A) \[\frac{125}{367}\]
B) \[-\frac{125}{367}\]
C) \[15\]
D) \[-15\]
Correct Answer: D
Solution :
Given equation is \[6{{x}^{2}}+11xy-10{{y}^{2}}+x+31y+c=0\] Here, \[a=6,\,b=-10,\,c=c,\] \[g=\frac{1}{2},\,f=\frac{31}{2},h=\frac{11}{2}\] This equation represents a pair of straight lines, if \[\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} 6 & 11/2 & 1/2 \\ 11/2 & -10 & 31/2 \\ 1/2 & 31/2 & c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[6\left( -10c-{{\left( \frac{31}{2} \right)}^{2}} \right)-\frac{11}{2}\left( \frac{11}{2}c-\frac{31}{4} \right)\] \[+\frac{1}{2}\left( \frac{31\times 11}{4}+\frac{10}{2} \right)=0\] \[\Rightarrow \] \[-60\,c-\frac{961\times 3}{2}-\frac{121}{4}c+\frac{341}{8}\] \[+\frac{341}{8}+\frac{10}{4}=0\] \[\Rightarrow \] \[\frac{-361\,c}{4}=\frac{5415}{4}\] \[\Rightarrow \] \[c=\frac{-5415}{4}\times \frac{4}{361}=-15\]
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