Some points are plotted on
plotted on a plane such that any three of them are non collinea. Each
d with all remaining points by line segments. Find the number of points in the
umber of line segments are 10.
Answers
Number of points = 5 if Number of line segments are 10 and points are such that any three of them are non collinear
Step-by-step explanation:
Let say there are n Points
we need two points to create a line segment
Number of line segments = number of way we can select 2 points out of n points
order does not matter here as line from point A to point B is same as Point B to Point A
Hence number of way we can select 2 points out of n points = ⁿC₂
= n!/(n-2)!2!
= (n)(n-1)/2
= (n² - n )/2
Number of line segments are 10.
=> (n² - n )/2 = 10
=> n² - n = 20
=> n² - n -20 = 0
=> n² - 5n + 4n - 20 = 0
=> n(n - 5) + 4(n - 5) = 0
=> (n + 4)(n - 5) = 0
n = 5 ( as n can not be -ve)
number of points = 5
Another Way to explain :
Let say there are n point
then number of line segments from 1st point = n - 1 ( one line segmnet with all other n-1 point)
then number of line segments from 2nd point = n - 1 ( as one line segmnet with all other n-1 point) but here line segment from 2 to 1 is already considered in line segment from 1 to 2
hence line segments from 2nd point = n - 2
and so on
Number of line segments from nth point = 0 ( as all line segment has been counted already)
total number of line segments
= (n-1) + (n-2) + ................................+ 1 + 0
This is an AP
a = First term = n-1
L = Last term = 0
n = number of terms = n
d = common difference = - 1
Sum = (n/2)(a + L)
= (n/2)(n-1 + 0)
= n(n-1)/2
= (n² - n )/2
solution from here is same as above
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