Question

Mark the following points on a sheet of a paper. Tell how many line segments can be obtained in each case:

(i) Two points A, B.
(ii) Three non-collinear points A, B, C.
(iii) Four points such that no three of them boelong to the same line.
(iv) Any five points so that no three of them are collinear.

Answer 1

If there are n points in a plane and no three of them are collinear, the number of line segments obtained by joining these points is equal to $\frac{n(n-1)}{2}$.
On applying the above formula, we get:

(i) For two points A and B:
Number of line segments = $\frac{2(2-1)}{2}=1$

(ii) For three non-collinear points A, B and C:

Number of line segments = $\frac{3(3-1)}{2}=\frac{3\times 2}{2}=3$

(iii) For four points such that no three of them belong to the same line:

Number of line segments = $\frac{4(4-1)}{2}=\frac{4\times 3}{2}=6$

(iv) For any five points so that no three of them are collinear:

Number of line segments = $\frac{5(5-1)}{2}=\frac{5\times 4}{2}=10$