UNIT 3 – GRAPHICAL REPRESENTATION

Exercises:   [1]   [6]   [7]   [8]   [9]

INTRODUCTION

By now you should be keenly aware of the value of a pictorial approach to problem solving. This unit is about graphical representation, that is plotting points on graphs and reading values from graphs. Once you can identify points on a graph you will want to know more about them. How far apart are they? Do they lie in straight lines? And so on. These questions are crucial in map reading and you will find that mastery of this unit will improve your geographical skills.

OBJECTIVES

1. To plot points and lines on graphs.
2. To use the Cartesian and polar co-ordinate systems.
3. To calculate the distance between a pair of points.
4. To find the equation of a straight line.
5. To find where (if at all) a pair of lines intersect.

REFERENCES

If you wish to learn more about the mathematics of map making look at D.H. Malings Co-ordination systems and map projections [10]. For an interesting, historical development of the Theorem of Pythagoras read Chapter 5 of J. Bronowskis The ascent of man [4].

ADVICE

You will find it helpful to use a calculator while you are working through this unit.

MAP REFERENCES

Historically, a place (represented by a point on a map) is identified by a pair of numbers, for example:

Brisbane: Lat. 27° 28'S, Long. 153° 2'E

Latitude and longitude are angular distances measured from the centre of the earth. For many purposes it is more convenient to refer to a position using actual distances (in metres) along the earths surface from some reference point. Nowadays, this is done using standard grid systems.

On the Australian Map Grid (A. M. G. ), Capital Hill in Canberra is described by:

Easting 6932

Northing 60906

This means that Capital Hill is 6932 metres east and 60906 metres north of the standard reference point (called the origin) of this grid. On the map of the Canberra region in the National Topographic Map Series (1:1000000) all the places are at least 6000 metres east and 60000 metres north of the origin. Thus the first digit of the easting and the first two digits of the northing are omitted from the identification and the grid reference for Capital Hill is 932906: the easting first and then the northing.

Figure 3.1

In mathematical notation, the position of Capital Hill would be written as (932, 906)

POINT ON A PLANE

Draw two perpendicular lines.

The point where they intersect is the origin (denoted by 0). The horizontal line is called the x-axis and is treated like the real number line with the left of 0 (i.e. west) associated with negative numbers and the right (east) with positive numbers. The vertical line is called the y-axis; the part of the line above 0 (north) is associated with positive numbers and the part below 0 (south) with negative numbers (Figure 3.2).

Figure 3.2

The position of any point on the plane is identified by two number which are called its co-ordinates. In Figure 3.3, the point marked P is associated with the pair of numbers (2, 3) and the point Q with (-1, 1).

Figure 3.3

By convention, the first number in the pair is the position with respect to the x-axis (the easting) and the second number is the position with respect to the y-axis (the northing). Thus, P is 2 units to the right of the origin and 3 units above it.

Exercises 3.1 Solutions

1. Plot in a plane the following points;

A (1, 3); B (-2, -4); C (, -p ) and D (-4, -2).

2. What are the pairs of numbers associated with each of the points P, Q, R and S in Figure 3.4.

Figure 3.4

STRAIGHT LINE TROUGH THE ORIGIN

The table below shows the total amounts of interest paid on loans of various sizes borrowed at an interest rate of 7% reducing and repaid in monthly instalments over 25 years.

(Note that all the interest amounts are greater than the amount borrowed that $9000 loan will really cost you $20385.) Take the interest as the y co-ordinate and the loan as the x co-ordinate and plot the points on a graph(Figure 3.14).

Figure 3.14

Notice that a straight line can be drawn which goes through all the four points (and, incidentally, through the origin). Thus the line is a characteristic which is common to all four points. Is there some way that the information in the table or on the graph can be written more simply?

Calculate the ratio of the interest to the loan for each pair of values. It comes to the same value, 1.265, in each case

.

For any of the points, if its co-ordinates are (x, y) then

so the values in the table are summarised by this equation, or by the alternative form

y = 1.265x

If the interest rate were 8% reducing (over 25 years) the total interest for the various loan amounts would be:

Figure 3.15

For the higher interest rate, the points again lie on a straight line through the origin and the ratio of the interest to the loan is again the same for each point:

The equation in this case is

or y = 1.316x

If you know only the equation for the particular interest rate (and the repayment time), you can use substitution to find the interest payable on any loan.

e.g. At 7% over 25 years, if the loan is $7000, the interest is

y = 1.265 × 7000 = $8855.

e.g. At 8% over 25 years, if the loan is $8500, the interest is

y = 1.316 × 8500 = $11186.

Thus the equation characterises all the points on the line not only the ones in the table.

Any straight line which goes through the origin can be describe by an equation

y = ax, where a is some numerical constant.

The value of a gives the slope (or gradient) of the line. For example, the line with a = 1.316 rises more steeply than the one with a = 1.265. If a = 0 the line coincides with the x axis . If a > 0 the line rises from left to right and a is called the rate of increase of y with respect to x. If a < 0 the line falls from left to right and a is called the rate of decrease of y with respect to x.

Exercises 3.6 Solutions

Draw graphs of the lines whose equations are given below (i.e. choose some x values, calculate the corresponding y values and plot the points). In each case state whether the slope, a, is a rate of increase or decrease (or neither).

1. y = 3x 2. y = -2x 3. y = 0.15x

4. 5y = 2x

CAUTION ABOUT GRAPHS

A large value of a does not necessarily mean that the line is steep it all depends on the scales of the units on the x- and y-axes. The graphs in Figures 3.16 and 3.17 represent exactly the same functions, y = x and y = x , but in Figure 3.16 the impression is given that the rates of increase (slopes) are small and nearly the same whereas in Figure 3.17 the opposite impression is given.

Figure 3.16 Figure 3.17

The impression of steepness is highly subjective. It is a well known trick in advertising and politics to manipulate steepness by a suitable (or unsuitable) choice of units.

GENERAL EQUATION OF A STRAIGHT LINE

Consider the straight y = x which passes through the origin and another line parallel to it but 2 y-units above it (Figure 3.18). The new line will not go through the origin but will cut the y-axis at the point (0, 2). To find the y value on the second line corresponding to any x value, find the y value for the first line then add 2,

e.g. corresponding to x = 3,

y₁ = ½(3) = 1½ on the first line,

y₂ = 1½ + 2 = 3½ on the second line.

Thus any point on the second line can be calculated using the equation

y =½x + 2

In general, the equation of a straight line is

y = ax + b

where a and b are constants; a is the slope of the line and b is called the intercept of the line on the y-axis since the line cuts the axis at the point (0, b)

Figure 3.19

Example 3b. Draw the graph of y = x + 2.

Does the point (1, 3) lie on the line?

This line y = x + 2 has slope = 1 and intercept = 2 (Figure 3.20)

Figure 3.20

L.H.S. = y = 3 and R.H.S. = x + 2 = 1 + 2 = 3.

Exercises 3.7 Solutions

1. Draw the graphs of the following lines:

(a) y = 3x + 1 (b) y = 2 x (c) y = 1.6x 6.37

2. Find the slope and intercept for each of the following lines:

(a) 3y = 6 2x (b) x y = 3

3. Do the given points lie on the given lines?

FINDING THE EQUATIONOF A STRAGIHT LINE

If several points lie on a straight line, you can use their co-ordinates to find the equation of the line. The following example shows how.

Take any tow points, for example (-1, -1) and (4, 3) as in Figure 3.21

Figure 3.21

The slope of the line is the ratio of the change in the y value between the points to the change in the x value;

The equation of the line must be:

y = + b

where b is the intercept. To find b, substitute the co-ordinates of any one of the points into the equation and solve for b

e.g. using the point (4, 3)

3 = 4 + b so b =

(Check that you get the same answer if you use the point (-1, -1) instead.)

Check back to the examples about interest rates and notice that the equations were found using this method and taking the origin (0, 0) as one of the points used to calculate the slope.

There area couple of special cases of this method which should be mentioned. First consider the pair of points (2, 3) and (-1, 3) in Figure 3.22. The slope of the line joining them is

;

the line is parallel to the x-axis and has equation y = 3.

Figure 3.22

In general, if tow points have the same y coordinate, y₁, the line joining them is parallel to the x-axis and has the equation y = y₁.

Second, consider (3, 2) and (3, -1) also in Figure 3.22. In this case

which is undefined.

The line joining these points is vertical and has the equation x = 3. In the general case when two points have the same x co-ordinate, x₁, the line joining them is parallel to the y-axis and has the equation x = x₁.

Exercises 3.8 Solutions

1. Find the equations of the lines through the following pairs of points:

(a) (1, 1) and (2, 4) (b) (-1, -1) and (2, 4)

(c) (0, 6) and (0, 3) (d) (0, 0) and (3, 3)

2. Find the slope of the land between points A and B in Figure 3.23. The altitude of A is 250m (above sea level) and the altitude of B is 200m. The distance between them is 300m.

PARALLEL AND INTERSECTING LINES

You will already have noticed that if two lines have the same slopes but different intercepts, they are parallel, as in Figure 3.24.

Figure 3.24

A pair of different lines which are not parallel must meet somewhere (Figure 3.25)

Figure 3.25

Let (p, q) be the co-ordinates of the point P where the lines y = x + 2 and y = 1 x meet. Since P is on both lines (p, q) must satisfy both equations,

i.e. q = ½p+ 2

and q = 1 – p

therefore ½p + 2 = 1- p.

Solve this equation to find p :

Add p to both sides p + 2 + 1

Subtract 2 from both sides p = -1 so p = .

Now substitute this value into the equation for either one of the lines to find q,

,

hence P is the point .

(Check that substituting the x co-ordinate of P into the other equation y = 1 x would have given the same value, q = .)

Exercises 3.9 Solutions

For each of the following pairs of lines check if they are parallel, and if they are not, find their point of intersection.

1. y = 2x + 3 and 2y = 4x – 1

2. 2y = x + 1 and y = 2x – 1

3. y 2 = 2x and 2y = 3 x