MATH 2.2: Introducing co–ordinate geometry	PPLATO @
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1 Opening items

1.1 Module introduction

You probably think of geometry as a subject that involves pictures of points, lines, curves, surfaces and so on; and indeed the origins of the topic are intimately related to the figures that one might draw on a piece of paper. However, if you stick to this classical approach to the subject, you very soon find that the pictures become exceedingly complicated and difficult to visualize; and particularly so in three dimensions. A break–through came in the 17th century when René Descartes (1596–1650) showed that algebraic methods could be applied to geometry, and formulated the subject that we know today as coordinate (or Cartesian) geometry. Of course we continue to use diagrams, but we are no longer bound by them, and most of our arguments are entirely algebraic. You have almost certainly met the idea of a Cartesian coordinate system in school (although you may not have called it that), and so our first task in this module will be to review some of the fundamental ideas. In Sections 2 and 3 we are concerned solely with geometry in two dimensions, but in Section 4 we discuss briefly how some of the ideas can be extended to three dimensions.

The essence of the Cartesian approach to geometry is to specify the position of a point by distances, in mutually perpendicular directions from a given fixed point. In a Cartesian coordinate system two distances are needed to specify the location of a point on a flat surface (known as a plane) and these distances are called coordinates. The given fixed point is called the origin (of coordinates) and the two directions are called axes: one of these is known as the x–axis and the other as the y–axis and the coordinates are written as an ordered pair of numbers. Thus the point represented by the pair (2, 3) has x–coordinate equal to 2 and y–coordinate equal to 3.

Having specified the points in a plane by a Cartesian coordinate system, it is natural to ask how we can determine the distance between two points in terms of their coordinates. We answer this question in Subsection 2.6.

Apart from a point, a straight line is the simplest geometric entity, and we show, in Subsection 2.2 that it can be represented by a simple equation relating values of y for points on the line to corresponding values of x. There is just one equation for a given line, but in Subsection 2.3Subsections 2.3 and Subsection 2.42.4 we will see that this equation can be manipulated into various equivalent forms in order to provide the information that we require.

In Section 3 we take an introductory look at the equation of a circle. In Subsection 3.1 we first derive the equation for a circle with its centre at the origin; then we generalize this equation to cover the cases where the centre is at any point in the plane. In Subsection 3.2 we derive equations for the tangents to a circle, in cases where the tangent passes through a given point outside the circle and in cases where the point of contact on the circle is specified. Subsection 3.3 deals briefly with an important alternative coordinate system – the polar coordinate system.

Section 4 extends some of the earlier results to three dimensions. Subsection 4.1 deals with points in three dimensions and the calculation of the distance between any two of them. Subsection 4.2 presents the equation of a plane surface which has similarities with the general form of the equation of a line in two dimensions. Finally, Subsection 4.3 considers the specification of a line in three dimensions.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to theSubsection 1.3Ready to study? Subsection.

1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

1.3 Ready to study?

2 Straight lines

2.1 Cartesian coordinates

The framework that we use to draw pictures of curves is provided by a Cartesian coordinate system, consisting of a pair of straight lines at right angles, called coordinate axes, as shown in Figure 1. The two lines are infinitely long, but we draw only a portion of each near the point where they cross, which is known as the origin. The horizontal line is called the x–axis and the vertical line the y–axis, and both lines are scaled to indicate distance from the origin along the line. Distances to the right of the origin along the x–axis are positive while those to the left are negative; along the y–axis, distances above the origin are positive while those below are negative. It is usual to place arrowheads on the axes to show the direction of increasing x and y (since occasionally we may wish to change the orientation of the axes).

Any pair of values for x and y can be used to determine a point in the coordinate system. Thus, for example, the ordered pair (2, 1) is shown as the point A in Figure 1; a vertical line from A to the x–axis meets it at x = 2, while a horizontal line from A to the y–axis meets it at y = 1. Similarly, the points B and C in Figure 1 represent the ordered pairs of numbers (1, 1) and (−1, 2), respectively.

The points where a given line crosses the x– and y–axes are known as the x– and y–intercepts, respectively (although some authors refer to the y- intercept as the intercept).

2.2 Equation of a straight line - gradient and y–intercept

Table 1 A table of
values for f(x) = 2x + 1.

x	y
−3	−5
−2	−2
−1	−1
0	1
1	3
2	5
3	7

A linear function is a function of the form f(x) = mx + c where m and c are constants. For example, f(x) = 2x + 1 when m = 2 and c = 1, and, if we let y denote the values of f(x), we may plot the graph of y = 2x + 1 as in Figure 2. It should come as no surprise that the graph is a straight line; this is after all why such functions are called linear.

You may have used a table of values (as in Table 1) to construct such a graph.

In fact, once you know that the graph must be a straight line, it is quite unnecessary to tabulate all these values. You just need to know two well spaced points, such as (−3, −5) and (3, 7), then join them with a straight line (although it is wise to calculate a third point and check that it lies on the line that you have drawn).

It is essential to realize that the equation y = 2x + 1 is a true statement if, and only if, we choose a point (x, y) that lies on the line. For example, the values x = 2.5 and y = 6 correspond to the point (2.5, 6) which lies on the line because 6 = 2 × 2.5 + 1; on the other hand, the point (3.5, 7) does not lie on the line because 7 ≠ 2 × 3.5 + 1. i (Notice that we do not need to draw the graph in order to reach these conclusions.)

The two constants, m and c, which characterize a linear function are called the gradient (or sometimes the slope) and the y–intercept, respectively. i

Look again at Figure 2. Note that the straight line y = 2x + 1 crosses the y–axis at the point where y = 1, which is the value of c (i.e. the y–intercept). Also, from Table 1, we see that a unit increase in the value of x gives rise to an increase of two units in the value of y. This factor of 2 corresponds to the value of m, the gradient of the linear function, and, in graphical terms, it is this value that determines how steep the line will be. In the case of m = 2, one step to the right implies two steps up. i

Figure 3 represents an arbitrary line with equation y = mx + c. The constant c is known as the y–intercept because the point (0, c) on the y–axis lies on this line.

Suppose now that (x₁, y₁) and (x₂, y₂) are any two points lying on the line, then

y₁ = mx₁ + c

andy₂ = mx₂ + c

Subtracting these equations we have

y₂ − y₁ = mx₂ − mx₁

If we let x₂ − x₁ = h, as in Figure 3, then y₂ − y₁ = mh and tanθ = m.

Figures 4a and 4b show, respectively, the effects on the line of different values of m and of c.

Varying the value of m, and keeping c fixed, will rotate the line about a fixed point (the intercept) on the y–axis. If m is positive then the graph slopes upwards from left to right, whereas if m is negative then the graph slopes downwards from left to right. Note that the graph is horizontal when m = 0.

Varying the value of c, and keeping m fixed, produces lines with the same gradient (they are parallel to one–another). If c = 0 the line passes through the origin; increasing c raises the line, while decreasing c lowers it.

Equations of lines need not be in gradient–intercept form, for example the equation x = 5y + 3 represents a line, but its gradient and y–intercept are not immediately obvious. However, if we rearrange this equation into the form $y = \dfrac15x - \dfrac35$ it becomes clear that the gradient is 1/5 while the y–intercept is −3/5.

2.3 Straight line with a given gradient through a given point

Suppose that we wish to find the equation of the line through the point (1, −2) with gradient 0.5. We know the gradient, so the equation must be of the form

y = 0.5x + c

and it remains to find the value of c. Using the fact that the point (1, −2) lies on the line we have −2 = 0.5 + c from which it follows that c = −2.5, and so the required equation is y = 0.5x − 2.5.

Alternatively, we could use Equation 2, letting the known point (1, −2) be the first point on the line, and an arbitrary point (x, y) on the line be the second, then

$0.5 = \dfrac{y-(-2)}{x-1}$

which can be rearranged to give y = 0.5x − 2.5 as before.

To obtain a general formula we can let the given point have coordinates (x₁, y₁) and let the gradient of the line be m, then for an arbitrary point (x, y) on the line, using Equation 2, we have $m = \dfrac{y-y_1}{x-x_1}$ which can be rearranged into

i.e. y = mx + y₁ − mx₁, from which it is easy to see that the y–intercept is y₁ − mx₁. i

While it is quite acceptable to find the equation of the line by calculating the value of c, it is perhaps slightly easier to use Equation 2.

$m = \dfrac{y_2-y_1}{x_2-x_1}$(Eqn 2)

2.4 Other forms of the equation for a straight line

Given two points on the line

Suppose that we are given two points, say (2, −5) and (7, 3), and that we are asked to construct the equation of the line joining them. A straightforward method would be to first use Equation 2 to find the gradient, and then to find c, the y–intercept. From Equation 2 we have

$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{3-(-5)}{7-2} = \dfrac85$

If the equation of the line is y = mx + c, and using either one of the points on the line, (7, 3) say, we have $3 = \dfrac85 \times 7 + c$ so that c = −41/5.

Finally we can write the equation of the line as $y = \dfrac85x - \dfrac{41}{5}$

The above method is quite acceptable, but it is slightly easier to use Equation 2 twice.

First we find the gradient of the line to be 8/5 as before, but then we choose an arbitrary point (x, y) on the line, and use Equation 2 again with say (7, 3) as the first point and (x, y) as the second, which gives

$m = \dfrac{y-3}{x-7}$

Since we know the value of m to be 8/5, this gives

$\dfrac85 = \dfrac{y-3}{x-7}$

which can be rearranged to give $y = \dfrac85x - \dfrac{41}{5}$.

We will now generalize this result, using the second of these methods.

Let the coordinates of two given points A and B be (x₁, y₁) and (x₂, y₂) respectively, i and let P(x, y) be a general point on the line, as in Figure 5.

Using Equation 2 and the coordinates of points A and B we have

$m = \dfrac{y_2-y_1}{x_2-x_1}$

but using Equation 2 with the coordinates of points A and P gives us

$m = \dfrac{y-y_1}{x-x_1}$

Therefore equating these two expressions for m gives us

In physics we are often confronted with relationships which are certainly not linear, for example the periodic time T for small vibrations of a simple pendulum is given by the formula $T=2\pi\sqrt{l/g\os}$ where l is the length of the pendulum and g is the magnitude of the acceleration due to gravity.

Such equations can sometimes be simplified by a change of variable, and in this case if we let y = T and x = l then the equation becomes $y = (2\pi/\sqrt{g\os})x$ which is the equation of a line through the origin with gradient $2\pi/\sqrt{g\os}$.

Such an expression could form the basis for an experiment to determine the value of g. We could perhaps vary the length l of the pendulum and measure the corresponding times T, then plot y against x and measure the gradient. Once we have estimated this value it would be then a simple matter to estimate the value of g, as the following exercise shows.

Given the intercepts on the axes

The form of the above equation makes it particularly easy to find the two intercepts; but the reverse is also true, if the intercepts are given it is straightforward to write down the equation of the line.

In Figure 6 the intercepts of the line on the x–and y–axes are A(a, 0) and y B(0, b), respectively, and the equation of the line may be written using

We can illustrate the use of Equation 5 with the following example.

At time t, the velocity υ_x of an object moving in a straight line with constant acceleration a_x is given by υ_x = u₀ + a_xt, where u₀ is its initial velocity in the x–direction. i

General form of the equation of a straight line

The previous forms of the equation of a straight line have a deficiency which is not immediately apparent, as the following example is intended to show.

Any line parallel to the y–axis has an equation x = b for some constant b.

It is often useful to have a general form for the equation of a line that includes all possible lines. Thus

Note that by putting b = 0 we obtain the equation ax + c = 0 from which we find that x = −ac which is a line parallel to the y–axis.

2.5 Intersection of two lines

Parallel lines and perpendicular lines

If two lines have the same gradient then they are parallel. For example, the lines with equations y = 2x + 3 and y = 2x − 1 are parallel.

The essential point to notice from Example 2 is that we have two lines through the origin, y = mx and y = −x/m, which must be perpendicular to each other for any value of m. This means, for example, that we can see instantly that the two lines y = 2x and y = −0.5x are perpendicular because the product of their gradients is −1.

We quote without proof the general result that if the equations of two lines are y = m₁x + c₁  and  y = m₂x + c₂ then:

Note that the lines x = a, y = b, where a and b are constants, are perpendicular. We cannot use Equation 7 in this case (but we do not need to).

Intersection of two lines

Two lines that are not parallel intersect at a point. In order to find the coordinates of the point of intersection we could certainly draw the graphs of the lines on the same set of axes, and find where they cross. The lines y = x − 1 and x + y = 3, which appear to intersect at (2, 1), are shown in Figure 8.

A graphical approach is sometimes useful, but it can be laborious and rather inaccurate, so we would prefer an algebraic method.

The equation y = x − 1 is a true statement only for points (x, y) lying on the first line, while the equation x + y = 3 is a true statement only for points (x, y) lying on the second line. If we are told that both equations are true, then we must be referring to the particular point (x, y) that lies on both lines.

A pair of equations such as

$\left. \begin{align}& &y=x-1\\ \text{and}& &x+y=3\\ \end{align}\right\}$ i

which are true for the same values of x and y are known as simultaneous equations, and they can be solved to find these values of x and y. In this instance we already suspect (from the graph) that the values x = 2 and y = 1 are the solution, and we have merely to verify that this is so by substituting these values into the equations. We do indeed find that both equations are satisfied, because 1 = 2 − 1 (the first equation) and 2 + 1 = 3 (the second equation).

Algebraic solution

There are several methods of solving a pair of simultaneous equations, but they all reduce to essentially the same idea – we use one of the equations to eliminate one of the variables, then solve the resulting equation in the remaining variable.

We will use the equations

 y = x + 1(i)

 2x + y = 3(ii)

to illustrate the general methods. Notice that we have numbered the equations in order to clarify the subsequent algebraic steps, and you should do likewise in your own work.

Method 1 - solving by direct substitution

We substitute for y using Equation (i) into Equation (ii) to obtain

2x + (x + 1) = 3

so that3x + 1 = 3

i.e.3x = 2

and therefore   $x = \dfrac23$

Substituting this value for x into Equation (i) we find that $y = \dfrac53$.

We then check that these values also satisfy Equation (ii),

and we find $2x + y = 2 \times \dfrac23 + \dfrac53 = \dfrac43 + \dfrac53 = \dfrac93 = 3$.

Method 2 - solving by indirect substitution

We re–write the two equations as

−x + y = 1(i)

2x + y = 3(ii)

We eliminate y by subtracting Equation (i) from Equation (ii) to obtain

2x − (−x) + y − y = 3 − 1

i.e.3x = 2 as before.

Then we can use either Equation (i) or Equation (ii) to calculate y and use the other equation as a check.

Alternatively, we could begin by eliminating x, and multiply Equation (i) by 2 and add the result to Equation (ii).

 −x + y = 1(i)

 2x + y = 3(ii) i

−2x + 2x + 2y + y = 2 × 1 + 3

i.e.3y = 2 + 3 = 5

so that$y = \dfrac53$

Either Equation (i) or (ii) yields x = 2/3 and the other equation can be used as a check.

The choice of method is often a matter of taste, although sometimes the numbers involved ensure that one method is quicker than another. It can be helpful to sketch the graphs of the two lines in order to get a rough idea of where they intersect, and perhaps to give an extra check on the results.

Strange cases

There are some particular cases that may lead you to some strange results. Suppose we were asked to find the point of intersection of the lines

x − 2y = 3  and  2x − 4y = 6

and let us suppose that inspiration strikes and you guess the solution x = 5 and y = 1. You try these values in the equations, and you find that they work; so everything is fine and you have found the solution. Right?

However, the values x = −1, y = −2 also satisfy both equations.

How can this be? How can a pair of lines intersect at more than one point?

In fact, the two equations both represent the same straight line. (You can obtain the second equation by multiplying the first equation by 2). Any point on the common line has values of x and y which satisfy both equations.

Ill-conditioned equations

Perhaps another example will make things clearer.

In such cases, where the lines are almost parallel, very slight changes in the given values of the constants can lead to massive changes in the calculated values of the solution. Such systems of equations are said to be ill–conditioned.

2.6 Distance between two points

We have already seen in Example 2 how Pythagoras’s theorem can be used to find distances, but here we discuss the distance between two arbitrary points P and Q, as shown in Figure 10. The point R has the same y–coordinate as that of P and the same x–coordinate as that of Q.

The distance PR is the difference in the x–coordinates of P and R, i.e. (x₂ − x₁). The distance RQ is the difference in the y–coordinates of R and Q, i.e. (y₂ − y₁). Applying Pythagoras’s theorem to the triangle PQR gives the equation (PQ)² = (PR)² + (RQ)², i.e. (PQ)² = (x₂ − x₁)² + (y₂ − y₁)².

Hence the distance we require is given by

3 Circles

3.1 Equation of a circle, given its centre and radius

Our objective in this section is to determine the general equation of a circle, but first we consider a simple case, when the centre of the circle is at the origin. The point P lies on the circumference of the circle as shown in Figure 11a. Since OP is the radius of the circle its length will be the same whatever the choice of P(x, y ) on the circumference, i.e. OP = R where R is the radius of the circle. But, using Equation 8, (OP)² = (x−0)² +(y−0)² = x² +y² and therefore the equation of the circle is

In the general case where the centre of the circle, C, is the point (x₀, y₀) in Figure 11b, it is the distance CP which is constant and equal to R. Therefore, for an arbitrary point P(x, y) on the circle, (CP)² = R² and so:

The standard equation of a circle with centre (x₀, y₀) and radius R is

Suppose that we wish to find the points of intersection of the circles

x² + y² = 1(i)

and (x − 1)² + (y +1)² = 4(ii)

As a first step we may expand Equation (ii) to give

x² − 2x + y² + 2y = 2(iii)

We can now subtract (i) from (iii) to give

−2x + 2y = 1(iv)

which is the equation of a line. These equations are all true provided that we are referring to the points (x, y) where the circles meet. This means that the points where the circles meet must also lie on the line (iv). Instead of completing the calculation by finding these points we now invite you to carry out some similar calculations for yourself.

3.2 Tangents to a circle

In this subsection we are mainly concerned with circles whose centres are at the origin.

We consider the intersections of a straight line QP with such a circle. As depicted in Figure 12, the line could

• cut the circle at two points A and B as in diagram (a),
• not intersect the circle as in diagram (b),
• or touch the circle at one point T as in diagram (c).

In this last case the line is a tangent_to_a_curvetangent to the circle.

Let the equation of the straight line be

y = mx + c(i)

and the equation of the circle be

x² + y² = R²(ii)

To find where the line meets the circle we solve Equations (i) and (ii) simultaneously.

Substituting the expression for y from Equation (i) into (ii)

x² + (mx + c)² = R²

which can be rearranged to give a quadratic equation in x:

(1 + m²)x² + 2mcx + (c² − R²) = 0(iii)

Now, we know that the general quadratic equation αx² + βx + γ = 0

• has two real roots if β² > 4αγ
• no real roots if β < 4αγ
• one repeated real root if β² = 4αγ

So, for the particular case of Equation (iii) there will be one root and therefore one point of contact if (2cm)² = 4(1 + m²)(c²− R²)

i.e.4c²m² = 4c² + 4m²c² − 4R²(1 + m²)

i.e.c² = R²(1 + m²)(11)

which can be rearranged to give

$m^2 = \dfrac{c^2-R^2}{R^2}$(12)

From Equation 11,

c² = R²(1 + m²)(Eqn 11)

we see that for any given value of m there are two values of c ; c = $R\sqrt{1 + m^2}$ and c = −R −$\sqrt{1 + m^2}$. These values of c together with the given value of m determine, via Equation (i), the two straight lines of gradient m that are tangential to a circle of radius R centred on the origin.

From Equation 12 we see that there are two values of m for any value of c provided that c < −R or c > R. These values, m = $\sqrt{c^2 - R^2}/R$ and m = − $\sqrt{c^2 - R^2}/R$, together with the given value of c, determine, via Equation (i), the two straight lines passing through y = (0, c) that are tangential to a circle of radius R centred on the origin. (Note that the condition c < −R or c > R ensures that the point (0, c) is outside the circle and thus guarantees that there are two tangents passing through that point.)

Equation of the tangent at a given point on a circle

We can obtain an interesting result if we consider an arbitrary point (x₁,y₁) and the points of intersection of the line

 x₁x + y₁y = R²(i) i

with the circlex² + y² = R²(ii)

The equation of the line (i) can be rewritten as:   $y=-\dfrac{x_1}{y_1}x+\dfrac{R^2}{y_1}$

(so that $m=-\dfrac{x_1}{y_1}$ and $c = \dfrac{R^2}{y_1}$ ) and this line will be a tangent to the circle if Equation 11 is satisfied,

i.e. if   $ \left(\dfrac{R^2}{y_1}\right)^2 = R^2\left[1+\left(-\dfrac{x_1}{y_1}\right)^2\right]$, which reduces to x₁² + y₁² = R²

In other words:

So far we have only considered circles centred on the origin. However, by introducing a new system of coordinates, with axes X and Y and a different origin from that of the usual (x, y) system, we can investigate the equation of the tangent to a circle that is not centred on the origin. This is what we do in the next example.

3.3 Polar coordinates

Cartesian coordinates provide a valuable method of specifying the location of a point relative to a given origin, but they are not the only way in which such information can be specified. Other kinds of coordinate system exist and may be better suited to specific problems. A case in point is that of two–dimensional polar coordinates, which are particularly useful when distance from a fixed origin is of paramount importance.

The polar coordinate system is illustrated in Figure 14. The origin is represented by the point O and the point of interest by P. The line segment from O to P (sometimes called the radius vector of P) is of length r and is inclined at an angle θ, measured in the anticlockwise direction, from an arbitrarily chosen ray, called the polar axis, that emanates from O. The length r is called the radial coordinate and the θ the polar angle of P. The polar coordinates r and θ of the point P and are usually combined so that P is represented by the ordered pair (r, θ).

Figure 15 shows four points P₁, P₂, P₃ and P₄ together with their polar coordinates.

Notice that an anticlockwise rotation is taken to be positive so that, for example, the point P₄ has polar coordinates (4, 315°). Since clockwise rotation is negative, and we could reach P₄ by a clockwise rotation of 45° from the polar axis, its polar coordinates could be written (4, −45°).

The polar coordinates of a given point are not uniquely determined, because we can add or subtract any multiple 360° to the angle. We sometimes prefer polar coordinates to be completely specified, in which case the angle θ may be restricted to lie in the range −180° < θ ≤ 180°; (although there are occasions when the range 0 ≤ θ < 360° is more convenient).

Linking Cartesian and polar coordinates

If the origin of polar coordinates O coincides with the origin of Cartesian coordinates, and if the polar axis is chosen as the positive x–axis, we may link the Cartesian and polar coordinates as in Figure 16.

Since r is the hypotenuse of a right–angled triangle, it is clear that

and these equations enable us to convert from polar to Cartesian coordinates.

It is also possible to deduce from Figure 16 that

and these equations enable us to convert from Cartesian to polar coordinates. Since r is a positive quantity, its determination presents no difficulty; the same is not true for θ.

You may have noticed that the two equations

sinθ = y/r   and   cosθ = x/r

can be combined to give tanθ = y/x, but specifying the tangent of an angle does not uniquely determine the angle, so this does not resolve the difficulty.

The message is clear:

When determining the polar angle θ it is wise to sketch a diagram to show the position of the point.

4 Three–dimensional geometry

4.1 Points in three dimensions

We need two pieces of information to determine the position of a point in a plane, but in three dimensions we need to provide three independent pieces of information.

Figure 18 shows a Cartesian coordinate system in three dimensions. The point N is ‘vertically’ below the point P so that the line NP is parallel to the z–axis.

A point O is chosen as the origin of coordinates and three mutually perpendicular rays x, y and z are drawn as shown. The position of P is specified by three coordinates: x, y and z as indicated; and we refer to the point as P(x, y, z). Having decided on the x– and y–axes, there are two possible directions, opposite to one another, in which the z–axis can be directed, so that it is normal (i.e. perpendicular) to both the x– and y–axes.

To remove any ambiguity about the choice of direction of the z–axis, a right–handed Cartesian coordinate system is almost invariably used. There are various ways of describing this system, but one simple method only is included here.

If the thumb and first two fingers of the right hand are arranged so that they are approximately mutually perpendicular as in Figure 19, then, if the first and second finger point along the x– and y–axes, respectively, the thumb points along the z–axis in a right–handed system.

Distance between two points

The distance OP in Figure 18 can be found by applying Pythagoras’s theorem twice. In the horizontal plane which contains the x– and y–axes,

(ON)² = (OL)² + (LN)²

In the vertical plane which contains the lines ON, NP and OP,

(OP)² = (ON)² + (NP)² = (OL)² + (LN)² + (NP)²

Hence(OP)² = x² + y² + z²

Now we consider the distance between two general points.

Let the two points be P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂). We can ‘move’ the point P₁ to the origin by subtracting x₁ from its first coordinate, y₁ from its second coordinate and z₁ from this third coordinate (this has the same effect as moving the origin to the point P₁). The same transformation will move P₂ to a point whose coordinates are (x₂ − x₁, y₂ − y₁, z₂ − z₁). If we call this point P we can apply the result just obtained to determine the distance between P₁ and P₂. Thus the distance d between the points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by

It is worth noting that the formula for the distance between two points in two dimensions (Equation 8)

${\rm PQ} = \sqrt{\smash[b]{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$(Eqn 8)

is the special case of the formula for three dimensions, i.e. Equation 16, with z₁ and z₂ both put equal to zero.

4.2 Planes in three dimensions

A plane is a flat surface which has the property that a straight line joining any two points in the plane lies entirely in the plane itself.

where a, b, c and d are constants. Note that if we choose a = 1 with b = c = d = 0 the equation becomes x = 0 which is the (y, z) plane.

Other planes can be described by choosing suitable combinations of the constants a, b, c and d. The plane z = 2 is parallel to the plane z = 0 (the (x, y) plane) and cuts the z–axis at z = 2, i.e. the point (0, 0, 2). The equation z = 2 corresponds to Equation 17 with a = 0, b = 0, c = 1 and d = 2.

Note that if d = 0 in Equation 17 then the plane will pass through the origin, since x = y = z = 0 will satisfy the equation of the plane.

Intersection of two planes

If two planes are not parallel then they meet in a line. Imagine two consecutive pages of a book. Hold the pages taut and you see that they meet in the line of the binding connecting them. For example the (x, y) plane meets the (y, z) plane on the y–axis, this means that the y–axis corresponds to the points where both x and z are zero; in other words, we can see instantly that a point such as (0, 5, 0) must lie on the y–axis.

If we keep the values of a, b and c fixed in Equation 17,

ax + by + cz = d(Eqn 17)

but double the value of d, we obtain a plane that is parallel to the original plane (but twice as far from the origin). Compare the results of the following exercise with those of the previous exercise.

4.3 Lines in three dimensions

In three dimensions, straight lines are often specified by the intersection of two planes. Of course we try to choose the planes so that the specification is as neat as possible; for example, we could specify a line as the intersection of the planes

3x − 2y = 5

and5y − 3z = −11

but this specification is not as neat as it might be. The first of these equations can be rewritten in the form $\dfrac{x-1}{2} = \dfrac{y+1}{3}$ while the second can be rewritten in the form $\dfrac{y + 1}{3} = \dfrac{z-2}{5}$, so that the two equations can be combined into the particularly neat form

$\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-2}{5}$(18) i

In this form we can see instantly that the point (1, −1, 2) must lie on the line. The values 2, 3 and 5 are also significant, since they can be used to determine the direction of the line (but it would take us beyond the scope of this module to explain how this is done).

If we put each of the fractions in Equations 18 equal to s, then they can be written in the equivalent form

$\left.\begin{align} x&=1+2s\\y&=-1+3s\\z&=2+5s\\ \end{align} \right\}$(19)

The variable s is then a parameter that moves the point (x, y, z) up and down the line as we change its value. It is often this form of the equations of a line which is most useful, as the following exercise illustrates.

In three dimensions only lines that lie in a plane will meet in a point. Indeed, in three dimensions, it is more likely that a pair of lines chosen at random will not meet, in which case they are said to be skew lines.

5 Closing items

5.1 Module summary

1

In two dimensions the Subsection 2.1Cartesian system of coordinates consists of two perpendicular lines, called axes, which meet at the origin of coordinates; one of these lines, usually drawn horizontally, is designated the x–axis, the other, which is usually drawn vertically, is the y–axis. The x–coordinate of a point is the perpendicular distance of that point from the y–axis, with a negative sign attached if the point lies to the left of the y–axis. The y–coordinate of the point is its perpendicular distance from the x–axis, with a negative sign attached if the point lies below the x–axis.

2

The equation of a straight line can appear in a variety of forms; the one chosen depends on the information available about the line and the information likely to be required about the line.

3

If the line has a known gradient m and a known intercept (on the y–axis) c then the equation relating the x– and y–coordinates of a point on it is

y = mx + c(Eqn 1)

4

If the line has a known gradient m and passes through the point (x₁, y₁) then the equation of the line is

y − y₁ = m(x − x₁)(Eqn 3)

5

If the line meets the x–axis at x = a and the y–axis at y = b then the equation of the line is

ax + by = 1(Eqn 5)

6

The so–called general form of the equation of a line is

ax + by + c = 0(Eqn 6)

7

If two lines are parallel then their gradients are equal, and if two lines are perpendicular then the product of

their gradients is −1.

8

If two lines intersect then the coordinates of the point of intersection can be found by solving

simultaneously the equations of the lines.

9

The distance between the points P(x₁, y₁) and Q(x₂, y₂) is given by

${\rm PQ} = \sqrt{\smash[b]{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$(Eqn 8)

10

The equation of a Section 3circle centred at the origin and of radius R is

x² + y² = R²(Eqn 9)

11

The equation of a circle centred at the point (x₀, y₀) and of radius R is

(x − x₀)² + (y − y₀)² = R²(Eqn 10)

12

The equation of the Subsection 3.2tangent to the circle x² + y² = R² at a point (x₁, y₁) on the circle is given by

x₁x + y₁y = R²(Eqn 13)

13

In the Subsection 3.3polar coordinate system a point is specified by its distance r from the origin and the polar angle θ which it makes with a given line through the origin. The link between the two systems is provided by the pair of equations

x = rcosθ  and  y = rsinθ(Eqn 14)

andr² = x² + y²,   sinθ = y/r,   cosθ = x/r(Eqn 15)

where we must take care to identify the correct value of θ. (This can often be done by plotting the point on a diagram, and then using the fact that tanθ = y/x.)

14

The distance d between the points (x₁, y₁, z₁) and (x₂, y₂, z₂) in three dimensions is given by

d² = (x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²(Eqn 16)

15

The general equation of a plane in three dimensions is

ax + by + cz = d(Eqn 17)

If two planes intersect, they do so in a line.

16

The general equations of a line in three dimensions are

$\dfrac{x-a}{l} = \dfrac{y-b}{m} = \dfrac{z-c}{n}$(Eqn 20)

where (a, b, c) is a point on the line and l, m, n are constants, known as direction ratios lmn (or direction cosines) of the line.

5.2 Achievements

Having complete this module, you should be able to:

A1

Define the terms that are emboldened and flagged in the margins of the module.

A2

Set up a Cartesian coordinate system and plot points in it.

A3

Recognize the different forms of equation of a straight line and produce the appropriate equation from the information given about the line, then derive the required information from it.

A4

Convert one form of the equation of a line into another.

A5

Write down the equation of a circle, given its centre and radius, and deduce the centre and radius of a circle from its equation.

A6

Produce the equation of tangents to a circle, both from a given point outside the circle and at a given point of contact on the circle.

A7

Set up the polar coordinate system, plot points in it and convert from Cartesian coordinates to polar coordinates and vice versa.

A8

Calculate the distance between two points in three dimensions.

A9

Recognize the equation of a plane, and use it to indentify points that lie in the plane.

A10

Recognize the equations of a line in three dimensions and use them to identify points that lie on the line and the point at which a line intersects a given plane.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

5.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions, each of which tests one or more of the Achievements.

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