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Ques. 1JEE Main 2024, 27 January (Morning Shift)
If S=\z :|z-i|=|z+i|=|z-1|\, then, n(S) is :
- 1
- 2
- 3
- 0
Answer
(1) 1
Solution
|z-i|=|z+i| z is equidistant from i and -i z lies on the real axis y=0
|z-i|=|z-1| z lies on the perpendicular bisector of i and 1, i.e. y=x
only z=0 (the circumcentre of the triangle with vertices i, -i, 1)
n(S)=1
Answer: (1)