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Given 3 lines in the plane such that the points of
intersection form a triangle
with sides of length 20, 20 and 30, the number of points
equidistant from all the 3
lines is
1
4
3
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Answer Posted / anil pattanaik
ans is 4 (1 in-centers + 3 ex-centers)
sorry for the above ans plz remove it..ans is 4 but i wrote
the wrong logic
| Is This Answer Correct ? | 26 Yes | 4 No |
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