Question

If $3x^2 +xy -y^2 -3x + 6y +k = 0$ represents a pair of lines, then $ k =$

• A. 0
• B. 9
• C. 1
• D. -9

Answer 1

Answer: (D)

Given equation is
$3x^2 + xy -y^2 - 3x +6y +k = 0 $
Here, $a = 3, b = -1 , h = \frac{1}{2} . g = - \frac{3}{2}, $
$f= 3, c - k$
To represent a pair of lines,
$abc +2fgh -af^{2} -bg^{2} -ch^{2} = 0 $
$\therefore 3\left(-1\right)\left(k\right)+2\times3 \times\left(- \frac{3}{2}\right) \times\frac{1}{2} -3 \left(3\right)^{2} + 1 \left(\frac{-3}{2}\right)^{2} -k \left( \frac{1}{2}\right)^{2} = 0 $
$\Rightarrow -3\,k - \frac{9}{2} -27 + \frac{9}{4} - \frac{k}{4} = 0$
$ \Rightarrow \frac{-13\,k}{4}- \frac{117}{4} = 0 $
$\Rightarrow k = -9 $