ABC is a variable triangle such that A is (1,2), and B and C lie on the line y=x+λ (λ is a variable). Then the locus of the orthocenter of △ABC is
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a
x+y=0
b
x-y=0
c
x2+y2=4
d
x+y=3
answer is D.
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Detailed Solution
Equation of line perpendicular to x-y+λ=0 and passing through (1,2) is x+y=3As the altitude from A is fixed and the orthocenter lies on the altitude, x + y = 3 is the required locus.