A) \[3{{y}^{2}}+8xy-3{{x}^{2}}=0\]
B) \[3{{x}^{2}}+8xy-3{{y}^{2}}=0\]
C) \[3{{y}^{2}}-8xy+3{{x}^{2}}=0\]
D) \[3{{x}^{2}}+8xy+3{{y}^{2}}=0\]
Correct Answer: B
Solution :
\[{{m}_{1}}=3,\ \ {{m}_{2}}=-\frac{1}{3}\]. Hence, the lines are\[y=3x,\ y=-\frac{1}{3}x\]. Multiplying both the lines, we get \[(y-3x)(3y+x)=0\Rightarrow 3{{x}^{2}}+8xy-3{{y}^{2}}=0.\]
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