Answer :
Let n = the number of points
(x-1) + ... (x-x)
The last term will always be 0, when you reach that, stop.
ex. 1pt: 1-1=0
2pt: (2-1) + (2-2) = 1
3pt: (3-1) + (3-2) + (3-3) = 3
(x-1) + ... (x-x)
The last term will always be 0, when you reach that, stop.
ex. 1pt: 1-1=0
2pt: (2-1) + (2-2) = 1
3pt: (3-1) + (3-2) + (3-3) = 3
⇒Number of Points in the Plane = n
→There are two Possibility
Either All Points are Collinear, that is Lie Along a Line.
Or, They are, Non- Collinear.
To Determine a segment we need two distinct points.
If All "n" Points are Collinear,Distinct Number of Segment=1
If there are two points in plane, number of Distinct Segment
[tex]=_{2}^{2}\textrm{C}\\\\=\frac{2!}{(2-2)! \times 2!}\\\\=1[/tex]
If there are three points in plane, number of Distinct Segment
[tex]=_{2}^{3}\textrm{C}\\\\=\frac{3!}{(3-2)! \times 2!}\\\\=3[/tex]
If there are Four points in plane, number of Distinct Segment
[tex]=_{2}^{4}\textrm{C}\\\\=\frac{4!}{(4-2)! \times 2!}\\\\=6[/tex]
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⇒So,If Points are Not Collinear,that is there are "n" points in the plane, then Distinct number of Segment
[tex]=_{2}^{n}\textrm{C}\\\\=\frac{n!}{2!\times (n-2)!}\\\\=\frac{n \times(n-1)}{2}[/tex]
