This topic will enlighten the mind of students who are not aware how to solve for the maximum number of intersections are there with a given points. I will give different insights how to deal with the problem. Let us start treating the problem geometrically.
Remember: A line can be drawn if we have two points
As shown in the figure we can tell that the number of intersections we can draw if we have 3 points is also 3. To investigate this further let’s put another point D and again make sure to maximize their number of intersections. As seen in the figure below we have 6 intersections.
Now, by observation the next number of intersections we can obtain if we draw another point is the number of intersections in 4 points which is 6 plus the number of intersection of 1 line to 4 lines which is 4 that is a total of 10. Add another point and the number of intersections will be 10+5 since there are already 5 lines. We can continue with the process. But to do it quickly let’s derive a formula for that.
Result by analysis:
Observe the table above. Let n be the number of points and S for the number of intersections.
Having a knowledge in generating of equation, by doing quick subtraction for S(number of intersection) we can see that the constant appeared in second degree. Meaning, the formula will follow a quadratic equation.
If n=3, S=3.
By direct substitution,
Since we have 3 unknowns we need to have 3 equations.
If n=4, S = 6
If n =5, S= 10
Solving the equations simultaneously we have a=1/2, b= -1/2, c=0
Going back to our equation
by plugging the values of a,b, and c we have.
That is the required formula. To test the formula you can try to plug in n=3, n=4, n=5, n=6. It will give you a quick answer.
Is there a quicker way to do this? The answer is yes! You don’t need to generate that formula. The concept is through combinations. That will be in a separate topic.