Q63 / 180 · KCET 2008 · Mathematics
If $3 x^{2}+x y-y^{2}-3 x+6 y+k=0$ represents $a$ pair of lines, then $k$ is equal to
- A0
- B9
- C1
- Correct answerD$-9$
Solution
Given equation is
$$
3 x^{2}+x y-y^{2}-3 x+6 y+k=0
$$
Here $\mathrm{a}=3, \mathrm{~b}=-1, \mathrm{~h}=\frac{1}{2}, \mathrm{~g}=-\frac{3}{2}$,
$$
\mathrm{f}=3, \mathrm{c}=\mathrm{k}
$$
To represent a pair of lines,
$$
\begin{aligned}
&\mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^{2}-\mathrm{bg}^{2}-\mathrm{ch}^{2}=0 \\
&\therefore 3(-1)(\mathrm{k})+2 \times 3 \times\left(-\frac{3}{2}\right) \times \frac{1}{2} \\
&\Rightarrow \quad-3(3)^{2}+1\left(\frac{-3}{2}\right)^{2}-\mathrm{k}\left(\frac{1}{2}\right)^{2}=0 \\
&\Rightarrow \quad-3 \mathrm{k}-\frac{9}{2}-27+\frac{9}{4}-\frac{\mathrm{k}}{4}=0 \\
&\Rightarrow \quad \frac{-13 \mathrm{k}}{4}-\frac{117}{4}=0 \\
&\mathrm{k}=-9
\end{aligned}
$$
Hence, the correct answer is option: 'D'.