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2.

THE x-y PLANE

2.1 The Real Line
When we plot quantities on a graph we can plot not only integer values like 1, 2 and −3 but
also fractions, like 3½ or − 4¾ . In fact we can, in principle, plot any real number. Roughly
speaking real numbers are positive or negative numbers that can be represented by (possibly
infinite) decimal expansions like 3.14159… If you can’t think of any other sort of number that’s
O.K. There are other numbers, called complex numbers, but probably you haven’t come across
these yet.
Real numbers correspond to points on an infinite line. We choose one point, called the
origin, to represent 0. Then we step out to the right in equal steps and call these points 1, 2, 3, …
Going left from the origin, using the same size steps, we mark off −1, −2, −3, …

−4 −3 −2 −1 0 1 2 3 4
We can represent fractions by dividing these intervals into equal parts. In particular if we divide the
interval from 3 to 4 into 10 equal parts we can plot 3.1, 3.2, 3.3, … , 3.9.

3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0
By dividing each of these tiny intervals again into tenths we can plot points corresponding to
numbers like 3.14. By dividing and subdividing we can, in principle, find points that represent any
real number.

Perhaps this is a convenient moment to talk about the way real numbers came about. They
didn’t come all at once any more than you learnt all about real numbers in kindergarten. Just as you
first learnt about the counting numbers 1, 2, 3, … and only later extended your number universe to
include fractions, negative and irrational numbers so mankind gradually built the system of real
numbers over a period of many centuries. In fact the order in which you learnt about numbers at
school roughly corresponds to the order in which they developed throughout history.
The earliest numbers were 1, 2, 3, … and these were used to measure numbers of objects. A
man in ancient times might have had 3 wives, 27 children and 155 camels. These numbers were
satisfactory until it came to division. If the man had 155 camels and 10 sons he couldn’t leave each
son the same number of camels. Now camels could be cut into halves to facilitate the division but
half camels are nowhere near half as useful as whole ones! But if a man had 155 cheeses and 10
sons he could easily give each son 15½ cheeses. So fractions were invented.
Then it was discovered that numbers could be used to measure distances, and fractions
seemed to cope with this rather well. You could divide any interval length into any number of
equal portions and so fractions could be used to represent the resulting lengths. But then the Greeks
discovered that fractions aren’t enough to represent all distances. The well-known theorem of
Pythagoras about right-angled triangles (the square on the hypotenuse is the sum of squares on the
other two sides) led to a distance whose square is exactly 2.

h 1 h2 = 12 + 12 = 2

1
The problem was that they were able to show that no fraction could give exactly 2 when squared.
Their argument went something like this.

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m2 2 2 2
If n = 2 then m = 2n . But when factorised into prime factors m must have an even
number of 2’s and 2n2 must have an odd number. Since there’s a length whose square is exactly 2,
some numbers can’t be written as fractions or ratios of whole numbers. They became known as
irrational numbers. But it took many hundreds of years before mathematicians were really happy
The next number to be invented was zero. Originally there had seemed to be no need to
have a number for nothing. While it might be handy to have the number zero so that we could say
that the human population of Mars is 0 we could just as well say that nobody lives on Mars.
It was the development of the system of notation for representing whole numbers that made
it reasonable to have such a number. The ancient world had many different number systems but all
of them were very cumbersome. The Roman system was used widely throughout the first
millennium and well into the second. But numbers like CMLXXVIII were not only cumbersome to
write (978 is much easier), they were a nightmare when it came to performing calculations. If you
can remember the Roman numeral system, try multiplying CMLXXVIII by CCCLXVII using only
The so-called Arabic system, the one we use today, seems to have arisen in India around the
11th century and was soon used widely throughout the Arab world. Europe took somewhat longer
to be weaned off the venerable Roman system. But the place-value system where 908 means 9
hundreds, no tens and 8 units could not have existed if there had been no symbol to represent 0.
Even after it was introduced 0 wasn’t considered a number in its own right. It was
considered a piece of punctuation rather like the commas we use in 1,000. The number 908 could
have been written as 9 8 but using a blank to represent no tens would have been confusing. So the
zero symbol was used as a separator. It wasn’t until the invention of negative numbers, around the
seventeenth century, that zero was really accepted as a number in its own right.

It was around that time that numbers were represented on a “number line”:

½ 1 √2 2 3π 3¾ 4 …
| | | | | | | |

It’s perfectly reasonable to ask what numbers should represent points in the left-hand portion of the
number line. So negative numbers were invented and of course 0 had to put onto the number line to
separate the positives from the negatives:

… −4 −3¾ −π −3 −2 −√2 −1 −½ 0 ½ 1 √2 2 3π 3¾ 4 …
| | | | | | | | | | | | | | | | |

2.2 Two Axes Are Better Than One

In the 17th century René Descartes (1596 – 1650) discovered that geometry could be done
algebraically by introducing coordinates. We take it for granted in reading maps that we represent
points on a map by a pair of letters or numbers where one gives the position in a left-right sense and
the other gives the position in an up-down sense. So to find C7 we look where the C vertical and 7
horizontal intersect.
C

7 C7

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We navigate around a computer spreadsheet in a similar way. But in the 17th century it was quite a
novel idea.
Because we want arbitrary precision we use two numbers, instead of a letter and a number.
We take two copies of the number line, one horizontal and one vertical, and measure every point in
the plane against them. The horizontal axis, called the x-axis, has the numbers increasing from left
to right. The vertical axis, called the y-axis, has numbers increasing from bottom to top. The whole
plane is called the x-y plane and the point where the axes cut is called the origin.

y-axis 3

1
−3 −2 −1 1 2 3

−1
origin x-axis
−2

−3

The two numbers that describe the position of a point P are the x-coordinate (measuring the
horizontal position of P) and the y-coordinate (measuring the vertical position). We write the
coordinates as an ordered pair (x, y) with the horizontal coordinate first.

P (x, y)
y

Sometimes it’s convenient to use different scales on the axes, but it must be remembered
that this will make the graph appear more steep or less steep. Here we shall generally assume that
we’re using the same scale on both axes.
The two axes divide the x-y plane into four quadrants. These are numbered, starting with
the top right quadrant and moving around anticlockwise.

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The points on the two coordinate axes could be considered to be in more than one quadrant
but it’s more usual to exclude them from the quadrants altogether. If we do this then we can say
that the points in the 1st quadrant have positive x and y coordinates. Those in the 2nd quadrant have
a negative x coordinate but a positive y coordinate, and so on. We could summarise this as follows:

−, +)
(− (+, +)

− , −)
(− (+, −)
Example 1: Plot the following points on the x-y plane.
A = (0, 3), B = (4, 3), C = (−2, −2), D = (1, 1), E = (5, 1), F = (−2, 1), G = (1, −2), H = (5, −2).
Now join AB, BE, DE, AD, AF, CF, DG, EH, CG, FD, GH by straight lines.

A B

F D E

C G H

2.3 Functions and Their Graphs

Some graphs appear much smoother than others. These are usually graphs of functions that
can be expressed by simple formulae. A function (here we limit ourselves to describing functions
of a real variable) is a rule that associates with every real number x some real number y. In
principle the rule could be so complicated that it would take several pages to describe but here we
only consider functions where the rule can be expressed by a single formula. The formula y = x2 is
a very compact way of expressing the rule that we take a real number x, multiply it by itself, and
call the resulting square y. So if x = 3 then y = 9.
It’s not always the case that the relationship between two variables can be expressed by a
simple formula. If there was a simple formula that gave the value of certain shares in terms of the
number of months since it was listed on the Stock Exchange we would all be millionaires. But in
physics many phenomena behave according to reasonably simple formulae. Even in econometrics,
while it might not be possible to describe particular events, it’s possible to set up useful economic
models that describe the whole economy in terms of systems of formulae. By programming these
on a computer, economists can experiment with a mathematical model of the economy in a way that
wouldn’t be possible with the real economy. Of course we must point out that an economic model
consists of many equations connecting a very large number of variables while we are only going to
consider one equation connecting two variables. But we have to start somewhere!
Every function, where y is given as a formula in x, can be represented as a graph, in the
following way:

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(1) Substitute some values of x and come up with the corresponding values of y.
(2) Set these out in a table of values.
(3) Plot the corresponding points on the x-y plane.
(4) Join the points by a smooth curve.

x2 – 2x
Example 2: y =
5

x −3 −2 −1 0 1 2 3 4 5
y 3 1.6 0.6 0 −0.2 0 0.6 1.6 3

6
Example 3: y = + 3
x
x −4 −3 −2 −1 0 1 2 3 4 5
y 1.5 1 0 −3 × 9 6 5 4.5 1.2
Notice that there is no y value corresponding to x = 0.

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Example 4: y = ½x − 1
x −4 −3 −2 −1 0 1 2 3 4
y −3 −2½ −2 −1½ −1 −½ 0 ½ 1

2.4 Straight Lines

The equation y = 2x − 1 represents a straight line. We’ll see shortly that the general
equation for a straight line is y = mx + b, where m and b are constants. But first let’s consider a
segment, that is, the part of a straight line that joins two points (x1, y1) and (x2, y2).

Subscripts
Now we could have used a, b, c and d as the coordinates rather than x1, y1, x2 and y2. But
the subscript notation helps to remind us which are the x- coordinates and which are the y-
coordinates, as well as which coordinates belong to the first point and which belong to the second.
The important thing to remember in all this is that these are double-barrelled names, just like people
who have a family name and a given name. Here the x and the y don’t stand for anything by
x2 x
themselves and if we ever have y we certainly can’t cancel the 2’s to get y . The little subscripts
2
1 and 2 must never be separated from their x or y.

Midpoint
x1 + x2 y 1 + y2
The midpoint, M, of the segment from P1(x1, y1) to P2(x2, y2) is clearly 2 , 2 .

MIDPOINT
x 1 + x 2 y1 + y2
2 , 2

Remember this as the average of the x coordinates and the average of the y coordinates.
x2
P2

(x1+x2)/2
M
y2
y1 (y1+y2)/2
P1

x1

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1+9 6+4
Example 5: The midpoint of the line segment joining (1, 6) to (9, 4) is 2 , 2 = (5, 5).

Slope
Rise y2 − y1
The slope of the interval P1P2 is Run = .
x2 − x1

SLOPE
y2 − y1
x2 − x1
.
x2
P2

y1 − y2
y2
y1
P1 x1 − x 2

x1

8−5 3
Example 6: The slope of the line segment joining (3, 5) to (7, 8) is = .
7−3 4

Distance
The distance between the points P1 and P2 is (x1 − x2)2 + (y1 − y2)2 .

DISTANCE
(x1 − x2)2 + (y1 − y2)2
Take the horizontal difference and the vertical, square and add and then take the square root of this
sum. The reason for this is Pythagoras’ Theorem (the square on the hypotenuse is the sum of the
squares on the other two sides). So (P1P2)2 = (x1 − x2)2 + (y1 − y2)2.

Example 7: The distance between the points (3, 5) and (8, 4) is (3 − 8)2 + (5 − 4)2 = 25 + 1 =
26 .

One Point Plus Slope Equation of a Line

We’re now in a position to work out the equation of a straight line. If we know the slope of
the line and one point on the line we can find the equation as follows. Suppose the line passes
through P1(x1, y1) and has slope m. If P(x, y) is a typical point on the line then the slope of the line
interval P1P is equal to m.
P2
P
slope = m
P1

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y − y1
So = m. We can multiply both sides by x − x1 to get
x − x1
POINT + SLOPE EQUATION OF A LINE
y − y1 = m(x − x1)
This is just one of many forms for the equation of a straight line but it’s the most useful for our
purposes.

Example 8: The equation of the line through (1, 5) with slope 3 is y − 5 = 3(x − 1), ie y = 3x + 2.

Two Point Equation of a Line

Sometimes we don’t know the slope but we’re given two points on the line. Then all we do
is to work out the slope and substitute into the y − y1 = m(x − x1) formula.
P2(x2, y2)
y2 − y1
P(x, y) slope = =m
x2 − x1
P1(x1, y1)
y2 − y1
The equation is thus y − y1 = (x − x1).
x2 − x1
But rather than having an extra formula to learn it’s better to work out the slope first.
.
2 − (−1) 3
Example 9: The slope of the line through the points (3, −1) and (5, 2) is = 2 . So the
5−3
3 3 11
equation of the line through these points is y − 2 = 2 (x − 5), ie y = 2 x − 2 which can be written as
2y = 3x − 11 or 3x − 2y − 11 = 0.

Slope Intercept Equation of a Line

If we know that the slope of a line is m and the line cuts the y-axis at y = b then the
equation of the line simplifies to
SLOPE + INTERCEPT EQUATION OF A LINE
y = mx + b
This is because the line passes through (0, b) and so its equation is
y − b = m(x − 0) which, on simplification becomes y = mx + b.

Example 10: The equation of the line with slope 5 that cuts the y-axis at y = −2 is y = 5x − 2.

General Equation of a Line

All of the above methods rely in some way on the line having a slope. But a vertical line
doesn’t have a slope. We might say that it has infinite slope but this is not very useful since we
can’t use “infinity” as if it was a number. If this line cuts the x-axis at x = a then the x-coordinate
of every point on the line will be “a” and so the equation of this vertical line will be x = a.

x=a

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The general equation of a straight line is px + qy + r = 0 (with not both of p, q equal to 0)
r
If p ≠ 0 and q = 0 this becomes x = − p which represents a vertical line.
p r p r
If p ≠ 0 and q ≠ 0 we can rewrite px + qy + r = 0 as y = − q x − q . If we put m = − q and b = − q
this becomes the standard y = mx + b form for the equation of a line.

EQUATION OF A STRAIGHT LINE

ax + by + c = 0
Clearly it doesn’t matter which symbols we use. We used p, q, r at first so as not to confuse this “b”
with the “b” in the y = mx + b form.

Parallel Lines
If two lines are parallel they must have the same slope. So lines with slopes m1 and m2 are
parallel if and only if m1 = m2.

PARALLEL LINES
m1 = m2
Perpendicular Lines
Suppose we have two perpendicular lines of slopes m1 and m2 respectively. The
relationship between these two slopes is given by:
PERPENDICULAR LINES
m1 m2 = − 1
We can see this by using similar triangles.
slope = m2
slope = m1
P

m1

Q R
1 S n

PS RS
Since ∆PSQ is similar to ∆RSP (corresponding angles are equal) then QS = PS (corresponding
m1 n m1
sides are proportional). So 1 = m and hence n = m12. Now from triangle RSP, m2 = − n
1
m1 1
= − m 2 = − m which gives m1m2 = −1.
1 1

Example 11: The slope of any line parallel to y = 4x + 7 is 4. The slope of any line perpendicular
to this line is −¼.

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Example 12: If A = (1, 7), B = (3, 7) and C = (5, 3) we can find the area of the triangle ABC using
the “half base times perpendicular height” formula. Taking the base to be AC we need to calculate
the distance between A and C. This is 16 + 16 = 32 .
But what is the perpendicular height? It’s BM where M is the foot of the perpendicular from B to
the base AC. B
A

M
C
3−7
The slope of AC is = −1. So the equation of AC is y − 7 = − (x − 1), ie y = − x + 8.
5−1
Using the m1m2 relationship we find that the slope of BM is 1. So the equation of M is
y − 7 = x − 3, ie y = x + 4.
Solving y = − x + 8 and y = x + 4 simultaneously we get − x + 8 = x + 4, so 2x = 4 and x = 2.
This gives y = 6, so M is (2, 6). Hence the perpendicular height is the length of BM = 1 + 1 = 2.
The area is therefore ½ 32 2 = 4.
[There are other methods for working out the area of a triangle that require less calculation but we
did it this way to illustrate these techniques.]

EXERCISES FOR CHAPTER 2

Exercise 1:
Plot the following points: A(1, 3), B(−2, 4), C(5, −3), D(−2, −1).

Exercise 2: In which quadrants do the following points lie?

(i) (3, 2); (ii) (−3, −2); (iii) (2, −5); (iv) (−1, 7).

Exercise 3:
(a) Draw up a table of values for y = x2 + x + 1, using the values x = −3, −2, −1, 0, 1, 2, 3.
(b) Draw the graph of this function.
(c) Use the graph to estimate the slope at x = 1.
(d) Use the Counting Squares Method to estimate the area under the graph from x = 0 to x = 2.

Exercise 4:
2(1 − x)
(a) Draw the graph of y = 1 + x , from x = −5 to x = 5.
(Warning: It comes in two separate pieces.)
(b) Use the graph to estimate the slope of the tangent at x = 1.
(c) Find the equation of the line through (1, 0) which is perpendicular to this tangent. This is called
the normal at the point. Draw this normal on your graph.
(d) It can be shown that the shortest distance between two curves occurs when the line joining the
points is a normal (perpendicular) to both curves. Use this fact to work out the shortest distance
between the two halves of this curve.

Exercise 5: If A = (1, 3) and B = (4, 7) find:

(i) the midpoint of AB;
(ii) the slope of AB;
(iii) the length of AB;
(iv) the equation of AB.

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Exercise 5: Find the equation of the line with slope 2 that passes through the point (1, 3).

Exercise 6: A certain curve has slope 2 at the point (1, 3). Find the equation of the tangent at that
point.

Exercise 7: Find the equation of the line that cuts the x-axis at x = 3 and the y-axis ay y = 2.

Exercise 8:
(a) Find the slope of the line 5y − 2x + 7 = 0.
(b) Find the slope of any line that’s perpendicular to the above line.

Exercise 9: Let A, B, C, D be the points in exercise 1.

(a) Find the length of CD.
(b) Find the slope of AB.
(c) Find the slope of any line that is perpendicular to AB.
(d) Find the midpoint of AC.
(e) Find the equation of AB.

Exercise 10: Let A = (2, 5), B = (3, 8), C = (0, 4).

(a) Find the slope of AC.
(b) Find the equation of the line AC.
(c) Find the equation of the line through B that is perpendicular to AC.
(d) Find the point, M, where the two lines obtained above intersect. (Solve the two equations
simultaneously.)
(e) Find the lengths of AC and BM.
(f) Use the formula “half base times perpendicular height” to find the area of the triangle ABC.

SOLUTIONS FOR CHAPTER 2

Exercise 1:

Exercise 2:
(i) 1st (ii) 3rd (iii) 4th (iv) 2nd
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Exercise 3:
(a)
x -3 -2 -1 0 1 2
y 7 3 1 1 3 7
(b)

(d) The area under the graph from x = 0 to x = 2 is approximately 7.

Exercise 4:
(a)
x -5 -4 -3 −2 −1 0 1 2 3 4 5
y −3 −3.3 −4 −6 × 2 0 −0.7 −1 −1.2 −1.3

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(b) The slope of the normal at x = 1 appears to be 1. (This is in fact the exact value.)

(c) Using this value, the equation of the normal at x = 1 is y − 0 = (x − 1), which can be simplified
to y = x − 1.

(d) The normal at x = 1 passes through (−3, −4) and appears to be the normal at that point. (In fact
calculus can tell us that it is.) So the shortest distance between the two halves of our graph is the
distance between (1, 0) and (−3, −4). This distance is (−3 − 1)2 + (−4 − 0)2 = 16 + 16 = 32 =
4 2.

Exercise 5:
1+4 3+7 7−3 4
(i) Midpoint = 2 , 2 = (5/2, 5); (ii) Slope = = ;
4−1 3
(iii) Distance = (4 − 1)2 + (7 − 3)2 = 5

4
(iv) y – 3 = 3 (x – 1) which can be written as 4x – 3y + 5 = 0

Exercise 6: y– 3 = 2 (x – 1) which can be written as 2x – y + 1 = 0.

2 2
Exercise 7: The slope is = − 3 . It passes through (0, 2) so the equation is y − 2 = − 3 (x − 0). This
can be simplified to 2x + 3y – 6 = 0.

2 7
Exercise 8: Writing 5y – 2x + 7 = 0 in the y = mx + b form we get y = 5 x − 5 . The slope is
2 5
therefore 5 . For perpendicular lines m1m2 = −1 so the slope of perpendicular is − 2 .

Exercise 9:
(a) CD = (−2 − 5)2 + (−1 + 3)2 = 49 + 4 = 53 .
4−3 1
(b) slope of AB = =−3.
−2 − 1
(c) slope of a perpendicular to AB is 3.
1+5 3−3
(d) midpoint of AC is 2 , 2 = (3, 0).
1
(e) equation of AB is y − 3 = − 3 (x − 1), which can be simplified to x + 3y = 10.

Exercise 10:
5−4 1
(a) Slope = =
2−0 2
1
(b) y – 5 = (x – 2) which can be written as x – 2y + 8 = 0
2
(c) The slope of perpendicular is –2, so the equation is y – 8 = −2 (x − 3) which can be written as
y = −2x + 14
x − 2y + 8 = 0
(d) Solving simultaneously we get x = 4. Substituting back into the first equation
y = − 2x + 14
we get y = 6. So M is point (4,6).

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(e) AC = (0 − 2)2 + (4 − 5)2 = 5.
BM = (4 − 3)2 + (6 − 8)2 = 5.

5
(f) The area of the triangle ABC = ½. 5 . 5 = 2 .

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