    Question

# The real value of λ for which the image of the point (λ,λ−1) with respect to the line 3x+y=6λ is the point (λ2+1,λ) is

A
1
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B
0
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C
1
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D
2
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Solution

## The correct option is D 2Let (λ,λ−1) be P, (λ2+1,λ) be Q and R be the point of intersrection of PQ and L Here Q is the image of the point P So PQ⊥L and PR=QR ⇒ slope of PQ × slope of L=−1 ⇒(λ−(λ−1)λ2+1−λ)×(−3)=−1 ⇒λ2−λ+1=3 ⇒λ=−1,2 But this is not sufficient, we need PR=QR For λ=2 P≡(2,1), Q≡(5,2) R≡(72,32) Putting value of R in equation of L R satisfies the equation of L For λ=−1 P≡(−1,−2), Q≡(2,−1) R≡(12,−32) Putting value of R in equation of L R does not satisfies the equation of L So λ=2  Suggest Corrections  1      Similar questions  Related Videos   Basic Concepts
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