A) 1
B) \[-1\]
C) 3
D) \[-3\]
Correct Answer: D
Solution :
The pair of lines is\[6{{x}^{2}}-xy+4c{{y}^{2}}=0\]. On comparing with\[a{{x}^{2}}+2\text{ }hxy+b{{y}^{2}}=0,\]we get \[a=6,2h=-1,b=4c\] \[\therefore \]\[{{m}_{1}}+{{m}_{2}}=-\frac{2h}{b}=\frac{4}{4c}\]and\[{{m}_{1}}{{m}_{2}}=\frac{a}{b}=\frac{6}{4c}\] Since, one line of given pair of lines is \[3x+4y=0\] Slope of line\[=-\frac{3}{4}={{m}_{1}}\] (say) \[\therefore \] \[-\frac{3}{4}+{{m}_{2}}=\frac{1}{4c}\] \[\Rightarrow \] \[{{m}_{2}}=\frac{1}{4c}+\frac{3}{4}\] \[\therefore \] \[\left( -\frac{3}{4} \right)\left( \frac{1}{4c}+\frac{3}{4} \right)=\frac{6}{4c}\] \[\Rightarrow \] \[-\frac{3}{4}\left( \frac{1+3c}{4c} \right)=\frac{6}{4c}\] \[\Rightarrow \] \[1+3c=\frac{-6\times 4}{3}\] \[\Rightarrow \] \[1+3c=-8\]\[\Rightarrow \]\[3c=-9\] \[\Rightarrow \] \[c=-3\] Alternate Solution Since,\[3x+4y=0\]is one of the two lines. Then, \[y=-\frac{3x}{4}\]will satisfy the equation\[6{{x}^{2}}-xy+4c{{y}^{2}}=0.\] \[\therefore \] \[6{{x}^{2}}-x\left( \frac{-3x}{4} \right)+4c{{\left( \frac{-3x}{4} \right)}^{2}}=0\] \[\Rightarrow \] \[6{{x}^{2}}+\frac{3x}{4}+4c\frac{9{{x}^{2}}}{16}=0\] \[\Rightarrow \] \[{{x}^{2}}(24+3+9c)=0\]\[\Rightarrow \] \[9c=-27\] \[\therefore \] \[c=-3\]
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