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

The correct option is D $-9$

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

Condition for pair of lines:

$\Delta =\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

comparing the equation with standard equation

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$\Rightarrow a=3;b=-1;c=k;h=\frac{1}{2};g=\frac{-3}{2};f=3$

So, substituting the values in the above condition:

$\Delta =\left|\begin{array}{ccc}3& 1/2& -3/2\\ 1/2& -1& 3\\ -3/2& 3& k\end{array}\right|=0$

$\Rightarrow 3(-k-9)-\frac{1}{2}(\frac{k}{2}+\frac{9}{2})-\frac{3}{2}(\frac{3}{2}-\frac{3}{2})=0$

$3(-k-9)-\frac{1}{4}(k+9)=0$

$\Rightarrow k=-9$