# Number of Intersections of Lines with Given Points

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

Now, let’s add point C and draw lines and make sure that the number of intersections is maximized.

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.

Derivation,

$S = an^2 + bn +c$    If n=3, S=3.

By direct substitution,

$3 = a(3)^2 + b(3) +c$

$3 = 9a+ 3b +c$     -> Eqn.1

Since we have 3 unknowns we need to have 3 equations.

If n=4, S = 6

$S = an^2 + bn +c$ $6 = a(4)^2 + b(4) +c$

$6 = 16a + 4b +c$  -> eqn.2

If n =5, S= 10

$10= a(5)^2 + 5b +c$

$10= 25a+ 5b +c$     -> eqn.3

Solving the equations simultaneously we have  a=1/2, b= -1/2, c=0

Going back to our equation

$S = an^2 + bn +c$ by plugging the values of a,b, and c we have.

$S = \displaystyle\frac{n(n-1)}{2}$

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.

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

### 4 Responses

1. Can you give me quick way to solve sum of series or sequences example sum of multiples of 9 from 500 to 700.tnx

2. Techie says:

first term = 504
Last term = 693

S_n = (504+693)(22)/2

S_n = 13,167

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