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,
$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$
$\therefore 3\left(-1\right)\left(k\right)+2\times 3\times \left(-\frac{3}{2}\right)\times \frac{1}{2}-3{\left(3\right)}^2+1{\left(\frac{-3}{2}\right)}^2-k{\left(\frac{1}{2}\right)}^2=0$
$\Rightarrow -3k-\frac{9}{2}-27+\frac{9}{4}-\frac{k}{4}=0$
$\Rightarrow \frac{-13k}{4}-\frac{117}{4}=0$
$\Rightarrow k=-9$