A) \[4\]
B) \[2\]
C) \[1\]
D) None of these
Correct Answer: A
Solution :
Key Idea: If the equation \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] represents a pair of lines, then \[abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\] or \[\left| \begin{matrix} a & h & f \\ h & b & g \\ f & g & c \\ \end{matrix} \right|=0\] Given equation is \[2{{x}^{2}}+7xy+3{{y}^{2}}-9x+7y+k=0\] On comparing with \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] We get \[a=2,\,\,h=\frac{7}{2},\,\,b=3,\,\,g=-\frac{9}{2},\,\,f=-\frac{7}{2},\,\,c=k\] Since, the given equation represents, a pair of straight lines. \[\therefore \] \[abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\] \[\Rightarrow \]\[2\times 3\times k+2\left( -\frac{7}{2} \right)\left( -\frac{9}{2} \right)\left( \frac{7}{2} \right)-2{{\left( -\frac{7}{2} \right)}^{2}}\] \[-3{{\left( -\frac{9}{2} \right)}^{2}}-k{{\left( \frac{7}{2} \right)}^{2}}=0\] \[\Rightarrow \] \[6k+\frac{441}{4}-\frac{98}{4}-\frac{243}{4}-\frac{49k}{4}=0\] \[\Rightarrow \] \[\frac{24k+100-49k}{4}=0\] \[\Rightarrow \] \[25k=100\Rightarrow k=4\] Alternative Solution: Since, the given equation represents, a pair of straight lines \[\therefore \] \[\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\] \[=\left| \begin{matrix} 2 & \frac{7}{2} & -\frac{9}{2} \\ \frac{7}{2} & 3 & -\frac{7}{2} \\ -\frac{9}{2} & -\frac{7}{2} & k \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[2\left( 3k-\frac{49}{4} \right)-\frac{7}{2}\left( \frac{7k}{2}-\frac{63}{4} \right)\] \[-\frac{9}{2}\left( -\frac{49}{4}+\frac{27}{2} \right)=0\] \[\Rightarrow \] \[6k-\frac{98}{4}-\frac{49k}{4}+\frac{441}{8}+\frac{441}{8}-\frac{243}{4}=0\] \[\Rightarrow \] \[\frac{-25k}{4}+\frac{200}{8}=0\] \[\Rightarrow \] \[k=4\]
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