In a plane there are totally 8 points (no three points are collinear), how many lines can be drawn ?
Answers
Answer:
26
Step-by-step explanation:
Although I have presented 3 (seemingly different) solutions herein, each solution simply uses the Fundamental Counting Principle, which basically says that if event A can be peformed in ‘a’ ways and event B can be performed in ‘b’ ways, then the number of ways in which both events can be performed sequentially is given by the product, ‘ab’. In this problem, we are simply drawing a line, which involves two points—a starting point, and an end point. The line AB is understood to go to infinity in both directions through the given points. We simply must count carefully, and avoid both duplications and omissions of lines.
Let the 3 collinear points be denoted by A, B and C; the five other points, no three of which are collinear, are denoted by P1 to P5. Assume that no two of P1 to P5 AND any one of A or B or C are collinear (that is, the only 3 collinear points are A, B and C as is given in the problem.
The number of lines determined by any two of P1 to P5 is given by 5C2 = 10
(or (5 X 4)/2
The number of lines determined by exactly one point from (P1 to P5) AND exactly one of (A or B or C) = 5C1 X 3C1 = 15
The number of lines determined by A, B and C = 1.
The total number of lines determined is 10 + 15 + 1 = 26.
Alternate solution:
The number of lines determined by any 2 or the 8 given points is 8C2=28.
(or (8 X 7)/2 = 28)
Considering A, B and C, there are 3 duplicate lines, and only 1 unique line
The number of lines determined is 28 - 3+1 = 26.
Personally, I think this is the simplest solution…
Alternate solution:
This is just another way to present the first solution:
There are 3 types of lines that can be drawn:
Type 1: Lines drawn between any 2 of (P1 to P5): that is, we can draw (5 x 4)/2 = 10 lines. We divide by 2, b/c the line drawn from, say P1 to P5, duplicates the line drawn from P5 to P1.
Type 2: Lines drawn from one of Pn to one of A, B or C; that is, we can draw
5 x 3 = 15 lines of this type. There is no need to divide by 2 here, because of the words … FROM and …TO. Besides, if I had divided by 2, we would have 7.5 lines, which is inconsistent…
Type 3: Lines using only the collinear points A,B and C: because A, B and C are said to be collinear, only 1 such line can be drawn.
Total distinct lines = 10 + 15 + 1 =26 lines.
It’s nice when one obtains the same answer using different methods. The optimist would claim that the answer is correct; the pessimist will claim that the same error (probably an omission or a duplication in such a problem) has occurred in all solutions. Today, hopefully, I am the optimist!
I hope this is of some help to you…