A) 1/2
B) -1/2
C) \[1/\sqrt{2}\]
D) 0
Correct Answer: D
Solution :
Making the equation of curve homogeneous with the help of line \[x+y=1\], we get \[{{x}^{2}}+{{y}^{2}}-2y(x+y)+\lambda {{(x+y)}^{2}}=0\] \[\Rightarrow {{x}^{2}}(1+\lambda )+{{y}^{2}}(-1+\lambda )-2yx=0\] Therefore the lines be perpendicular, if \[A+B=0\]. \[\Rightarrow 1+\lambda -1+\lambda =0\Rightarrow \lambda =0\].
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