Polynomials

These are a very common sight in almost any study of mathematics. It’s worth taking a moment to define them properly and think about how they can be used.

A polynomial is a mathematical expression involving a whole, positive power of x. The type of polynomial depends on the highest power of x in the expression. The different bits of the expression are known as terms, and the polynomial as a whole is the sum of the terms it contains.

Take a quadratic expression: 4x² + 2x - 2. It has three powers of x: 2, 1 and 0. These are whole numbers and positive (except 0). This is therefore a polynomial. The same applies to cubic expressions: x³ + 4x² + 5x + 2 is a polynomial too.

In fact, the same applies, whatever the power of x, as long as it is not a fractional or negative number. A polynomial does not need to have several terms, like most cubics and quadratics do. So, 1 is also a polynomial, being a term in x raised to the power of zero. The same applies to any constant (ie any number). 4x + 1 is a polynomial too, as x is x raised to the power of one.

Polynomials are written in descending powers of x, though they do not have to be.

The numbers in front of the x variables are called coefficients, and can take any value, including fractional or negative values, as can constants.

The highest power of x in a polynomial defines the degree of that polynomial. So 1 is a polynomial of degree 0, because the highest power of x here is 0. 4x + 1 is a polynomial of degree 1, and quadratics are polynomials of degree 2.

Polynomials can be created by expanding brackets. (x+2)² gives x² + 4x + 4 when expanded. In fact, both expressions here are polynomials. If the brackets were to give us an expression with a fractional power, then of course this would not be the case.

When you make a polynomial like a quadratic equal to 0, then you have a polynomial equation. In the case of a quadratic, such as 4x² + 2x – 2, the resulting equation is usually written as: 4x² + 2x – 2 = 0. As an equation, it is now a description of a parabolic curve which intersects the x-axis at (-1,0) and (½, 0).

Polynomials can be added, subtracted, multiplied and divided. The Factor and Remainder Theorems can be used to find out additional information about them. For example, if a number assigned to the variable x causes the polynomial to equal zero, that number is a factor of the polynomial (a root, a value of x where the curve of the equation crosses the x-axis). So we can demonstrate that -1 is a factor of 4x² + 2x – 2 by putting it in place of x. 4(-1)² + (-1x2) – 2 = 0. Therefore -1 is a factor of this polynomial. This is the essence of the Factor Theorem. The Remainder Theorem is a little more involved but states that if we divide the polynomial by x-a then the remainder is f(a) – in other words, the remainder is the sum of the polynomial when a is put in place of the variable x.

These are the absolute basics. Of course it does get a lot more complicated! But it is worth familiarising yourself with these principles before moving on.

Integration, Differentiation

This is quite tricky to get your head round, I think.

Differentiation is what you do to a curve to obtain its gradient. It works by splitting the curve into infinitesimally tiny chunks and tending towards a limit (ie where you started from). I think I put the proof of it on an earlier post.

Basically it gives you the rate of change in the curve as a formula. Because curves are always changing, the gradient will be different at each point. You plug the x-coordinate into the differentiated gradient formula and there's your gradient at x=whatever.

But if you integrate that curve you get the area under the curve. It works in a similar way to differentiation: it takes the space under the curve and chops it up into tiny rectangles to approximate the area.

It seems to work out that in terms of a formula, integration is the opposite of differentiation (ie raise the power by one and divide by the value of the new power, instead of dropping the power by one and timesing by the new power).

The curve bounding an area with the x-axis is now the value of the area formula differentiated - ie it is the rate of change of the area itself.


Differentiation and integration are inverse processes: that's the theorem of calculus (which means stones used for counting, or something).

If you have a derivative, you can integrate it to get the indefinite integral - which you can then find the full equation for if you know a set of co-ordinates. So by knowing the derivative, you can find the equation of that curve.

If you have a derivative, often you can differentiate again to find a stationary point on the curve - the second derivative test.

I'm still not sure I get this, you know. I still find the connection between the two loose and hard to actually articulate. I can do it with the formulae no problem at all - but understanding the linkage and how rates of change relate to curves is much harder.

General AS Advice

Know your volumes and areas from GCSE.

I don't.

On The Importance Of Factorising

Without it, even linear equations become extremely difficult to re-arrange.

Take 6xy + 3x - 2y = 7

Re-arrange this for x.

MOVE the non x bit.

6xy + 3x = 7- 2y.

SPOT the common factor

x(6y + 3) = 7 - 2y.

DIVIDE by (6y+3)

x = 7-2y
       ----
       6y+3

PIECE of the proverbial.

Geometry of the Line (4): Circles

Time to go back to straight line geometry for a bit.

With knowledge of how to calculate the gradient, the perpendicular gradient, the mid-point and the length of the line, you've also done quite a lot of co-ordinate geometry of the circle. This is an interesting concept. A circle on a set of cartesian co-ordinates.

You start from the premise of a circle with centre (0,0). The equation of this circle is x² + y² = r². If you study a diagram of it, or even draw one yourself, the reason for this becomes clear. Like a lot of co-ordinate geometry, it's based on Pythagoras. In this case, drawing a triangle inside the circle to a point somewhere on the circumference. Say that this circle with centre (0,0) passes through the point (3,4).

The line from the centre to the point functions as the hypoteneuse of the right angled triangle, while line x measures the distance travelled horizontally to the point, and the line y measures the vertical distance to the point.

The equation of this particular circle then tells us the radius.

3² + 4² = r²

or 9 + 16 = r²

= 25

r = 5.

The radius of the circle is 5.

Things get a little more complicated if the circle's centre is not (0,0). Say we have a circle with centre (2,3). We still devise the equation of the circle based on its distance from (0,0).

This circle's centre has moved 2 units x-wards and 3 y-wards (across and up) from the origin. Any calculation of the radius needs to take into account the fact that we are already some way from the origin.

So if the circle passes through the point (5,8), the distance from the centre to the point will be 5-2 and 8-3 (otherwise we're talking about including space outside the circle).

So the equation will be: (x-2)² + (x-3)² = r².

In this case: 9 + 25 = 34

So r = √34

It is perfectly acceptable to leave the answer as a surd at C1, indeed often you have to.

Usually you are given the equation, say (x-4)² + (y-7)² = 4

or something.

The centre of the circle is ALWAYS THE OPPOSITE OF THE NUMBERS IN BRACKETS, in this case (4,7). Worth remembering. If it is (x-a)² + (y-b)² = r² the centre is always (a,b).

Dividing Polynomials (ii)

Alright then, let's have a crack at this properly.

We are armed with some fraction revision and a little bit of knowledge of polynomials, including factorising using the system of comparing coefficients.

We only need to divide a polynomial by a linear factor for C1.

Division has this nasty habit of not being exact, but of leaving remainders. Unfortunately, 5 does not go easily into 24 but leaves a remainder. Polynomials are also subject to this problem. A linear expression might not be a factor of P(x) as such, but might go into something else, leaving factors and a remainder when asked to be slotted into P(x).

Remember that QUADRATIC EQUATIONS sometimes factorise into TWO LINEAR expressions,
while CUBICS factorise into LINEAR + QUADRATIC (and sometimes three linear).

So if you divide a quadratic by a linear, this might happen:

x² + 2x + 4 divided by x-2.

If we are to use the concept of comparing coefficients, which is very efficient, we first rewrite the equation into the form it would take expressed as factors and remainders:

x² + 2x + 4 = (x+p)(x+q) + r.

Got it?

We need to divide it by x-2 though, so we already have one desired factor (THIS DOES NOT MEAN THAT x-2 WILL BE A FACTOR - just that we are trying to express the quadratic in terms of how much x-2 goes into it.

So:

x² + 2x + 4 = (x-2)(x+q) + r.

Then we collect the different terms in this form. We are not completely, expanding the brackets, we are re-arranging them in terms of their own factors in order to look for coefficients which will tell us the other factors and remainders.

x² + 2x + 4 = x² + (-2 + q)x + (-2q+r)

F(x) is in this case equal to x² and (-2+q) lots of x, and then the two constants, with no x variable, (-2q and r, the possible remainder).

Therefore 2= (-2+q) so q= 4.

4 = (-2x4 + r)

4=8+r

r = -4

x² + 2x + 4 = (x-2)(x+4) -4.

This only expresses the quadratic as factors and a remainder. The division isn't over yet.

Now let's actually divide the quadratic by (x-2).

x² + 2x + 4
-------------
    x-2

=

(x+4) - 4
            ----
            (x-2)



I'll explain it later. Proper work beckons.

Dividing Polynomials

As promised....

One of the key elements of this is the ability to split off fractions effectively.


Think of a normal, but improper fraction, such as 4/3. This could be expressed as 3/3 + 2/3 (or any other combination but this is clean and tidy and useful for us now). 3/3 is of course 1. We are left with 1 + 1/3.

The same thing can be done algebraically, so as to solve basic division questions.

Take (x + 6)/x. As above, we can think of this as x/x + 6/x. x/x = 1, so the answer is 1 + 6/x.

We use this property, making part of the numerator match the denominator, in dividing polynomials.

Rounding Up Factorisation

...so after all that you now have the technology (all right, the skills) to factorise a polynomial completely. You first do trial and error to find the first linear factor, then you multiply it by ax² + bx + c and compare coefficients to find the quadratic. Then you check if the quadratic factorises further.

Onto dividing polynomials proper....

Factor Theorem (ii)

Well that's all very neat and tidy, you might be thinking, but what good is this theorem?

It gives you a lovely and easy way of trying to find linear factors by substituting values into the polynomial you are given. Sometimes this does not work easily but it is often worth a try for a minute or so.

It also enables you to do slightly trickier things with polynomials.

You often get questions like this (we'll use the cubic from the previous post for this). A polynomial, P(x) = x³ + ax² + bx + 6, has (x+1) and (x+2) as factors. Find the values of a and b.

This looks tricky but it is where the factor theorem comes in.

P(-1) = 0. You know this from the theorem.

that means that (-1) + a(1) + b(-1) + 6 = 0

so -1 + a - b + 6 = 0

therefore a - b = -5

P(-2) = 0.

So -8 + 4a -2b + 6 = 0

therefore 4a - 2b = 2

You now have two equations to solve SIMULTANEOUSLY!!

a - b = -5
4a - 2b = 2

Times the first equation by 2:

2a - 2b = -10
4a - 2b = 2


SUBTRACT THE TWO TO ELIMINATE b

-2a = -12

a = 6

Plug this back into one of the equations:

2(6) - 2b = -10
-2b = -22

b = 11

a = 6, b = 11

There is nothing that hard here and yet in just writing this post I have made LOADS of mistakes, which is why it's taken me an hour or so to write.

Here are the errors I made:

1) Using the wrong constant value - ie reading the cubic wrong;
2) Forgetting to cube or square the values in the cubic equation;
3) subtracting with minus signs incorrectly;
4) multiplying incorrectly.

All because I was surfing the blogs while writing this post!

The Factor Theorem

Well that stuff in the previous post leads us nicely on to the factor theorem. Although this sounds like a low budget 70s sci-fi drama, it is in fact a key part of A Level maths.

The factor theorem is dead simple. If you have a polynomial x³ + 6x² + 11x + 6, it has factors, namely a linear and a quadratic.

Only this time, the quadratic itself factorises, so that the cubic has 3 linear factors, namely (x+1)(x+2)(x+3).

If I now did something a bit sneaky, I could prove that these are factors of this cubic.

Watch. P(-1) = -1³ + 6(-1²) + 11(-1) + 6
= -1 + 6 - 11 + 6
= 0

When x=-1 the cubic equation = 0.

This proves that (x+1) is a factor of x³ + 6x² + 11x + 6

In fact, if (x-a) is a factor of P(x) then P(a) = 0. Always and everywhere.

Why?

Well, look at (x+1)(x+2)(x+3).

These are all factors of x³ + 6x² + 11x + 6.

So if you put x= -1 into the first factor, it will equal 0 (-1+1) and therefore the whole polynomial will equal 0. The same goes for all the other factors.

If a value, a, put into P(x) does not equal 0, then (x-a) IS NOT a factor.

Dividing Polynomials (1)

This can actually be a bit tricky but it's not impossible.

Any polynomial of order 3 - which is what you deal with at C1 - is the product of either linear factors, or a linear and a quadratic factor.

Finding a factor like this, you don't need to worry about remainders but these crop up soon enough, and we'll come onto them in due course.

A basic example of what I mean is: x(x² + 2x - 4). Multiplying these two factors gives the cubic equation x³ + 2x² - 4x.

But what do you do if you know one factor and the polynomial?

Say you know the linear factor to be (x-2) and the polynomial to be x³ + x² - 11x + 10

The other factor must be a quadratic.

(x-2)(ax²+ bx + c) = x³ + x² - 11x + 10

expand the two brackets as you would normally:

ax³ + bx² + cx - 2ax² -2bx - 2c

and then find common factors to tidy this expression up and make it usable:

= ax³ + (b-2a)x² + (c-2b)x -2c

That bit is worth checking. You're simply collecting together the different values of x² or whatever, as you would normally in an equation. You're only putting brackets round them because you don't know their values and you need to understand them as together making x².

This expression is now completely identical to our original polynomial:
= ax³ + (b-2a)x² + (c-2b)x -2c
x³ + x² - 11x + 10


We now go through the fairly simple but again easy to cock up process of comparing the co-efficients. Co-efficients are of course the numbers in front of variables (3 is the coefficient of x in the expression 3x).

Comparing the co-efficients here means we look at our original polynomial and compare it with the unknown result - which we can see is a cubic - of our algebraic multiplication. The co-efficients in the known polynomial will tell us the missing co-efficients of our unknown quadratic.

In this case: a = 1

(b-2a) = 1
b - 2 = 1
b= 3

(c-2b) = -11
c-6 = -11
c = -5

We can check this last one by seeing -2c at the end of our algebraic expression. If c is right it will match -2 x -5 = 10, so yes it's right.

This process has given us the three co-efficients for our missing quadratic: 1, 3 and -5

so the two factors of our polynomial are (x-2) and (x² + 3x -5).

Just be careful at this point: can the quadratic factor be factorised itself, to give three linear factors? Give it a quick b² - 4ac test: 9 - (4x1x-5) = 9 - (-20) = 29. NO. It has to stay as a quadratic.

Often it will. You'll find that you're often - but not always - asked to find the three factors of p(x) or to factorise p(x) completely when this is the case.

Multiplying polynomials

This is fairly easy too: it is, at its basic level, expanding brackets, though you can make use of the distributive law of algebra to help you do it more systematically.

For example.

(x+2)(x-5) can either be expanded bit by bit (ac+ad+bc+bd) or you can write it like this: x(x-5) + 2(x-5). Instead of adding four separate products individually, you are adding two sets of them. It's basically the same thing, just a little more organised. You could also write it x(x+2) -5(x+2).

You can use this principle to help you multiply polynomials.

(x+9)(3x³ -4x² + 3)

x(3x³ -4x² + 3)
+ 9(3x³ -4x² + 3)

= (3x⁴ - 4x³ + 3x)
+ (27x³ - 36x² + 27)

using the knowledge of collecting polynomials from the previous post:

= 3x⁴ + 23x³ - 36x² + 3x + 27.


It gets fiddlier than this but usually the problem is signs. Pay really close attention to them because they are an easy ways of losing lots of marks quickly.

Collecting Polynomials

This is not difficult in principle, but it can become messy if you are not careful with your signs.

You can only add or subtract values with the same powers of x.

Be careful: this means that you CANNOT add 3x⁴ and 3x³ because they have different powers of x and are therefore completely different values.

You also CANNOT add 3x² and 3r² because they are different variables and so different values.

Here is a fairly basic example: add 3x³ + 5x to 2x³ -4x

It helps to collect the two expressions into brackets first:

(3x³ + 5x) + (2x³ -4x)

Now you can sort out the signs more easily:

3x³ + 5x + 2x³ -4x

As it happens there is no sort out needed here!

Now collect the like terms:

(3x³ + 2x³) + (5x - 4x)

= 5x³ + x

It does get a lot more fiddly than this sometimes:

(4x⁴ + 5x³ - 3x²) - (2x⁴ - 6x³ + x² + 9)

But use the same principles. Here the two expressions are already in brackets so we can SORT THE SIGNS OUT:

4x⁴ + 5x³ - 3x² - 2x⁴ + 6x³ - x² - 9

What we've done here is simply say "all the values in the second polynomial are being subtracted and we all know what to do with a - and a + or a - and another -". We've effectively subtracted the two expressions here by reversing the signs in the second expression.

You might want to look back at that again.

OK. Now we can collect the like terms.

4x⁴ - 2x⁴ + 5x³ + 6x³ - 3x² - x² - 9

= 2x⁴ + 11x³ - 4x² - 9

Polynomials - The Basics

I've got fed up with doing straight line geometry, as it needs diagrams, and I am having real difficulty drawing & importing them.

So to polynomials. Firstly, what is a polynomial? Well it is an expression involving numbers and powers of x. Such as:

2x + 1

3x² + 4x - 6

2x⁴ - 4x³ + 2x² +2


These are all polynomials. They are usually written, as here, in descending powers of x, from highest to lowest.

Sometimes they are called polynomials of degree x, depending on the highest power of x. So a cubic expression is a polynomial of degree 3, a quadratic of degree 2, a linear of degree 1 and a constant (ie just a number on its own) of degree 0 (because the highest power of x here is 0, ie x⁰ - ie 1).

Polynomials DO NOT have fractional or negative powers.

2x^-2 + 1

is not a polynomial.


That's the basics of it.

Gradients of Perpendicular Lines

If a line intersects another one such that the two are perpendicular,and the gradient of the other is known, it is a piece of cake to work out the unknown gradient.

The gradients of perpendicular lines have a product of -1.

Our line from the previous post has gradient 12-6/8-3 = 6/5 Its perpendicular bisector will therefore have gradient of (using the common letter m to stand for unknown gradient):

m₁xm2=-1

6/5xm₂ = -1

m₂= -1 ÷ 6/5

At this point use the fraction division rule:

m₂= -1 x 5/6

= -5/6

The gradient of the perpendicular bisector is -5/6.

The really cool bit is this. If you have two co-ordinates of the first line, say in the post below (3,6) and (8,12); then you can now easily find the equation of the perpendicular bisector.

If it bisects the first line it must pass through the mid-point: x1+x2/2, y1+y2/2

or 3+8/2, 6+12/2

which is: (5 1/2, 9)

Plug what you know of this line into y - y₁ = m(x - x₁).

Or: y - 9 = -5/6(x - 11/2)

I've turned 5 1/2 into 11/2 for ease of calculation.

y -9 = -1/3x + 55/12

y = -1/3x + 163/12

This is a bit messy so multiply through by 12:

12y = -4x + 163



Lovely.

NOTE: I updated this post because it was wrong: it is still a bit more tricky than most of the questions you will get at C1. Most of these type questions resolve into a line with a constant of no more than about 20. But theoretically one like this could come up.

Geometry of the Line (3): Length

It is a fairly simple job to calculate the length of a line. So far I've shown the mid-points, gradients, and equations. This is similar stuff, and based on the principle of Pythagoras, as for gradients.

Take a line with two points (3,6) and (8,12) lying on the line.

We can see any straight line on a graph as being equivalent to a hypotenuse of a right angled triangle.

Now Pythagoras says that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Looking at the diagram:




The line we want to measure is the hypotenuse and the other two sides are clearly horizontal and vertical. So the length of the horizontal line is easy - just x₂ - x₁ and the length of the vertical line the same but with ys. y₂-y₁

This is where Pythagoras comes in. We square the lengths of the two lines and add them together. This gives us the square of the hypotenuse.

We write it like this:

√(x₂-x₁)² + (y₂-y₁)²

In this case:

√ (8-3)² + (12-6)²

√ 25 +36

√ 61

This will be a surd so we can leave the answer as √61 for the purposes of C1.

Beware of Surds

Surds are right at the start of C1.


They are easy to forget.

As I found out tonight, trying a C1 practice paper.

Surds are basically this. They are an exact representation of irrational numbers, where a fraction or a decimal will not do. Normally we have rational numbers, like 1, 2, 3.5, 4.54, and so on. These numbers can be expressed as a fraction or as a decimal. But sometimes we get numbers like √2 or π which would go on forever if we tried to express them as a decimal (3.14159265 is only the start of π) so a decimal or a fraction would only ever give us an approximation.

1.41.... will never really describe √2 exactly, because it has an infinite number of digits after the decimal point. Apparently there is a proof of this but you don't need to prove it for C1.

So we leave these numbers expressed as a square root, or combination of square root and integer or fraction, and we call these surds (because 3 or 5 times the irrational square root of 2 will be irrational itself, usually). They enable us to do exact calculations of numbers which cannot be represented exactly as fractions or decimals.

There are a few rules of surds.

1) You can only add or subtract like surds. ie 3√7 - 2√7 = √7

2) You can multiply surds but there are a couple of rules to remember:

√5 x √6 = √30

3 x √5 = 3√5

5√3 x √4 = 5√12

3) Sometimes a surd will reveal itself to be a rational number: watch out for this:

√3 x √3 = 3

3√4 x 2√4 = 6√16 = 24

4) surds multiply out of brackets just like anything else does.

5)The trickiest thing about surds is fractions with a surd as a denominator. You need to get rid of the surd.

If you think about it, there is a way to make rational numbers out of surds that uses a rule of brackets. (a-b)(a+b) always gives you a²-b².

So if you have 3+√5 as the denominator, multiply it by 3-√5 and you rationalise the denominator into 9-5 = 4.

Of course whatever you times the denominator by you need to times the numerator by.

This means you will often end up with your answer being a surd, but you will have eliminated the irrational denominator and so you will have been able to simplify your fraction.


I know I'm meant to be doing a series on geometry of the line but I keep getting distracted...

Why Doesn't Completing the Square Always Work?

I was meant to be going to bed but then I saw that someone had googled the above question and come to me.

The honest answer is "it does always work" but I imagine the problem my putative reader has is that some quadratics are really inconsiderate and don't have real roots.


Well imaginary and complex numbers are Further Pure nowadays but I will hazard an answer.

A quadratic is lovely because it has its famous formula: -b +/- (sqt b²-4ac)/2a

The b² - 4ac bit is called the discriminant and determines the kinds of solutions the equation can have.

I can't really draw a graph tonight but we see it in terms of the points the quadratic crosses the x-axis of the graph.

It goes a little bit like this:

b² - 4ac = a square number (36, say) - quadratic factorises (ie can be reduced to (x+1)(x-2) say).

b² - 4ac = +ve numbers, including fractions - quadratic cuts x-axis twice but can't factorise.

b² - 4ac = 0 - quadratic hits x-axis once (x-axis is a tangent to quadratic)

b² - 4ac = -ve number, quadratic does not touch x-axis at all.


If b² - 4ac is less than 0 it means we need a square root of a negative number. At c1 we can't do that at all. Later on one learns that numbers which square to give -ve numbers are called imaginary numbers and have all kinds of key roles. For now we just don't really want them,

so for our A/S Level purposes the quadratic does not have real roots (ie the values would be complex numbers - an imaginary with a real number) and we simply cannot calculate it at all.

You just end up with a complex number statement like (-2+sqrt -1)/2 or something. As I say, probably great for FP1, no use for C1.

However, that does not mean that completing the square doesn't work - it can ALWAYS be used to solve a quadratic. It just means that it can produce imaginary numbers.

Geometry of the Line (2) The Equation of the Line

The equation of the (straight) line isn't difficult. In fact it's dead easy.

The reason is this. Any curve or line has a statement that describes what happens to a y-co-ordinate when you plug in a given x-value. This is why we talk of f(x), because the y co-ordinate is basically the function of the x under certain conditions, which differ for various lines and curves. With this statement you have the potential to describe the co-ordinates of the curve or line completely knowing only the x-value.

The equation is simply the statement of what happens to the y co-ordinate when you select any x co-ordinate you like. It contains the gradient of the line within it, because clearly where the y-value is will depend on how steep the line is.

We often hear about y=mx + c , where m=gradient, and c= y-intercept (or where the line crosses the y-axis).

It is fine and dandy, to be sure.

But for MPC1 purposes the following equation is also useful, not least because solving it leads to y=mx+c.

(y - y₁) = m(x - x₁)

Where (x,y) is any, unknown point on the line and (x₁, y₁) is a known point.



For example: take a line passing through (2,4) and (6,10).

Firstly, work out m. 6/4. or 3/2.

Plug that into the equation: (y-4) = 3/2(x-6) or y-4 = 3/2x - 9 or y= 3/2x -5.

Suddenly you have your y-intercept (-5) and your y =mx+c.

The only thing is, you need the gradient, so you need to know two points on the line from the start.

Geometry of the Line (1) The MidPoint of the Line

Well I'm not sure that I'll bother to start with a definition of a line, although Barry Mazur, in his book that tied up my guts, Imagining Numbers, quotes Euclid: "one and only one line passes through any two distinct points on the plane". That makes sense to me.

For the purposes of this post the line will be assumed to be straight, and two dimensional. We are not going to muck about with curves and stuff.

So.

We have already seen that if you know two co-ordinates on the line you can calculate its gradient. This is very difficult to do with only one set of co-ordinates, unless you have a perpendicular line handy (which we don't, at the moment - that might come later).




If your line begins at 2,2 and ends at 8,4, as above, you need to think "How far has it gone x-wise and y-wise?" Well it has gone 6 units in the x direction and 2 in the y direction.

To find the midpoint of this line you can either just say well mid-way through a movement of 6 is 3, and mid-way through a movement of 2 is 1. So 2+3 = 5 (the x co-ordinate of your mid-point) and 2+1=3 (the y co-ordinate). Your midpoint is (5,3).

This isn't practical for a lot of lines which are (3 1/2, 5 1/2) and so on. So instead we use a simple formula:

The midpoint is, (x₁ + x₂)/2, (y₁+y₂)/2

So: (2+8)/2, (2+4)/2

Thus: 5, 3

The midpoint is (5,3), as above.


The midpoint is useful for all sorts of knotty problems involving perpendicular bisectors and stuff.