Question

(a) How many line segments do you get?
(b) The points which are collinear?
c) The points which are non-collinear?​

Answer 1

Answer:

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

Explanation:

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Answer 2

Answer:

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 n(n−1)2.

On applying the above formula, we get:

(i) For two points A and B:

Number of line segments = 2(2−1)2=1

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

Number of line segments = 3(3−1)2=3×22=3

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

Number of line segments = 4(4−1)2=4×32=6

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

Number of line segments = 5(5−1)2=5×42=10

Explanation:

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