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The starting point I have been given for this investigation is three lines which intersect each other, as shown below.

R I

L

The lines, which we will refer to as L, are of infinitive length, though I will only draw them long enough so that we can see the diagrams clearly. The points that the lines cross each other (intersections) I will refer to as I. The areas that are bounded by the lines, I will call regions and refer to them as R. To help me find a relationship between I, L and R, I will have to set a number of boundaries for my lines, if I did not do this then I would have an endless investigation. In this investigation we wish to investigate the maximum number of intersections (I) and regions (R) that a certain number of lines (L) can make. To do this I will start the investigations with simple diagrams and then build them up to complicated sketches, I will then put the results of the findings of the sketches in tables. If I find it necessary I will also draw graphs to show my results clearer, when I have done this, using the tables I will look for a relationship between the no. of lines, intersections and regions.

**Write your GCSE Lines investigation research paper**

I am now going to set boundaries for the lines so as my investigation will have a limit and so I will be able to find a pattern between the lines, intersections and regions.

1) The lines must be straight.

e.g.

1 4 5 6

If the lines weren’t straight, and they were curved, as above, I could have 1 line with I=5 and R=7.

If this rule wasn’t used, then I couldn’t find any relationship because 1 line could intersect itself numerous times.

) Lines must be of a standard length.

e.g.

As you can see, it would be very difficult to find a pattern if the lines were not of similar length. To cancel this out I will make all the diagrams of similar length, so as I don’t have diagrams as above.

) Lines must not be parallel. I will now do a mini investigation to see what happens if the lines are parallel.

1 L=1 1 L=

R= R=

I=0 I=0

1 4 L=

R=4

I=0

I will now put these results into a table, so I can understand them easier

L I R

1 0

0

0 4

From looking at the table I can now say, that for parallel lines, there is no intersections and the no. of regions equals the no. of lines plus 1.

I can now say that for all parallel lines I=0 and R=L+1.

I can now predict that for 5 lines, there will be no intersections and 6 regions. I will now sketch this out to see if I am correct.

L=5

1 4 5 6 R=6

I=0

I can now see from this diagram that my prediction is correct, L=5 R=6 and I=0.

In this mini investigation I found that R=L+1 and I=0.

As I=0, I now know parallel lines don’t intersect. In the main investigation I need the lines to intersect, so I will make the boundary that no lines can be parallel.

4) I will now investigate, to see what happens, if all the lines intersect at 1 point. I will do this by drawing diagrams.

I=0 1 I=1

1 R= R=4

L=1 L=

4

I=1

1 R=6

L=

6 4

5

I can now say that I will have the constraint that only two lines can cross at one intersection. I have done this because I need to find to maximum no. of intersections.

5) All the lines must cross each other.

e.g.

As you can see, all the lines don’t intersect each other. I want to maximise the no. of intersections so I will say that all the lines must cross each other.

I am now ready to start the main part of the investigation, I will do this by using all of the boundaries that I have found which are

1) All the lines must be straight

) Lines must be of a standard length

) Lines must not be parallel

4) Only lines can intersect at 1 point

5) All lines must cross each other

Using the boundaries I have found, I will start this investigation with simple diagrams and then work to harder diagrams.

I=0 I=1

L=1 L=

R= R=4

I=

L=

R=7

From these diagrams we can see that when I use the boundaries, we can make a clear diagram and I can see now that R=I+L+1. I will now put these results into a table, using the formula so I can see the results more clearly.

I L +1 = R

0 1 =

1 4 = 4

7 = 7

From this table I can see that the formula is correct. I can now predict that if I have 4 lines and 6 intersections, I will have 11 regions.

I=6

L=4

R=11

I can now see that my prediction is correct. I will now try to find a relationship between L & I. First I will put my results into a table and find the 1st and nd differences and work from there.

L I 1st Diff nd Diff

1 0 1 1

1 1

4 6

From this I can now say that because the nd difference is constant, the relationship must be quadratic.

The general equation for a quadratic equation is

ax +bx +c

I know from research that the x coefficient is half of the nd difference. I can now say, knowing this that

a= ½ x L= ½

The equation now becomes I= ½ L + bL + c

I will now put the results of I, ½ L & L into another table, and I will see from that if I can find a relationship between L & I.

L I ½ L Error in ½ L & I

1 0 ½ ½

1 1

4 ½ 1 ½

4 6 8

From this table, I can see that to get from ½ L to I each time we need to subtract ½ L.

We can now say that I= ½ L � ½ L

Predict

I can now say using this equation, that if I have 5L, I will equal;

I= ½ 5 � ½ 5

= 1.5 � .5 =10

I will now draw a diagram to show that this equation is correct

I can now say that my prediction is correct, because I=10 and I have used 5L. This diagram shows that when we have 5L, I will equal 10 as long as we use the boundaries set.

An alternate method can also be used to find a similar equation, which is by putting the results on the table for L & I, onto a graph and then work from there. I will now plot the results onto a graph and study my result.

From the graph I can see that, when I join the points together, it becomes a curved line, & from my mathematical knowledge I know that a curved line means that the relationship between L & I must be quadratic, and the general quadratic equation is

ax + bx + c

In this case the equation now becomes I= aL + bL + c

Instead of using the second difference, I will fill the results of the table into simultaneous equations to work out what a, b & c are.

1) I=1, L= I=aL + bL + c

1=4a + b + c

In this equation I have used results from my table and subbed them into this equation, I will now do the same for more equations, and then I will subtract them from each other to find out the results.

) I=, L= =a + b + c

) I=6, L=4 6=16a + 4b + c

Using these equations I will now be able to find out the values for a, b, & c.

If I now subtract 1) from ) then I will be able to cancel out the c term.

) =a + b + c

- 1) 1=4a + b + c I will call the new

4) =5a + 1b + c equation 4)

I will now do the same as before, for & .

) 6=16a + 4b + c

- ) = a + b + c I will call the new equation,

5) =7a + 1b + c 5)

I will now take 4) away from 5), so I can find out what a is,

5) =7a + b

- 4) =5a + b

1=a

a= ½

I will now fill a into equation 4), so as I can find out what b is,

= 5a + b

=.5 + b

-.5=b

b= -0.5

Using what I have found I can fill a & b into equation 4) to find c

=5a + b + c

=.5 + (-0.5) + c

= +c

c= 0

I can now sub a, b, & c into the general quadratic equation.

I= aL + bL + c

I= ½ L + (-0.5) b + 0

I now will use the example L=5 to test out this equation

I= ( ½ 5) + (-0.5 5) + 0

I=10

I now know that this equation is correct because as I found out in the previous experiment, when L=5, I will equal 10, and there is a diagram to prove that this is correct.

I will now try to find a relationship between L & R, using the methods that I previously used. I will use the results of the diagrams that I drew at the start of this investigation and put them onto a table and try o find a relationship.

L R 1st Diff nd Diff

1 1

4 1

7 4

4 11

From this I can say that the equation is quadratic, because the nd diff is constant.

The general quadratic equation is ax + bx + c

From my research, I can say that the x coefficient is ½ of the second difference, knowing this I can say that a= ½ L= ½

The equation now becomes R= ½ L + bL + c

I will now put the results of R, ½ L & L into another table, and I will be able to see from it, if there is a relationship between L & I.

L R ½ L Error in ½ L & R

1 ½ 1 ½

4

7 4 ½ ½

4 11 8

From the table, I can see that to get from ½ L to R, each time we need to add ½ L and I

We can now say that R= ½ L + ½ L + 1

Predict

I can now say, using this equation, that if I have 5L, R will equal

R= ½ L + ½ L + 1

R= ½ 5 + ½5 + 1

R=1.5 + .5 + 1

R=16

I will now draw a diagram to see if my prediction is correct..

I can now say that my prediction is correct because from the diagram we can see that when we have 5L, R will equal 16 which it does.

I now will use an alternate method to solve this equation

First I need to plot these results between R & I onto a graph.

L R

1

4

7

4 11

I can see from the graph that when the points are joined together, it produces a curved line, & from my research I know that a curved line on a graph, means that the relationship must be quadratic.

The general quadratic equation is ax + bx + c, which in our case becomes R= aL + bL + c

As I did before, I will fill values from my table into the equations, to find the values for a, b, & c.

1) L=, R=4 R= aL + bL + c

4=4a + b + c

In this equation I have used results from my table and subbed them into this equation, I will now do the same for more equations, and then I will subtract them from each other to find out the results.

) L=, R=7 7=a + b + c

) L=4, R=11 11=16a + 4b +c

If we subtract 1 from we will then be able to cancel out the c term.

) 7=a + b + c I will call the new

1) 4=4a + b + c equation 4, I will now do

4) =5a + b the same for &

) 11=16a + 4b + c

- ) 7=a + b + c I will now take 4 away

5) 4=7a + b from 5, so as to find out a.

5) 4=7a + b I will now fill my result for a into

4) =5a + b equation 4, so as to find b

1=a

a= ½

4) =5a + b I will now fill my results for a &

=.5 + b b into equation 1, so as to find c

-.5 = b

b= ½

1) 4=4a + b + c

4= + 1 + c

4=+c

c=1

Now I will sub a, b, & c into the general quadratic equation so it becomes R= ½ L + ½ L + 1

Predict

If the number of lines were 5, then the number of regions would be R= (½ 5) + (½ 5) + 1

R= 1.5 + .5 + 1

R=16

I will now draw a diagram to prove that this equation works

I can now say that my equation & prediction were correct.

At the start of this investigation we where given lines and told to find a relationship between I, L & R, I did this by setting boundaries so I could find a relationship between them. First I drew out some diagrams using the boundaries set, and then I entered the results into a table. From there, using graphs and tables, I found equations that linked L, R, & I together. To further this investigation I would like to try and find a relationship between R & I, to see if it is quadratic or not.

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