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Question
Six points in a plane be joining in all possible way by staright lines, and if no two of them be coincident or parallel, and no three pass through the same point (with the exception of the original 6 points). The number of distinct points of intersection is equal to
Six points in a plane be joining in all possible way by staright lines, and if no two of them be coincident or parallel, and no three pass through the same point (with the exception of the original 6 points). The number of distinct points of intersection is equal to
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Solution
The correct option is C 51
Number of lines from 6 points = 6C2=15.
Points of intersection obtained from these lines = 15C2=105.
Now we find the number of times, the original 6 points come.
Consider one point say A1.
Joining A1 to remaining 5 points, we get 5 lines, and any two lines from these 5 lines give A1 as the point of intersection.
∴A1 come 5C2=10 times in 105 points of intersection.
Similar is the case with other five points.
∴ 6 original points comes 6×10=60 times in points of intersection. Hence the number of distinct points of intersection =105−60+6=51
Number of lines from 6 points = 6C2=15.
Points of intersection obtained from these lines = 15C2=105.
Now we find the number of times, the original 6 points come.
Consider one point say A1.
Joining A1 to remaining 5 points, we get 5 lines, and any two lines from these 5 lines give A1 as the point of intersection.
∴A1 come 5C2=10 times in 105 points of intersection.
Similar is the case with other five points.
∴ 6 original points comes 6×10=60 times in points of intersection. Hence the number of distinct points of intersection =105−60+6=51
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