Question

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

• A. $9$
• B. $1$
• C. $-9$
• D. $-1$

Answer 1

Answer: (C)

$-9$

Solution:

Given that:

$3x^2+xy-y^2-3x+6y+k=0$

Comparing with standard equation, we get

$a=3,2h=1,b=-1,2g=-3,2f=6,c=k$

Condition for a pair of straight lines:

$abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

Now,

$⟹3\times (-1)\times k+2\times 3\times \frac{-3}{2}\times \frac{1}{2}-3\times 3^2-(-1)\times {\left(\frac{-3}{2}\right)}^2-k\times {\left(\frac{1}{2}\right)}^2=0$

$⟹-3k-\frac{9}{2}-27+\frac{9}{4}-\frac{k}{4}=0$

$⟹-\frac{13}{4}k=\frac{117}{4}$

$⟹k=-9$

Hence, C is the correct option.