)(6x-2y+q)$. Comparing the coefficients
$6p+2q=11$, $-2p-q=-5$ and $pg=\lambda$
The first two give $p=\frac{1}{2},q=4$
$\therefore pq=2=\lambda$
and the lines are $2x-y+\frac{1}{2}=0$
and $6x-2y+4=0$.