A) \[2{{x}^{2}}=2y(2x+y)\]
B) \[{{x}^{2}}+{{y}^{2}}+3=0\]
C) \[2{{x}^{2}}=y(2x+y)\]
D) \[{{x}^{2}}=2(x-y)\]
Correct Answer: A
Solution :
Since\[2{{x}^{2}}=2y(2x+y)\Rightarrow {{x}^{2}}-2xy-{{y}^{2}}=0\]. Hence, coefficient of \[{{x}^{2}}+\]coefficient of \[{{y}^{2}}=1-1=0\]. Hence the lines are perpendicular.
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