MTH5P1: Co-ordinate geometry

This unit explains Co-ordinate geometry

Co-ordinates

Co-ordinates are a means of describing a position relative to some fixed point, or origin. In two dimensions you need two pieces of information; in three dimensions, you need three pieces of information.
In the Cartesian system (named after René Descartes), position is given in perpendicular directions: x, y in two dimensions; x, y, z in three dimensions (see figure 2.1). This chapter concentrates exclusively on two dimensions.

Plotting, sketching and drawing

In two dimensions, the co-ordinates of points are often marked on paper and joined up to form lines or curves. A number of words are used to describe this process.
Plot (a line or curve) means mark the points and join them up as accurately as you can. You would expect to do this on graph paper and be prepared to read information from the graph.
Sketch means mark points in approximately the right positions and join them up in the right general shape. You would not expect to use graph paper for a sketch and would not read precise information from one. You would however mark on the co-ordinates of important points, like intersections with the x and y axes and points at which the curve changes direction.
Draw means that you are to use a level of accuracy appropriate to the circumstances, and this could be anything between a rough sketch and a very accurately plotted graph.

The gradient of a line

In everyday English, the word line is used to mean a straight line or a curve. In mathematics, it is usually understood to mean a straight line. If you know the coordinates of any two points on a line, then you can draw the line.
The slope of a line is measured by its gradient. It is often denoted by the letter m. In figure 2.2, A and B are two points on the line. The gradient of the line AB is given by the increase in the y coordinate from A to B divided by the increase in the x coordinate from A to B.

When the same scale is used on both axes, m = tan θ (see figure 2.2). Figure 2.3 shows four lines. Looking at each one from left to right: line A goes uphill and its gradient is positive; line B goes downhill and its gradient is negative. Line C is horizontal and its gradient is 0; the vertical line D has an infinite gradient.

Parallel and perpendicular lines

The distance between two points

When the coordinates of two points are known, the distance between them can be calculated using Pythagoras’ theorem, as shown in figure 2.6.

This method can be generalised to find the distance between any two points,

The mid-point of a line joining two points

Look at the line joining the points A(2, 1) and B(8, 5) in figure 2.8. The point M(5, 3) is the mid-point of AB.

Notice that the coordinates of M are the means of the coordinates of A and B.

A and B are the points (2, 5) and (6, 3) respectively (see figure 2.9). Find:

(i) the gradient of AB
(ii) the length of AB
(iii) the mid-point of AB
(iv) the gradient of a line perpendicular to AB.

SOLUTION

Using two different methods, show that the lines joining P(2, 7), Q(3, 2) and R(0, 5) form a right-angled triangle (see figure 2.10).
SOLUTION
Method 1

⇒ Product of gradients = 1 × (−1) = −1
⇒ Sides RP and RQ are at right angles.

Method 2
Pythagoras’ theorem states that for a right-angled triangle whose hypotenuse has length a and whose other sides have lengths 

Conversely, if you can show that for a triangle with sides of lengths a, b, and c, then the triangle has a right angle and the side of length a is the hypotenuse.
This is the basis for the alternative proof, in which you use.

⇒ Sides RP and RQ are at right angles.

The equation of a straight line

The word straight means going in a constant direction, that is with fixed gradient.
This fact allows you to find the equation of a straight line from first principles.

Find the equation of the straight line with gradient 2 through the point (0, −5).
SOLUTION

Take a general point (x, y) on the line, as shown in figure 2.11. The gradient of the line joining (0, −5) to (x, y) is given by

Since we are told that the gradient of the line is 2, this gives

Since (x, y) is a general point on the line, this holds for any point on the line and is therefore the equation of the line.
The example above can easily be generalised (see page 50) to give the result that the equation of the line with gradient m cutting the y axis at the point (0, c) is
y = mx + c.
(In the example above, m is 2 and c is −5.)
This is a well-known standard form for the equation of a straight line.

Drawing a line, given its equation

There are several standard forms for the equation of a straight line, as shown in figure 2.12.
When you need to draw the graph of a straight line, given its equation, the first thing to do is to look carefully at the form of the equation and see if you can recognise it.

Lines parallel to the axes
Lines parallel to the x axis have the form y = constant, those parallel to the y axis the form x = constant. Such lines are easily recognised and drawn.

Equations of the form y = mx + c
The line y = mx + c crosses the y axis at the point (0, c) and has gradient m. If c = 0, it goes through the origin. In either case you know one point and can complete the line either by finding one more point, for example by substituting x = 1, or by following the gradient (e.g. 1 along and 2 up for gradient 2).

Equations of the form px + qy + r = 0
In the case of a line given in this form, like 2x + 3y − 6 = 0, you can either rearrange it in the form y = mx + c (in this example or you can find the co-ordinates of two points that lie on it. Putting x = 0 gives the point where it crosses the y axis, (0, 2), and putting y = 0 gives its intersection with the x axis, (3, 0).

Sketch the lines x = 5, y = 0 and y = x on the same axes.
Describe the triangle formed by these lines.

SOLUTION
The line x = 5 is parallel to the y axis and passes through (5, 0).
The line y = 0 is the x axis.
The line y = x has gradient 1 and goes through the origin.

The triangle obtained is an isosceles right-angled triangle, since OA = AB = 5
units, and ∠OAB = 90°.
EXAMPLE 2.5 Draw y = x − 1 and 3x + 4y = 24 on the same axes.
SOLUTION
The line y = x − 1 has gradient 1 and passes through the point (0, −1).
Substituting y = 0 gives x = 1, so the line also passes through (1, 0).
Find two points on the line 3x + 4y = 24.
Substituting x = 0 gives 4y = 24 so y = 6.
Substituting y = 0 gives 3x = 24 so x = 8.

The line passes through (0, 6) and (8, 0).

Finding the equation of a line

The simplest way to find the equation of a straight line depends on what information you have been given.

Given the gradient, m, and the co-ordinates of one point on the line
Take a general point (x, y) on the line, as shown in figure 2.15.

This is a very useful form of the equation of a straight line. Two positions of the point lead to particularly important forms of the equation.

When the given point is the point (0, c), where the line crosses the y axis, the equation takes the familiar form
y = mx + c
as shown in figure 2.16.

When the given point is the origin, the equation takes the form
y = mx
as shown in figure 2.17.

Find the equation of the line with gradient 3 which passes through the point (2, −4).
SOLUTION

Given two points,

Find the equation of the line joining (2, 4) to (5, 3).
SOLUTION

This can be simplified to x + 3y − 14 = 0.

Show that the equation of the line in figure 2.19 can be written                           

 

Different techniques to solve problems
The following examples illustrate the different techniques and show how these can be used to solve a problem.

Find the equations of the lines (a) − (e) in figure 2.20.

SOLUTION
Line (a) passes through (0, 2) and has gradient 1
⇒ equation of (a) is y = x + 2.
Line (b) is parallel to the x axis and passes through (0, 4)
⇒ equation of (b) is y = 4.
Line (c) is parallel to the y axis and passes through (−3, 0)
⇒ equation of (c) is x = −3.
Line (d) passes through (0, 0) and has gradient −2
⇒ equation of (d) is y = −2x.

Line (e) passes through (0, −1) and has gradient

⇒ equation of (e) is y =

This can be rearranged to give x + 5y + 5 = 0.

Two sides of a parallelogram are the lines 2y = x + 12 and y = 4x − 10. Sketch these lines on the same diagram.
The origin is a vertex of the parallelogram. Complete the sketch of the parallelogram and find the equations of the other two sides.

SOLUTION

The gradient of the line PQ is

and so the gradient of the perpendicular bisector is +3.
The perpendicular bisector passes through the mid-point, R, of the line PQ. The co-ordinates of R are

Using y − y1 = m(x − x1), the equation of the perpendicular bisector is
y − 4 = 3(x − (−1))
y − 4 = 3x + 3
y = 3x + 7.

The intersection of two lines

The intersection of any two curves (or lines) can be found by solving their equations simultaneously. In the case of two distinct lines, there are two possibilities:

they are parallel
they intersect at a single point.

Sketch the lines x + 2y = 1 and 2x + 3y = 4 on the same axes, and find the coordinates of the point where they intersect.

SOLUTION

The co-ordinates of the point of intersection are (5, −2).

Find the co-ordinates of the vertices of the triangle whose sides have the
equations x + y = 4, 2x − y = 8 and x + 2y = −1.
SOLUTION
A sketch will be helpful, so first find where each line crosses the axes.
1.  x + y = 4 crosses the axes at (0, 4) and (4, 0).
2.  2x − y = 8 crosses the axes at (0, −8) and (4, 0).

Since two lines pass through the point (4, 0) this is clearly one of the vertices. It has been labelled A on figure 2.24.
Point B is found by solving 2 and 3 simultaneously:

The line l has equation 2x − y = 4 and the line m has equation y = 2x − 3.
What can you say about the intersection of these two lines?

René Descartes was born near Tours in France in 1596. At the age of eight he was sent to a Jesuit boarding school where, because of his frail health, he was allowed to stay in bed until late in the morning. This habit stayed with him for the rest of his life and he claimed that he was at his most productive before getting up.
After leaving school he studied mathematics in Paris before becoming in turn a soldier, traveller and optical instrument maker. Eventually he settled in Holland where he devoted his time to mathematics, science and philosophy, and wrote a number of books on these subjects.
In an appendix, entitled La Géométrie, to one of his books, Descartes made the contribution to co-ordinate geometry for which he is particularly remembered.
In 1649 he left Holland for Sweden at the invitation of Queen Christina but died there, of a lung infection, the following year.

Drawing curves

You can always plot a curve, point by point, if you know its equation. Often, however, all you need is a general idea of its shape and a sketch is quite sufficient.

How are the curves for even values of n different from those for odd values of n?

Stationary points
A turning point is a place where a curve changes from increasing (curve going up) to decreasing (curve going down), or vice versa. A turning point may be described as a maximum (change from increasing to decreasing) or a minimum
(change from decreasing to increasing). Turning points are examples of stationary points, where the gradient is zero. In general, the curve of a polynomial of order n has up to n − 1 turning points as shown in figure 2.26.

Figure 2.26

There are some polynomials for which not all the stationary points materialise, as in the case of 

(whose curve is shown in figure 2.27). To be accurate, you say that the curve of a polynomial of order n has at most n − 1 stationary points.

Behaviour for large x (positive and negative)
What can you say about the value of a polynomial for large positive values and large negative values of x? As an example, look at

Is n even or odd?
Is the coefficient of positive or negative?

According to the answers, the curve will have one of the four types of shape illustrated in figure 2.28.
Intersections with the x and y axes

The constant term in the polynomial gives the value of y where the curve intersects the y axis. (0, 23). Similarly, crosses the y axis at (0, 0), the origin, since the
constant term is zero.

When the polynomial is given, or known, in factorized form you can see at once where it crosses the x axis. The curve y = (x − 2)(x − 8)(x − 9), for example, crosses the x axis at x = 2, x = 8 and x = 9. Each of these values makes one of the brackets equal to zero, and so y = 0.

Since the polynomial is of order 3, the curve has up to two stationary points. The term in has a positive coefficient (+1) and 3 is an odd number, so the general shape is as shown on the left of figure 2.29.

The actual equation

tells you that the curve:
− crosses the y axis at (0, 3)
− crosses the x axis at (−1, 0), (1, 0) and (3, 0).
This is enough information to sketch the curve (see the right of figure 2.29).

Note
This illustrates an important result. If f(x) is a polynomial of degree n, the curve with equation y = f(x) crosses the x axis at most n times, and the equation f(x) = 0 has at most n roots.

An important case occurs when the polynomial function has one or more repeated factors, as in figure 2.31. In such cases the curves touch the x axis at points corresponding to the repeated roots.

The curves for n = 3, 5, … are not unlike that for n = 1, those for n = 4, 6, … are like that for n = 2. In all cases the point x = 0 is excluded because is undefined.

An important feature of these curves is that they approach both the x and the y axes ever more closely but never actually reach them. These lines are described as asymptotes to the curves. Asymptotes may be vertical (e.g. the y axis), horizontal, or lie at an angle, when they are called oblique.
Asymptotes are usually marked on graphs as dotted lines but in the cases above the lines are already there, being co-ordinate axes. The curves have different branches which never meet. A curve with different branches is said to be
discontinuous, whereas one with no breaks, like is continuous.

The circle
You are of course familiar with the circle, and have probably done calculations involving its area and circumference. In this section you are introduced to the equation of a circle.
The circle is defined as the locus of all the points in a plane which are at a fixed distance (the radius) from a given point (the centre). (Locus means path.)
As you have seen, the length of a line joining is given by

This is used to derive the equation of a circle.
In the case of a circle of radius 3, with its centre at the origin, any point (x, y) on

This circle is shown in figure 2.33.

These results can be generalised to give the equation of a circle centre (0, 0), radius r as follows:

The intersection of a line and a curve

When a line and a curve are in the same plane, there are three possible situations.
All points of intersection are distinct (see figure 2.34).

The line is a tangent to the curve at one (or more) point(s) (see figure 2.35).

In this case, each point of contact corresponds to two (or more) co-incident points of intersection. It is possible that the tangent will also intersect the curve somewhere else.

The line and the curve do not meet (see figure 2.36).

The co-ordinates of the point of intersection can be found by solving the two equations simultaneously. If you obtain an equation with no real roots, the conclusion is that there is no point of intersection.

The equation of the straight line is, of course, linear and that of the curve non-linear. The examples which follow remind you how to solve such pairs of equations.

Find the co-ordinates of the two points where the line y − 3x = 2 intersects the curve

SOLUTION
First sketch the line and the curve.

You can find where the line and curve intersect by solving the simultaneous equations:

Substitute into the linear equation, y = 3x + 2, to find the corresponding y co-ordinates.
x = 2 ⇒ y = 8

So the co-ordinates of the points of intersection are (2, 8) and

Find the value of k for which the line 2y = x + k forms a tangent to the curve

Hence, for this value of k, find the co-ordinates of the point where the line 2y = x + k meets the curve.

SOLUTION

You can find where the line forms a tangent to the curve by solving the simultaneous equations:

When you eliminate either x or y between the equations you will be left with a quadratic equation. A tangent meets the curve at just one point and so you need to find the value of k which gives you just one repeated root for the
quadratic equation.

Notice that this is a repeated root so the line is a tangent to the curve.
Now substitute y = 2 into the equation of the line to find the x co-ordinate.
When y = 2: 2y = x + 2 ⇒ 4 = x + 2
x = 2
So the tangent meets the curve at the point (2, 2).

ASSIGNMENT : Co-ordinate geometry Assignment MARKS : 30  DURATION : 60 minutes

 

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