understand how to apply the concept of reflection of a shape in a mirror line
understand the concept of the rotation of a shape with a given angle, direction and centre
be confident in making and identifying enlargements of 2-D shapes with given scale factors (and centres of enlargement).
This Unit describes how shapes can be transformed by reflection, rotation and enlargement. You have three sections to work through and there are check up audits and fitness tests for each section.
Reflections
Rotations
Enlargements
0, Chapter 2, Reflections,
Reflections
Reflections are obtained when you draw the image that would be obtained in a mirror.
Every point on a reflected image is always the same distance from the mirror line as the original.
This is shown on the bottom.
Note:
Distances are always measured at right angles to the mirror line.
1, Chapter 2, Task 1, Worked Example 1,
Worked Example
Use the slider to explore worked examples.
Worked Example
Draw the reflection of the shape in the mirror line shown.
Solution
The lines added to the diagram below show how to find the position of each point after it has been reflected. Remember that the image of each point is the same distance from the mirror line as the original.
The position of each point after it has been reflected show how to draw the reflected shape. The points can then be joined to give the reflected image
If the construction lines have been drawn in pencil they can be rubbed out.
Worked Example
Reflect this shape in the mirror line shown in the diagram.
Solution
The lines are drawn at right angles to the mirror
The points which form the image must be the same distance from the mirror lines as the original points. The points which were on the mirror line remain there.
The points can then be joined to give the reflected image.
The final image is shown on the bottom.
1, Chapter 2, Task 2, Exercise 1,
Exercise
Here are some questions to check your progress; there are more practice questions if needed.
Exercise 1
Copy the diagrams below and draw the reflection of each object.
Exercise 2
Copy the diagrams below and draw the reflection of each object.
0, Chapter 3, Rotations,
Rotations
Remember
Rotations are obtained when a shape is rotated about a fixed point, called the centre of rotation, through a specified angle.
The diagram shows a number of rotations.
It is often helpful to use tracing paper to find the position of a shape after a rotation.
1, Chapter 3, Task 1, Worked Example 2,
Worked Example
Use the slider to explore worked examples.
Worked Example
Rotate the triangle \(ABC\) shown in the diagram through \(90^\circ\) clockwise about the point with coordinates \(\left(0, 0\right)\).
Solution
The diagram shows how each vertex can be rotated through \(90^\circ\) to give the position of the new triangle.
Worked Example
The diagram shows the position of a shape \(A\) and the shapes, \(B, C, D, E\) and \(F\) which are obtained from \(A\) by rotation.
Describe the rotation which moves \(A\) onto each other shape.
Solution
The diagram shows the centres of rotation and how one vertex of the shape A was rotated.
A to B: Rotation of 180 \(^\circ\) about the point (5, 6).
A to C: Rotation of 180 \(^\circ\) about the point (3, 2).
A to D: Rotation of 90 \(^\circ\) anti-clockwise about the point (0, 0).
A to E: Rotation of 180 \(^\circ\) about the point (0, 0).
A to F: Rotation of 90 \(^\circ\) anti-clockwise about the point (0, 4).
1, Chapter 3, Task 2, Exercise 2,
Exercise
Here are some questions to check your progress; there are more practice questions if needed.
1. Copy the axes and shape shown below.
Rotate the original shape through 90\(^\circ\) clockwise around the point (1, 2).
Rotate the original shape through 180\(^\circ\) around the point (3, 4).
Rotate the original shape through 90\(^\circ\) clockwise around the point (1, − 2).
Rotate the original shape through 90\(^\circ\) anti-clockwise around the point (0, 1).
2. The diagram shows the position of a shape labelled \(A\) and other shapes which were obtained by rotating \(A\).
Describe how each shape can be obtained from \(A\) by a rotation.
Which shapes can be obtained by rotating the shape \(E\)?
0, Chapter 4, Enlargements,
Enlargements
Remember
An enlargement is a transformation which enlarges (or reduces) the size of an image. Each enlargement is described in terms of a centre of enlargement and a scale factor.
The example shows how the original, \(A\), was enlarged with scale factors \(2\) and \(4\).
A line from the centre of enlargement passes through the corresponding vertex of each image.
Remember
The distances, \(OA′\) and \(OA′′\), are related to \(OA\): \(OA′ = 2 \times OA\) \(OA′′ = 4 \times OA\) The same is true of all the other distances between \(O\) and corresponding points on the images.
1, Chapter 4, Task 1, Worked Example 3,
Worked Example
Use the slider to explore worked examples.
Worked Example
Enlarge the triangle shown using the centre of enlargement marked and scale factor \(3\).
Solution
The first step is to draw lines from the centre of enlargement through each vertex of the triangle as shown on the right.
As the scale factor is \(3\), then the distances from the centre of enlargement to the vertices of the image will be:
The points \(A′, B′\) and \(C′\) have also been marked on the diagram. Once these points have been found they can be used to draw the enlarged triangle.
Worked Example
Enlarge the pentagon with scale factor \(2\) using the centre of enlargement marked on the diagram.
Solution
The first step is to draw lines from the centre of enlargement which pass through the five vertices of the pentagon.
As the scale factor is 2, then the distances from the centre of enlargement to the vertices of the image will be:
These points can then be marked and joined to give the enlargement.
Worked Example
The diagram shows the square \(ABCD\) which has been enlarged to give the squares \(A′B′C′D′\) and \(A′′B′′C′′D′′\) .
Find the centre of enlargement.
Find the scale factor for each enlargement.
Solution
To find the centre of enlargement draw lines through \(A, A′\) and\( A′′\) , then repeat for \(B, B′\) and \(B′′\) , \(C, C′\) and \(C′′\) and \(D, D′\) and \(D′′\).
These lines cross at the centre of enlargement as shown in the diagram.
The sides of the square \(ABCD\) are each \(3\;units\). The sides of the square \(A′B′C′D′\) are \(6\;units\). As these are twice as long as the original, the scale factor for this enlargement is 2.
The sides of the square \(A′′B′′C′′D′′\) are \(12\;units\), which is \(4\;times\) longer than the original square.
So the scale factor for this enlargement is \(4\).
When the scale factor of an enlargement is a fraction, the size of the enlargement can be reduced.
The image of the original is then between the centre of enlargement and the original.
Worked Example
The diagram shows three triangles. \(ABC\) was enlarged with different scale factors to give \(A′B′C′\) and \(A′B′C′\).
Find the centre of enlargement.
Find the scale factor for each enlargement.
Solution
a. To find the centre of enlargement, lines should be drawn through the corresponding points on each figure.
b.
To find the scale factors, compare the lengths of sides in the different triangles. First consider triangles \(ABC\) and \(A′ B′ C′\):
\(AC = 12\;units\) and \(A′ C′ = 6\;units\), \(A^′ C^′=\frac{1}{2} \times AC\) which means the scale factor is \(\frac{1}{2}\) For triangles \(ABC\) and \(A′′ B′′ C′′\)\: \(AC = 12\;units\) and \(A′′ C′′= 3\;units\), \(A"C"=\frac{1}{2} \times AC\)
which means the scale factor is \(=\frac{1}{4}\)
Worked Example
Enlarge the triangle shown with scale factor and centre of \(\frac{1}{4}\) enlargement as shown.
Solution
The first stage is to draw lines from each corner of the triangle through the centre of enlargement.
These lines cross at the centre of enlargement as shown in the diagram.
Then the corners of the image should be fixed so that:
Here are some questions to check your progress; there are more practice questions if needed
Exercise 1
Enlarge the shape with scale factor \(2\) using the point marked as the centre of enlargement.
Exercise 2
Enlarge the triangle shown with scale factor \(3\) and the centre of enlargement shown.
Exercise 3
For the diagram find the centre of enlargement and the scale factor when the smaller shape is enlarged to give the bigger shape.
The centre of enlargement is shown on the diagram above. The scale factor is 3.
Exercise 4
For the diagram find the centre of enlargement and the scale factor when the smaller shape is enlarged to give the bigger shape.
The centre of enlargement is shown on the diagram above. The scale factor is 2.
Exercise 5
In each example below, the smaller shape has been obtained from the larger shape by an enlargement. For each example, state the scale factor and the coordinates of the centre of enlargement.
a) \(\frac{1}{5}, (8,7)\)
b) \(\frac{2}{3},
(13,9)\)
c) \(\frac{1}{4}, (9,1)\)
0, Chapter 5, Summary,
Summary
Reflections
are obtained when you draw the image of a shape in a mirror line. An example is shown below.
Rotations
are obtained when a shape is rotated about a point, the centre of rotation, through a specified angle. For example,
Transformation
moving a shape so that it is in a different position but retains the same size, area, angles and line lengths.
An enlargement
is similar to a transformation but it alters (enlarges or reduces) the size of the image. An enlargement is described in terms of a scale factor and also a centre of enlargement which defines the exact location of the image. For example, the image on the left below has been enlarged by a scale factor of \(2\frac{1}{2}\), with aspecified centre of enlargement.
The enlargement opposite (actually a reduction) of scale factor \(\frac{1}{3}\) has no specified centre of enlargement, so the actual location of the smaller shape could be anywhere.
Interactive Exercises:
Reflections, Rotations and Enlargements Interactive Exercises, https://www.cimt.org.uk/sif/geometry/g7/interactive.htm
This unit is focused on congruent and similar shapes and you need to understand the differences between these two concepts. After completing this unit you should
understand the concept of congruent shapes and be able to use the congruence tests for triangles
understand the concept of similar shapes and be able to determine similar shapes
You have two sections to work through namely;
Congruence
Similarity
0, Chapter 2, Congruence and Similarity,
Congruence and Similarity
Two shapes are said to be congruent if they are the same shape and size: that is, the corresponding sides of both shapes are the same length and corresponding angles are the same.
The two triangles shown here are congruent.
1, Chapter 2.1, Similarity 1,
Similarity
Shapes which are of different sizes but which have the same shape are said to be similar. The two triangles shown here are similar.
If you double the lengths for the triangle on the left, its sides will match those of the triangle on the right.
1, Chapter 2.2, Congruence Tests for Triangles,
Congruence Tests for Triangles
There are four tests for congruence with triangles.
TEST 1 (Side, Side, Side)
If all three sides of one triangle are the same as the lengths of the sides of the second triangle, then the two triangles are congruent. This test is referred to as \(SSS\).
The single (double or treble) crosses on sides correspond to equal sides
TEST 2 (Side, Angle, Side)
If two sides of one triangle are the same length as two sides of the other triangle and the angle between these two sides is the same in both triangles, then the triangles are congruent. This test is referred to as \(SAS\)
TEST 3 (Angle, Angle, Side)
If two angles and the length of one corresponding side are the same in both triangles, then they are congruent. This test is referred to as \(AAS\).
The single or double arcs on angles correspond to equal angles
TEST 4 (Right angle, Hypotenuse, Side)
If both triangles contain a right angle, have hypotenuses of the same length and one other side of the same length, then they are congruent. This test is referred to as \(RHS\).
1, Chapter 2.2, Task 1, Worked Example 1,
Worked Example
Worked Example
Which of the triangles below are congruent to the triangle \(ABC\), and why?
Solution
Triangle \(DEF\) is congruent to triangle \(ABC\) because sides lengths are the same in both triangles as shown below. (\(SSS\))
\(AB = DF\)
\(BC = EF\)
\(AC = DE\)
Triangle \(GHI\) is congruent to triangle \(ABC\) because they have one side and two angles the same, as shown below. \(\left(AAS\right)\)
(Note that we indicate angles with a ^ symbol above the letter)
\(BC = HI\) \(ABC=GHI\) \(ACB=GIH\)
Triangle \(JKL\) is not congruent to triangle \(ABC\) because two sides are known and match in lengths but the angle between them is unknown.
0, Chapter 3, Similarity,
Similarity
Similar shapes have the same shape but may be different sizes. The two rectangles shown below are similar - they have the same shape but one is smaller than the other.
They are similar because they are both rectangles and the sides of the larger rectangle are three times longer than the sides of the smaller rectangle.
1, Chapter 3.1, Similarity of Triangles,
Similarity of Triangles
These two triangles are not similar. The sides lengths of the triangles are not in the same ratio and so the triangles are not similar.
For two triangles to be similar, they must have the same internal angles, as shown in the similar shapes below.
1, Chapter 3.2, Similarity and Area,
Similarity and Area
It is interesting to compare the area of two similar rectangles. The area of the smaller rectangle is \(6\;cm^{2}\) and the area of the larger rectangle is \(54\;cm^{2}\), which is nine times \(\left(32\right)\) greater.
Remember
In general, if the lengths of the sides of a shape are increased by a factor \(k\), then the area is increased by a factor \(k^{2}\).
1, Chapter 3.3, Similarity and Volume,
Similarity and Volume
The diagrams below show 3 similar cubes.
Comparing the larger cubes with the \(1\;cm\) cube we can note that:
For the \(2\;cm\) cube The lengths are \(2\) times greater. The areas are \(4 \left(2^{2}\right)\) times greater. The volume is \(8 \left(2^{3}\right)\) times greater.
For the \(3\;cm\) cube The lengths are \(3\) times greater. The areas are \(9 \left(3^{2}\right)\) times greater. The volume is \(27 \left(3^{3}\right)\) times greater.
Remember
If the lengths of a solid are increased by a factor, \(k\), its surface area will increase by a factor \(k^{2}\) and its volume will increase by a factor \(k^{3}\).
1, Chapter 3, Task 1, Worked Example 2,
Worked Example
Use the slider to explore worked examples.
Worked Example
Which of the triangles, \(A, B, C, D\), shown below are similar?
How do the areas of the triangles which are similar compare?
First compare triangles \(A\) and \(B\).
Here all the lengths of the sides are twice the length of the sides of triangle \(A\), so the two triangles are similar.
Then compare triangles \(A\) and \(C\).
Here all the angles are the same in both triangles, so the triangles must be similar.
Finally, compare triangles \(A\) and \(D\).
These triangles are not similar because if you halve the sides on the first triangle, two of the sides match, but not all three:
\(4 =\frac{1}{2}\times 8\) and \(4.52 =\frac{1}{2} \times 9.04\) but \(3.5 \neq \frac{1}{2}\times 6.13\)
So these triangles are not similar as the matching lengths on triangle \(D\) must all be half of those on \(A\).
Worked Example
Explain why triangles \(ABE\) and \(ACD\) are similar.
Find the lengths of \(x\) and \(y\).
Find the ratio of the area of triangle \(ABE\) to triangle \(ACD\).
As the lines \(BE\) and \(CD\) are parallel, \(AEB=ADC\) and \(ABE=ACD\) Also the vertex \(A\), is common to both triangles, so \(DAC = EAB\) So the three angles are the same in both triangles and therefore they are similar.
Comparing the sides \(BE\) and \(CD\), the lengths in the larger triangle are \(1.5\) times the lengths in the smaller triangle (alternatively, it can be stated that the ratio of the lengths is \(2 : 3\)). So the length \(AC\) will be \(1.5\) times the length \(AB\). \(AC = 1.5 \times 6 = 9\) So \(y = 3\)
In the same way, \(AD = 1.5 \times AE\), so \(4 + x = 1.5x\) \(4 = 0.5x\) \(x = 8\)
As the lengths are increased by a factor of \(1.5\), or \(\frac{3}{2}\) , for the larger triangle, the areas will be increased by a factor of \(1.52 = 2.25\) (or \(\left(\frac{3}{2}\right)^{2} = 2.25)\).
We can say that the ratio of the areas of the triangles is \(1 : 2.25\)
So the ratio of the area of triangle \(ABE\) to triangle \(ACD\) is, by multiplying both sides of the ratio by \(4\), \(4 : 5\)
Worked Example
The diagrams show two similar triangles.
If the area of triangle \(DEF\) is \(26.46\;cm^{2}\), find the lengths of its sides.
If the lengths of the sides of triangle \(DEF\) are a factor \(k\) greater thanthe lengths of the sides of triangle \(ABC\), then its area will be a factor \(k^{2}\) greater than the area of \(ABC\).
Area of \(ABC\) is \(\frac{1}{2} \times 4\times 3 = 6 cm\;^{2}\) So
So the height must be increased by a factor of \(1.5\), to give
\(\begin{align*}
height&=1.5 \times 10\\
&= 15 cm
\end{align*}\)
0, Chapter 4, Summary,
Summary
Congruent shapes
Congruent shapes are identical but can have different orientations. For example, these five shapes are all congruent.
Similar shapes
Similar shapes have corresponding angles equal and corresponding sides in the same ratio. For example, shapes \(A, B, C, D\) and \(E\) are all similar; \(B\) and \(D\) are also congruent.
Tests for congruency
There are four tests for congruency of triangles, namely
Side, Side, Side (\(SSS\))
Angle, Angle, Side (\(AAS\))
Side, Angle, Side (\(SAS\))
Right angle, Hypotenuse, Side (\(RHS\))
Interactive Exercises:
Congruence and Similarity Interactive Exercises, https://www.cimt.org.uk/sif/geometry/g4/interactive.htm
This is the first of three units based on analytic geometry, which brings algebra and geometry together. The other two units are Straight Lines and Using Graphs to Solve Equations. After completing this unit you should
be able to identify and illustrate points in two dimensions with positive coordinates
be able to identify and illustrate points in two dimensions
be confident in plotting straight lines
be confident in plotting curves
be able to find the mid-point of a line segment.
Coordinates are used to describe position in two dimensions. This Unit will show you how to use coordinates and apply them to some problems.
You have five sections to work through namely;
Positive Coordinates
Coordinates
Plotting Straight Lines
Plotting Curves
Mid-Points of Line Segments
0, Chapter 2, Positive Coordinates,
Positive Coordinates
Remember
Coordinates are pairs of numbers that uniquely describe a position on a rectangular grid.
These numbers are sometimes referred to as Cartesian coordinates.
The first number refers to the horizontal \(\left(x-axis\right)\) and the second the vertical \(\left(y-axis\right)\). Note that axes should always be labelled.
1, Chapter 2, Task 1, Worked Example 1,
Worked Example
Use the slider to explore worked examples.
Worked Example
Plot the points with coordinates \(\left(3, 8\right)\), \(\left(6, 1\right)\) and \(\left(2, 5\right)\).
Solution
For \(\left(3, 8\right)\) move \(3\) across and \(8\) up.
For \(\left(6, 1\right)\) move \(6\) across and \(1\) up.
For \(\left(2, 5\right)\) move \(2\) across and \(5\) up.
Worked Example
Write down the coordinates of each point in the diagram below.
Find the maximum volume of water the extinguisher could hold.
Solution
\(A\) is \(6\) across and \(5\) up, so the coordinates are \(\left(6, 5\right)\). \(B\) has no movement across and is straight up \(5\), so the coordinates are \(\left(0, 5\right)\). \(C\) is \(6\) across and \(3\) up, so the coordinates are \(\left(6, 3\right)\). \(D\) is \(8\) across and no movement up, so the coordinates are \(\left(8, 0\right)\).
0, Chapter 3, Coordinates,
Coordinates
Remember
The coordinates of a point are written as a pair of numbers, \(\left(x, y\right)\), which describe where the point is on a set of axes. The axes can include positive and negative values for \(x\) and \(y\).
The \(x-axis\) is always horizontal (i.e. across the page) and the \(y-axis\) always vertical (i.e. up the page).
The \(x-coordinate\) is always given first and the \(y-coordinate\) second.
1, Chapter 3, Task 1, Worked Example 2,
Worked Example
Use the slider to explore worked examples.
Worked Example
On a grid, plot the point A which has coordinates \(\left(–2, 4\right)\), the point \(B\) with coordinates \(\left(3, –2\right)\) and the point \(C\) with coordinates \(\left(– 4, –3\right)\).
For \(A\), begin at \(\left(0, 0\right)\), where the two axes cross. Move \(-2\) in the \(x\) direction. Move 4 in the y direction. Points \(B\) and \(C\) are plotted in a similar way: For \(B\), move \(3\) in the \(x\) direction and \(-2\) in the \(y\) direction. For \(C\), move \(-4\) in the \(x\) direction and \(-3\) in the \(y\) direction.
Worked Example
Write down the coordinates of each place on the map of the island.
Lighthouse \(\left(7, 5\right)\) Jetty \(\left(1, 3\right)\) Church \(\left(3, -2\right)\) Camp Site \(\left(1, -3\right)\) Shop \(\left(-4, 2\right)\) Telephone Box \(\left(-4, 1\right)\) Café \((-5 , -2)\) Lifeboat Station \((-2, -2)\)
0, Chapter 4, Plotting Straight Lines,
Plotting Straight Lines
Use the slider to explore worked examples.
By calculating values of coordinates you can find points and draw a graph for any relationship, such as \(y = 2x - 5\). This is best demonstrated by studying the following worked examples.
Worked Example
Copy and complete the following pairs of coordinates using the relationship \(y = x − 2\). \(\left(4, ?\right), \left(1, ?\right), \left(–1, ?\right)\)
Plot the points on a set of axes and draw a straight line through them.
Solution
With \(y = x - 2\), for the first point \(x = 4\) , so \(y = 4 - 2 = 2\) So the first point is \(\left(4, 2\right)\).
For the second point \(x = 1\) , so \(y = 1 - 2 = -1\) So the second point is \(\left(1, –1\right)\).
For the third point \(x = -1\) , so \(y = -1 - 2 = -3\) So the third point is \(\left(-1, -3\right)\).
The points can be plotted, as shown below.
Then a straight line can be drawn through these points, as below.
Worked Example
Draw the graph of \(y = 2x − 1\).
Solution
The first step is to find the coordinates of three points on the line. Choose any three x-values for the coordinates. Three possible values are given below: \(\left(4, ?\right)\), \(\left(2, ?\right)\), \(\left(–2, ?\right)\)
With \(y = 2x − 1\), for the first point \(x = 4\), \(y = 2\times 4 - 1 = 7\) So the first point is \(\left(4, 7\right)\).
For the second point \(x = 2\) , so \(y = 2\times 2 - 1 = 3\) So the second point is \(\left(2, 3\right)\).
For the third point \(x = -2\), so \(y = 2 \times -2 -1 = -5\) So the third point is \(\left(−2, −5\right)\).
So three coordinates which lie on the line are: \(\left(4, 7\right)\),\(\left(2, 3\right)\) and \(\left(−2, −5\right)\)
These points can now be plotted, as shown on the axes on the right. Then a straight line can be drawn through them.
0, Chapter 5, Plotting Curves,
Plotting Curves
Some relationships produce curves rather than straight lines when plotted. This is shown in the worked examples below
Use the slider to explore worked examples.
Worked Example
a. Complete the table below using the relationship \(y = x^{2} - 2\).
b. Write a list of coordinates using the data in the table.
c. Plot the points and draw a smooth curve through them.
Solution
a. With \(y = x^{2} - 2\), for the last column, with \(x = 3\) , then \(y = 32 - 2 = 9 - 2 = 7\)
So in the table, in the last column, the value for \(y\) is \(7\) You can start to fill in the blanks in the table:
Similarly, as another example, if \(x = 2 , y = 22 − 2 = 2\)
Note that it doesn’t matter what order you complete the table in. Another example: if \(x = −2 , y = \left(−2\right)2 − 2 = 4 – 2 = 2\)
You can fill in more blanks in the table:
Do the same for all \(x\) values. The complete table looks like this:
b. The coordinates of the points are \(\left(–3, 7\right), \left(–2, 2\right), \left(–1, –1\right), \left(0, –2\right), \left(1, –1\right), \left(2, 2\right)\) and \(\left(3, 7\right)\). These points plotted are shown below.
Worked Example
Draw the graph of \(y = x^{3} − 4x\) for values of \(x\) from \(–3\) to \(3\).
Solution
For example, if \(x = –3, y = \left(−3\right)^{3} - 4\times \left(–3\right) = −27 + 12 = −15\) and, if \(x = 2, y = 2^{3} − 4\times2 = 8 − 8 = 0\)
Each pair of values from the table represent coordinates: \(\left(–3, –15\right), \left(–2, 0\right), \left(–1, 3\right), \left(0, 0\right), \left(1, –3\right), \left(2, 0\right), \left(3, 15\right)\)
These coordinates can be plotted and a smooth curve can be drawn through them, as seen on the graph.
0, Chapter 6, Mid-Points of Line Segments,
Mid-Points of Line Segments
The coordinates of the mid-point between two other points may be found by drawing or by calculation The mid Point of a line segment is the point which is exactly in the middle of it.
The diagram below shows the mid-point of the coordinates \(\left(2, 2\right)\) and \(\left(6, 8\right)\); the coordinate of the mid-point is \(\left(4, 5\right)\).
You can see from the diagram that the \(x-coordinate\) is in fact the mean value of the two \(x-coordinates\) of the end points of the line segment. Similarly the \(y-coordinate\) is the mean value of the two \(y-coordinates\) of the end points of the line segment.
Remember
Given any two points with coordinates \(\left(a, b\right)\) and \(\left(c, d\right)\), the coordinates of the mid-point of the line segment joining the two points is given by;
\(\left(\frac{a+b}{2},\frac{c+d}{2}\right)\)
1, Chapter 6, Task 1, Worked Example 3,
Worked Example
Worked Example
Find the coordinates of the mid-point of the line segment:
\(AB\)
\(AC\)
\(BD\)
Solution
The coordinates of \(A\) are \(\left(2, 7\right)\). The coordinates of \(B\) are \(\left(5, 3\right)\). The coordinates of the mid-point \(AB\) are \(\left(3.5, 5\right)\), as calculated below: \(\left(\frac{2+5}{2},\frac{7+3}{2} \right) =\left(\frac{7}{2} ,\frac{10}{2} \right) = \left(3.5,\;5\right)\)
The coordinates of \(A\) are \(\left(2, 7\right)\). The coordinates of \(C\) are \(\left(–5, 3\right)\). The coordinates of the mid-point \(AC\) are \(\left(–1.5, 5\right)\), as calculated below. \(\left(\frac{2+\left(-5\right) }{2} ,\frac{7+3}{2} \right) =\left(\frac{-3}{2},\frac{10}{2} \right) =\left(-1.5, 5\right)\)
The coordinates of \(B\) are \(\left(5, 3\right)\). The coordinates of \(D\) are \(\left(−4, −5\right)\) The coordinates of the mid-point \(BD\) are \(\left(0.5, −1\right)\), as calculated below. \(\left(\frac{5+\left(-4\right) }{2} ,\frac{3+\left(-5\right) }{2} \right) =\left(\frac{1}{2},\frac{-2}{2} \right) =\left(0.5, -1\right)\)
0, Chapter 7, Summary,
Summary
Coordinates
\(\left(x, y\right)\) means that the point \(P\) with coordinates \(\left(a, b\right)\) is such that \(x = a\), \(y = b\)
Coordinate axes
These are shown on the diagram below.
Straight line
Is defined by any two points on the line \(P\) and \(Q\) in the diagram below.
Curves
These are lines that are not straight (see diagram below).
Line segment
This is any part of a straight line between two points.
Mid-point of line segment
This is the point \(X\) in the line segment \(PQ\), such that length \(PX = length\;XQ\). The coordinates of \(X\) are: \(\left(\frac{a+c}{2},\frac{b+d}{2} \right)\)
This is the second unit in this strand with the first being Coordinates. Here we focus on the straight line. After completing this unit you should:
understand the concept of gradient
be able to calculate the gradient of a straight line
understand how to use straight line graphs in practical problems such as velocity/time graphs
be able to calculate the equation of a straight line
You have four sections to work through namely;
Gradient
Gradients of Perpendicular Lines
Applications of Graphs
Equation of a Straight Line
0, Chapter 2, Gradient,
Gradient
Remember
The gradient of a line describes how steep it is.
The diagram below shows two lines, one with a positive gradient and the other with a negative gradient.
Top Tip
Think of a gradient as going up or down a hill. A positive gradient can be seen to go ‘uphill’ in moving from left to right and a negative gradient goes ‘downhill’
1, Chapter 2.1, Calculating Gradient,
Calculating Gradient
Remember
The gradient of a line between two points, \(A\) and \(B\), is calculated using the following formula
Find the gradient of the line shown in the diagram.
Solution
Draw any triangle under the line below to show the horizontal and vertical distances.
Here the vertical distance is \(10\) and the horizontal distance is \(6\).
\(Gradient= \frac{vertical\;change}{horizontal\;change}\) \(Gradient= \frac{10}{6} = \frac{5}{3}\) or \(1\frac{2}{3}\)
Worked Example
Find the gradient of the line joining the point \(A\) with coordinates \(\left(2, 4\right)\) and the point \(B\) with coordinates \(\left(4, 10\right)\).
The line segments have the same gradient and so are parallel. The results in this example mean that the quadrilateral \(ABDC\) is, in fact, a parallelogram.
0, Chapter 3, Gradients of Perpendicular Lines,
Gradients of Perpendicular Lines
Remember
The product of the gradients of two perpendicular lines will always be \(–1\), unless one of the lines is horizontal and the other is vertical.
This was found in the previous example with the perpendicular lines of \(AB\) and \(PQ\): the gradient of \(AB = 3\): the gradient of \(PQ = -\frac{1}{3}\) and \(3\times-\frac{1}{3}=-1\)
An equivalent way of describing this is by showing \(gradient\;of\;AC= \frac{-1}{gradient\;AB}\)
If the gradient of a line is \(m\), and \(m \neq 0\),then the gradient of a perpendicular line will be \(-\frac{1}{m}\)
1, Chapter 3, Task 1, Worked Example 2,
Worked Example
Use the slider to explore worked examples.
In this section we explore the relationship between the gradients of perpendicular (at right angles to each other) lines and line segments.
Worked Example
Plot the points \(A \left(1, 2\right)\) and \(B \left(4, 11\right)\), join them to form the line \(AB\) and then calculate the gradient of \(AB\).
On the same set of axes, plot the points \(P \left(5, 4\right)\) and \(Q \left(8, 3\right)\), join them to form the line \(PQ\) and calculate the gradient of \(PQ\).
Measure the angle between the lines \(AB\) and \(PQ\). What do you notice about the two gradients?
a. First draw the points, as shown here. The points are then joined to form the line \(AB\) Gradient of \(AB = \frac{11 - 2}{4-1} = \frac{9}{3} = 3\)
b. The points \(P\) and \(Q\) can now be added to the diagram as shown here. \(Gradient\;of\;PQ = \frac{3-4}{8-5}=-\frac{1}{3}\)
c. The line \(PQ\) has been extended on the diagram, so that the angle between the two lines can be measured. The angle is \(90^{\circ}\), a right angle. In this case, the gradient of \(AB = 3\) the \(gradient of PQ = -\frac{1}{3}\) the gradients multiply to give \(3\times -\frac{1}{3}=-1\)
Note that if two lines are perpendicular, when you multiply together the gradients, you always get \(-1\) as your answer.
Worked Example
Show that the line segment joining the points \(A\;\left(3, 2\right)\) and \(B\;\left(5, 7\right)\) is perpendicular to the line segment joining the points \(P\;\left(2, 5\right)\) and \(Q\;\left(7, 3\right)\).
So the line segments \(AB\) and \(PQ\) are perpendicular.
0, Chapter 4, Conversion Graphs,
Conversion Graphs
In this section some applications of graphs are considered, particularly conversion graphs and graphs to describe motion.
The graph below is a conversion graph: it can be used for converting US Dollars into and from Jamaican Dollars.
Note ,as currency exchange rates are continually changing, these rates might not be correct now.
1, Chapter 4.1, Distance-Time Graphs,
Distance-Time Graphs
The following graph is used to describe motion.
A distance-time graph of a car is shown below. The gradient of this graph gives the speed of the car. The gradient is steepest from \(A\) to \(B\), so this is when the car has the greatest speed. The gradient \(BC\) is zero, so the car is not moving.
Note that we sometimes use speed and other times velocity. In fact, velocity is the speed in a given direction.
1, Chapter 4.2, Velocity/Speed-Time Graphs,
Velocity/Speed-Time Graphs
The following graph is also used to describe motion.
The area under a speed/velocity-time graph gives the distance travelled. Finding the shaded area on the graph shown opposite would give the distance travelled.
The gradient of this graph gives the acceleration of the car. There is constant acceleration from \(0\) to \(20\) seconds, then zero acceleration from \(20\) to \(40\) seconds (when the car has constant speed), constant deceleration from \(40\) to \(50\) seconds, etc.
1, Chapter 4.2, Task 1, Worked Example 3,
Worked Example
Use the slider to explore worked examples.
Worked Example
A temperature of \(20^{\circ}\) is equivalent to \(68^{\circ} F\) and a temperature of \(100^{\circ}C\) is equivalent to a temperature of \(212 ^{\circ}F\). Use this information to draw a conversion graph. Use the graph to convert:
\(30^{\circ}\) to \(^{\circ}Farenheit\)
\(180^{\circ}\) to \(^{\circ}Celsius\)
Solution
Taking the horizontal axis as temperature in \(^{\circ} C\) and the vertical axis as temperature in \(^{\circ}F\) gives the two points, \(\left(20, 68\right)\) and \(\left(100, 212\right)\).
These are plotted on a graph anda straight line drawn through the points
Start at \(30 ^{\circ} C\), then move up to the line and across to the vertical axis, to give a temperature of about \(86 ^{\circ} F\).
Start at \(180 ^{\circ} F\) , then move across to the line and down to the horizontal axis, to give a temperature of about \(82^{\circ}\) .
Worked Example
The graph shows the distance travelled by a girl on a bike.
Find the speed she is travelling on each stage of the journey. Note that units are m/s (metres per second), as m are the units for distance and s the units for time.
Solution
For \(AB\) the gradient \(\frac{750}{200} = 3.75\) So the speed is \(3.75\;m/s\)
For \(BC\) the gradient \(\frac{500}{50} = 10\) So the speed is \(10\;m/s\)
For \(CD\) the gradient is zero. So the speed is \(0\;m/s\)
For \(DE\) the gradient \(=\frac{500}{100}=5\) So the speed is \(5\;m/s\)
Worked Example
The graph shows how the velocity of a bird varies as it flies between two trees. How far apart are the two trees?
Solution
The distance is given by the area under the graph. In order to find this area it has been split into three sections, \(A, B\) and \(C\).
So the trees are \(60\;m\) apart. Note that the units are \(m\) (metres) because the units of velocity are \(m/s\) and the units of time are \(s\) (seconds).
Worked Example
The velocity-time graph below, not drawn to scale, shows that a train stops at two stations, \(A\) and \(D\). The train accelerates uniformly from \(A\) to \(B\), maintains a constant speed from \(B\) to \(C\) and decelerates uniformly from \(C\) to \(D\).
Using the information on the graph,
Calculate, in \(ms^{−2}\), the train's acceleration
Show that the train took \(30\;secs\) from \(C\) to \(D\) if it decelerated at \(\frac{1}{2}ms^{−2}\) .
If the time taken from \(A\) to \(D\) is \(156\;seconds\), calculate the distance in metres between the two stations.
Solution
Using the information on the graph,
Acceleration \(=\frac{15}{15}=1\;ms^{-2}\)
If it takes \(t\) seconds from \(C\) to \(D\), then acceleration \(=\frac{1}{2} =\frac{15}{t}\) and so \(t=30\;seconds\)
Time from \(B\) to \(C=156-\left(15+30\right)=111\;seconds\)
Total distance = area under graph \(=\frac{1}{2} \times 15 \times 15 +15 \times 111 + \frac{1}{2} \times 15 \times 30 = \frac{225}{2} +1665+225= \frac{4005}{2} = 2002\frac{1}{2}\)
So the total distance is \(2002.5\;metres\)
0, Chapter 5, Equation of a Straight Line,
Equation of a Straight Line
Remember
The equation of a straight line is usually written in the form \(y=mx+c\) where \(m\) is the gradient and \(c\) is the \(y\) intercept.
Note - the intercept is the point where the line crosses the \(y\;axis\). See the diagram below.
1, Chapter 5, Task 1, Worked Example 4,
Worked Example
Use the slider to explore worked examples.
In this section we explore the relationship between the gradients of perpendicular (at right angles to each other) lines and line segments.
Worked Example
Find the equation of the line shown in the diagram.
Solution
The first step is to find the gradient of the line, the diagram on the right can be used. Drawing the triangle shown under the line gives
The \(y\) intercept is \(7\). So \(m=-2\) and \(c=7\) Therefore the equation of the line is \(y= -2x+7\) or \(y=7-2x\) (either is correct)
Worked Example
Draw the line \(𝑥 = 4\).
Draw the line \(𝑦 = 2\).
Write down the coordinates of the point of intersection of these lines.
Solution
For the line \(x = 4\) the \(x-coordinate\) of every point will always be \(4\).
So the points \(\left(4, 0\right) \left(4, 3\right) \left(4, 5\right)\) all lie on the line \(x = 4\).
For the line \(y = 2\) the \(y-coordinate\) of every point will always be \(2\).
So the points \(\left(0, 2\right) \left(3, 2\right)\) and \(\left(5, 2\right)\) all lie on the line \(y = 2\).
The graph above shows that the lines intersect at the point with coordinates (4, 2).
Worked Example
The point with coordinates \(\left(4, 9\right)\) lies on the line with equation \(y=2x+1\). Determine the equation of the perpendicular line that also passes through this point.
Solution
The line \(𝑦=2𝑥+1\) has gradient \(2\). The perpendicular line will have gradient, \(m=-\frac{1}{2}\), is and so its equation will be of the form \(y=-\frac{1}{2}x+c\)
As the line passes through (4, 9), we can use \(x = 4\) and \(y = 9\) to determine the value of \(c\): \(9=-\frac{1}{2} \times 4 + c\) \(9=-2+c\) \(c=11\)
We now have the values of \(m=-\frac{1}{2}\) , and \(c=11\) . These are the values we use in the equation \(𝑦 = m𝑥+c \).
The equation of the perpendicular line is \(y=-\frac{1}{2}x+11\)
The graph on the below shows both lines.
Worked Example
In the diagram below, not drawn to scale, AB is the straight line joining \(A \left(−1, 9\right)\) and \(B \left(3, 1\right)\).
Calculate the gradient of the line, \(AB\).
Determine the equation of the line, \(AB\).
Write the coordinates of \(G\), the point of intersection of \(AB\) and the \(y-axis\)
Write the equation of the line through \(O\), the origin, that is perpendicular to \(AB\).
Write the equation of the line through \(O\) that is parallel to \(AB\).
Solution
\(Gradient = \frac{9-1}{-1-3}=\frac{8}{-4}=-2\)
So the equation of line \(AB\) is \(y= -2x+c\) As it passes through \(\left(3, 1\right), 1=-2 \times 3 + c\) threfore \(c=7\) equation is \(y=-2x+7\)
\(x=0,\) so \(y=-2 \times 0 + 7 = 7.\) \(G\) is point \(\left(0,7\right)\) (as it passes through the origin)
This unit continues the angle geometry theme, with the important topic of bearings. After completing this unit you should
be able to identify and use bearings in practical problems.
0, Chapter 2, Bearings,
Bearings
A compass bearing tells us direction.
The 4 main directions are North, South, East and West. We can remember the order (starting from north, and going clockwise) as ‘Never Eat Shredded Wheat’:
Note that the angle between \(N\) and \(E\) is \(90^{\circ}\) and between \(N\) and \(NE\) is \(45^{\circ}\)
Notice that the bearings in between start with either North or South: North-East, South-East, South-West, North-West.
1, Chapter 2.1, In between directions,
In between directions
In between these bearings we have:
North-North-East (\(NNE\))*
East-North-East (\(ENE\))
East-South-East (\(ESE\))
South-South-East (\(SSE\))
South-South-West (\(SSW\))
West-South-West (\(WSW\))
West-North-West (\(WNW\))
North-North-West (\(NNW\))
Notice the bearings in between start with: North, East, South or West:
NNE, ENE, ESE, SSE, SSW, WSW, WNW, NNW
* NNE can be read as North of North-East
1, Chapter 2.2, Calculating a Bearing,
Calculating a Bearing
Three figure bearings can also be used as an alternative to compass bearings.
They are measured in a particular way:
Remember
How to calculate a bearing
We start measuring from North
Measure the angle in degrees \(^{\circ}\) clockwise
The bearing (angle) is given as three figures;often this means we may have to ‘add’ zeros at the start. For example: North-East is 045 \(^{\circ}\)
The advantage of 3-figure bearings is that they can describe direction uniquely.
Note: Sometimes 3-figure bearings are given with such accuracy that they have digits after a decimal point, e.g. 192.75\(^{\circ}\) (this is still called a 3-figure bearing, as it has 3 figures (digits) before the decimal point).
1, Chapter 2.3, 3-Figure Bearings,
3-Figure Bearings
Examples of 3-figure bearings:
Bearings can be especially useful in navigation, be it at sea or for people walking in the wild on open moorland or hills.
1, Chapter 2.3, Task 1, Worked Example,
Worked Example
Use the slider to explore worked examples.
Worked Example
Work out the bearing of \(B\) from \(A\).
Solution
First, draw (or look for) a line to \(N\) (pointing straight up from \(A\)). We now measure clockwise from north. From north to the dashed line is \(180^{\circ}\) (angles on a straight line). We now simply add on the extra \(40^{\circ}\). Answer: Bearing of \(B\) from \(A\) is: \(180^{\circ} + 40^{\circ} = 220^{\circ}\)
Worked Example
a) Write down the bearing of A from P.
b) b) Work out the bearing of B from P.
Solution
a) Draw, or look for, a line to N from P.
Then measure clockwise from north.
This gives the bearing of \(060^{\circ}\).
b) Using angles around a point equal \(360^{\circ}\), we can calculate the bearing by using:
\( 360^{\circ} - 140^{\circ} \)
This gives the bearing as \(220^{\circ}\).
Worked Example
The bearing of a ship (S) from a lighthouse (L) is \(050^{\circ}\)
Work out the bearing of the lighthouse from the ship.
Solution
The bearing of S from L is \(050^{\circ}\).
The bearing of L from S is \(180^{\circ}\)
We now simply add on the extra 50˚ (use alternate ‘Z’ angles)
Answer: Bearing of the lighthouse from the ship:
\( 180^{\circ} + 50^{\circ} = 230^{\circ} \)
0, Chapter 3, Summary,
Summary
Bearings
A bearing gives the direction or position of something, or the direction of movement, relative to a fixed point. Bearings are of the form of angles, expressed in degrees as three-digit numbers; they are measured from north in a clockwise direction.
In finding bearings, say \(A\) from \(B\), you must find the angle (measured clockwise) made by the \(N\) direction with the line from \(B\) to \(A\). For example, \(060^{\circ}\), \(210^{\circ}\).
This is the first of two units on angle geometry; the second is Angles, Circles and Tangents. Angle geometry underpins all the work on trigonometry in the set of units that follows. After completing this unit you should
be able to measure and construct accurately, using a protractor, angles up to \(360^{\circ}\)
be able to identify the rotational symmetry and line symmetry of \(2\) dimensional shapes
know and understand the angle facts for points, lines, triangles and quadrilaterals
know and understand the angle facts for parallel and intersecting lines
be able to determine the size of the interior and exterior angles for a regular polygon.
You have five sections to work through namely;
Measuring Angles
Line and Rotational Symmetry
Angle Geometry
Angles with Parallel and Intersecting Lines
Angle Symmetry in Regular Polygons
0, Chapter 2, Measuring Angles,
Measuring Angles
Remember
An angle is a measure of turn. Angles can be measured in degrees: \(180\) degrees is a half turn, \(360\) degrees a full turn and \(90\) degrees a quarter turn, known as a right angle.
Degrees are indicated with a small circle, e.g. \(90^{\circ}\) represents \(90\) degrees.
The angle around a complete circle is \(360^{\circ}\), a full turn.
The angle around a point on a straight line is \(180^{\circ}\), a half turn.
A protractor can be used to measure or draw angles which we show in the next example.
1, Chapter 2, Task 1, Worked Examples 1,
Worked Examples
Use the slider to explore worked examples.
Worked Example
Measure the angle \(CAB\) in the triangle shown.
Place a protractor on the triangle as shown. The angle is measured as \(47^{\circ}\).
Note: When measuring an angle, start from the \(0^{\circ}\) which is in line with an arm of the angle.
Worked Example
Measure the marked angle.
Using a protractor, the smaller angle is measured as \(100^{\circ}\).
Draw a horizontal line. Place a protractor on top of the line and draw a mark at \(120^{\circ}\).
Then remove the protractor and draw the angle.
To draw the angle of \(330^{\circ}\), first subtract \(330^{\circ}\) from \(360^{\circ}\): required angle is \(360^{\circ} - 330^{\circ} = 30^{\circ}\)
Draw an angle of \(30^{\circ}\).
The larger angle will be \(330^{\circ}\).
0, Chapter 3, Rotational Symmetry,
Rotational Symmetry
An object has rotational symmetry if it can be rotated about a point so that it fits on top of itself without completing a full turn.
The shapes below have rotational symmetry.
In a complete turn this shape fits on top of itself two times. It has rotational symmetry of order \(2\).
In a complete turn this shape fits on top of itself four times. It has rotational symmetry of order \(4\).
0, Chapter 4, Lines of Symmetry,
Lines of Symmetry
Shapes have line symmetry if a mirror could be placed so that one side is an exact reflection of the other. These imaginary 'mirror lines' are shown by dotted lines in the diagrams below.
This shape has \(2\) lines of symmetry.
This shape has \(4\) lines of symmetry.
1, Chapter 4, Task 1, Worked Example 2,
Worked Examples
Worked Example
For the given shape, state:
the number of lines of symmetry,
the order of rotational symmetry.
Solution
There are \(3\) lines of symmetry, as shown.
There is rotational symmetry with order \(3\) because the point marked \(A\) could be rotated to \(A'\) then to \(A''\) and fit exactly over its original shape at each of these points.
0, Chapter 5, Angle Geometry,
Angle Geometry
There are a number of important results concerning angles in different shapes, at a point and on a line.
There are 6 rules that you must remember :
Before we describe the rules let us remind you of the names for different angles
Name
Definition
ACUTE angle
Less than \(90\) degrees
OBTUSE angle
Between \(90\) degrees and \(180\) degrees
RIGHT angle
\(90\) degrees
REFLEX angle
Between \(180\) degrees and \(360\) degrees
1, Chapter 5.1, Rules at a point or on a line,
Rules at a point or on a line
Angles at a Point The angles at a point will always add up to \(360^{\circ}\). It does not matter how many angles are formed at the point - their total will always be \(360^{\circ}\).
Angles on a Line Any angles that form a straight line add up to \(180^{\circ}\).
1, Chapter 5.2, Rules in triangles and quadrilaterals,
Rules in triangles and quadrilaterals
Angles in a Triangle The angles in any triangle add up to \(180^{\circ}\).
Angles in an Equilateral Triangle
In an equilateral triangle all the angles are \(60^{\circ}\) and all the sides are the same length.
Note that the tick marks mean lines of equal length
Angles in an Isosceles Triangle
In an isosceles triangle two sides are the same length and two angles are the same size.
Angles in a Quadrilateral
The angles in any quadrilateral add up to \(360^{\circ}\).
1, Chapter 5.2, Task 1 , Worked Example 3,
Worked Examples
Use the slider to explore worked examples.
Worked Example
Find the sizes of angles \(𝑎\) and \(𝑏\) in the diagram on the right.
Solution
First consider the quadrilateral. All the angles of this shape must add up to \(360^{\circ}\) so ,
Then consider the straight line formed by the angles \(a\) and \(b\). These two angles must add up to \(180^{\circ}\) and we have just shown that \(a=100^{\circ}\) so
In the figure, not drawn to scale, \(ABC\) is an isosceles triangle with
\(\angle CAB = p^{\circ}\) and \(\angle ABC = \left(p+3\right)^{\circ}\).
Write an expression in terms of \(p\) for the value of the angle at \(C\)
Determine the size of each angle in the triangle
Solution
As \(ABC\) is an isosceles triangle, \(\angle ACB =p+3^{\circ}\)
For triangle \(ABC\)
\(\begin{align*} p+\left(p+3\right)+\left(p+3\right)&=180^{\circ}\\ 3p+6&=180^{\circ}\\ p&=58^{\circ} \end{align*}\) And so it follows that the angles at \(A, B\) and \(C\) are \(58^{\circ}, 61^{\circ}\) and \(61^{\circ}\), respectively.
0, Chapter 6, Angles with Parallel and Intersecting Lines,
Angles with Parallel and Intersecting Lines
Opposite Angles
When any two lines intersect, two pairs of equal angles are formed. The two angles marked \(a\) are a pair of opposite equal angles. The angles marked \(b\) are also a pair of opposite equal angles.
Corresponding Angles
When a line intersects a pair of parallel lines, \(angle\;a= angle\;b\); the \(angles\;a\) and \(b\) are called corresponding angles.
Alternate Angles
The angles \(c\) and \(d\) are equal, and are called alternating angles
Supplementary Angles
The angles \(𝑏\) and \(c\) add up to \(180^{\circ}\), and are called supplementary angles.
1, Chapter 6, Task 1, Worked Example 4,
Worked Examples
Use the slider to explore worked examples.
Worked Example
Find the sizes of angles \(𝑎\) and \(𝑏\) and \(c\) in the diagram on the right.
There are two pairs of opposite angles here so: \(b=100^{\circ}\) and \(a=c\).
The angle marked \(b\) is opposite the angle \(a\), so \(b=a=50^{\circ}\).
Now \(c\) and \(d\) can be found using corresponding angles.
The angle \(c\) and the \(70^{\circ}\) angle are corresponding angles, so \(c=70^{\circ}\)
The angle \(d\) and the \(60^{\circ}\) angle are corresponding angles, so \(d=60^{\circ}\).
Worked Example
In the diagram on the right, not drawn to scale, \(AB\) is parallel to \(CD\) and \(EG\) is parallel to \(FH\), angle \(IJL =50^{\circ}\) and angle \(KIJ =95^{\circ}\)
Calculate the values of \(x, y\) and \(z,\) showing clearly the steps in your calculations.
Find value of \(x\): Angles \(BIG\) and \(END\) are supplementary angles, so \(\begin{align*} 95+END&=180^{\circ}\\ END&= 85^{\circ} \end{align*}\)
But angles \(END\) and \(FMD\) are corresponding angles, so \(x=85^{\circ}\)
Find value of \(y\): Angles \(BCD\) (\(y\)) and \(ABC\) are alternate angles, so \(y= ABC\)
In triangle \(BIJ\) \(\begin{align*} y+95+50&=180^{\circ}\\ y&= 35^{\circ} \end{align*}\)
Find value of \(z\): Angles \(AKH\) (\(z\)) and \(FMD\) (\(x\)) are alternate angles, so
Regular polygons have sides which are all equal in length; all interior angles are equal too. Regular polygons will have both line and rotational symmetry.
This symmetry can be used to find the interior angles of a regular polygon.
1, Chapter 7, Task 1, Worked Examples 5,
Worked Examples
Use the slider to explore worked examples.
Worked Example
Find the interior angle of a regular dodecagon (\(12\) sides).
Solution
The diagram shows how a regular dodecagon can be split into \(12\) isosceles triangles. As there are \(360^{\circ}\) around the centre of the dodecagon, the centre angle in each triangle is \(\frac{360^{\circ}}{12} = 30^{\circ}\)
Next we will find the angle indicated on the diagram on the right. The other angles of each triangle will together add up to \(180^{\circ} - 30^{\circ} = 150^{\circ}\)
Therefore the angle on the diagram is \(\frac{150}{2} = 75^{\circ}\)
As two adjacent angles are required to form each interior angle of the dodecagon, each interior angle will be \(75^{\circ} \times 2=150^{\circ}\)
As there are \(12\) interior angles, the sum of these angles will be \(12 \times 150^{\circ} =1800^{\circ}\).
Worked Example
Find the sum of the interior angles of a regular heptagon.
Solution
Split the heptagon into \(7\) isosceles triangles. Each triangle contains three angles which add up to \(180^{\circ}\), so the total of all the marked angles will be
\( 7\times 180^{\circ}= 1260^{\circ}\)
However the angles at the point where all the triangles meet should not be included, so the sum of the interior angles is given by
\(1260^{\circ} -360^{\circ} = 900^{\circ}\)
Worked Example
Copy the octagon shown in the diagram and draw in any lines of symmetry.
Copy the octagon and shade in extra triangles so that it now has rotational symmetry.
Solutions
There is only one line of symmetry.
The original octagon has no rotational symmetry. After shading the extra triangle shown, it has rotational symmetry of order \(4\).
After shading all the triangles, it has rotational symmetry of order \(8\).
0, Chapter 8, Summary,
Summary
An object has rotational symmetry if it can be rotated about a point so that it fits on top of itself without completing a full turn. The shapes below have rotational symmetry.
In a complete turn this shape fits on top of itself two times. It has rotational symmetry of order \(2\).
In a complete turn this shape fits on top of itself four times. It has rotational symmetry of order \(4\).
Shapes have line symmetry if a mirror could be placed so that one side is an exact reflection of the other. These imaginary 'mirror lines' are shown by dotted lines in the diagrams below.
This shape has \(2\) lines of symmetry
This shape has \(4\) lines of symmetry.
Angles at a point
The angles at a point will always add up to \(360^{\circ}\). It does not matter how many angles are formed at the point – their total will always be \(360^{\circ}\).
Angles on a line
Adjacent angles that form a straight line add up to \(180^{\circ}\)
Angles in a triangle
The angles in any triangle add up to \(180^{\circ}\)
Angles in an equilateral triangle
In an equilateral triangle all the angles are \(60^{\circ}\) and all the sides are the same length.
Angles in an isosceles triangle
In an isosceles triangle two sides are the same length and the two base angles are the same size.
Angles in a quadrilateral
The angles in any quadrilateral add up to \(360^{\circ}\) .
Opposite angles
When any two lines intersect, two pairs of equal opposite angles are formed. The two angles marked \(a\) and \(c\) are a pair of opposite equal angles, so angle \(a\) is equal to angle \(c\). The angles marked \(b\) and \(d\) are also a pair of opposite equal angles, so angle \(b\) is equal to angle \(d\)
Corresponding angles
When a line intersects a pair of parallel lines, angle \(a\) is equal to angle \(b\). The angles \(a\) and \(b\) are called corresponding angles.
Alternate angles
The angles \(c\) and \(d\) are equal.
Interactive Exercises:
Interactive Exercises - Angles and Symmetry, https://www.cimt.org.uk/sif/geometry/g1/interactive.htm
Angles and triangles, https://www.bbc.co.uk/bitesize/guides/zrck7ty/revision/4, images/Images/bitesize-logo.png, The four types of angle you should know are acute, obtuse, reflex and right angles. When you are estimating the size of an angle, you should consider what type of angle it is first.
A graph is a diagram showing the relationship between some variable quantities.
This unit uses the concept of graphical functions in order to solve equations.
After completing this unit you should
be able to graph linear simultaneous equations in order to find the solution
understand the graphical form of well-known functions; these include linear, quadratic, cubic and reciprocal functions
be able to graph functions in order to solve equations, including the solution of quadratic and linear equations.
You have four sections to work through namely;
Solution of Simultaneous Equations by Graphs
Graphs of Common Functions
Graphical Solutions of Equations
Tangents to Curves
0, Chapter 2, Simultaneous Equations,
Simultaneous Equations - Introduction
Try solving: \(x + y = 24\) What could ‘\(x\)’ be? What could ‘\(y\)’ be?
There are simply too many possible solutions.
So if we have more than one ‘unknown’ (variable), we must have more than one equation.
With two ‘unknowns’, we must have at least two equations, and we must solve them at the same time, i.e. simultaneously.
Let us try again:
\(x+y=24\) \((1)\)
\(2x=20\) \((2)\)
Can you solve it now?
What could \(x\) be? What could \(y\) be?
First, label the equations as shown.
From the equation 2, we can work out \(x=10\)
So now we can use this solution and ‘plug’ it in (i.e. substitute) this value into the equation 1.
\(x+y=24\) Substitute \(x\) value into equation 1 \(10+y=24\) Now solve for \(y\): \(y=14\)
We can see therefore that the solutions are: \(x=10\) \(y=14\)
1, Chapter 2.1, Solving by Graphs,
Solving by Graphs
There are three ways to solve simultaneous linear equations
One method is by substitution which we have jst seen in the last section
Another method is by elimination
In this unit we will look at solving simultaneous linear equations by drawing graphs (and finding the coordinates of where the 2 lines cross)
Now lets show the graphical method in an example by find solutions for \(x\) and \(y\) that satifies these two equations \(x + y = 24\) \(2x = 20\)
First we rearrange the equations and write in the form \(y = …\)
From the first equation, \(x + y = 24\) \(y = 24 - x\)
From the second equation, \(2x=20\) \(x=10\)
These are both linear (straight line) equations. To draw a straight line, only two pairs of coordinates are required.
For: \(y = 24 - x\), if \(x = 0\), then \(y = 24\) and if \(x = 24\), then \(y = 0\)
We can see that \(\left(0,24\right)\) and \(\left(24,0\right)\) are both points on the line
For \(x=10\) for each and every \('y'\) coordinate \(x=10\)
This is the vertical line going through \(x=10\)
We can now draw the graphs defined by our two equations and idenitify the point where they cross as shown below.
1, Chapter 2.1, Task 1, Worked Example 1,
Worked Example
Worked Example
By drwawing graphs find solutions for \(x\) and \(y\) that satifies these two equations? \(x + y = 8\) \(2x + 3y = 21\)
First we rearrange the equations and write in the form \(y = …\)
From the first equation, \(x + y = 8\) \(y = 8 - x\)
From the second equation,
\(\begin{align*}
2x+3y&=21\\
3y&=21-2x\\
y&=21-2x\\
y&=\frac{21-2x}{3}\\
y&=7-\frac{2x}{3}
\end{align*}\)
These are both linear (straight line) equations. To draw a straight line, only two pairs of coordinates are required.
For: \(y = 8 - x\), if \(x = 0\), then \(y = 8\) and if \(x = 8\), then \(y = 0\) \(\left(0,8\right)\) \(\left(8,0\right)\) are on the line
For: \(y=7-\frac{2x}{3}\) if \(x=0\), then \(y=7\) and if \(x=3\), then \(y=5\) \(\left(0,7\right)\) \(\left(3,5\right)\) are on the line
These points are then plotted: \(\left(0,8\right)\) \(\left(8,0\right)\) and \(\left(0,7\right)\) \(\left(3,5\right)\)
The two lines intersect at the point \(\left(3,5\right)\), so the solution is: \(x=3\) and \(y=5\)
Note - Graphing software (such as DESMOS) can be used to plot and explore graphs effortlessly.
0, Chapter 3, Graphs of Common Functions,
Graphs of Common Functions
Remember
Linear Functions is the name given to equations whose graph is a straight line. Such equations as always of the form \(y +mx +c\)
The graph is shown below. \(c\) is the point where the graphs cross the \(y\;axis\) and is called the intercept.
1, Chapter 3.1, Quadratic Functions,
Quadratic Functions
Remember
Quadratic functions contain an \(x^{2}\) term as well as multiples of \(x\) and a constant.
The following graphs show three examples:
1, Chapter 3.2, Cubic Functions,
Cubic Functions
Remember
Cubic functions contain an \(x^{3}\) term as well as multiples of \(x^{2}, x\) and a constant.
Here are some examples below
1, Chapter 3.3, Reciprocal ufnctions,
Reciprocal Functions
Remember
Reciprocal functions have the form of a fraction with \(x\) as the denominator, for example \(y =\frac {5}{2x}\).
The following graphs show three examples:
1, Chapter 3.4, Expotential Functions,
Expotential Functions
Remember
Exponential functions contain a term with the variable (unknown) as the power (or index or exponent), for example \(y =2^{x}\).
The following graph shows the graph of \(y =2^{x}\).
1, Chapter 3.4, Task 1, Worked Example 2,
Worked Example
Use the slider to explore worked examples.
Worked Example
Write down the letter of the graph which could have the equation:
a. \(y = 3x - 2\)
a. H
b. \(y = 2x^{2} + 5x - 3\)
b. D
c. \(y = \frac {3}{x}\)
c. A
Worked Example
This sketch below shows part of the graph with equation \(y = pq^{x}\) where \(p\) and \(q\) are constants. The points with coordinates \(\left(0, 8\right)\), \(\left(1, 18\right)\) and \(\left(1.5, k\right)\) lie on the graph.
Work out the values of \(p\), \(q\) and \(k\).
Solution:
\(y=pq^{x} \rightarrow\) Using \(\left(0,8\right)\) \(8=pq^{0}\) \(p=8\)
Using \(\left(1,18\right)\) and \(p=8\) \(y=pq \rightarrow 18=8q^{1}\) \(q=\frac{18}{8}\) \(q=\frac{9}{4}\)
Using \(\left(1.5,k\right)\): \(k=8 \left(\frac{9}{4}^{1.5}\right)\) \(k=8\left(\frac{27}{8}\right) \rightarrow k=27\)
0, Chapter 4, Graphical Solutions of Equations,
Graphical Solutions of Equations
Lets explore solving more complex equations
In fact we can solve more complex equations in exactly the same way
We draw a graph of both equations and find the points where the graphs cross. This give us the solutions.
Ofcourse it is possible that the graphs may not cross in which case the orginal equations do not have any common solutions
This is best demonstated by some examples, which are shown in the next section
1, Chapter 4, Task 1, Worked Example 3,
Worked Example
Use the slider to explore worked examples.
Worked Example
How can we solve: \(x^{2}=x+\frac{1}{x}\) by a graphical method?
Solution
We can plot the graphs of: \(y=x^{2}\) and \(y=x+\frac{1}{x}\)
(Try plotting these graphs using the DESMOS software.) The solution will be where the graphs intersect (cross). This is because the \(y\) values will be equal at this point, and so \(x^{2}=x+\frac{1}{x}\)
The curves intersect at \(x \approx 1.45\) and so this is the approximate solution to the equation.
Worked Example
How would you find the solution to: \(2x - x^{2} = x^{3}\) ?
Solution
Plot the graphs of: \(y = 2x - x^{2}\) and \(y = x^{3}\)
(Try plotting these graphs using the DESMOS or other software.)
The solution will be where the graphs intersect (cross).
At this point the y values will be equal and so: \(2x - x^{2} = x^{3}\)
The curves intersect at \(x = 0, - 2, 1\) and so these are the solutions to the equation.
0, Chapter 5, Tangents to Curves,
Tangents to Curves
A tangent is a line that touches a curve at one point only, as shown here.
The gradient of the tangent gives the gradient of the curve at that point. A gradient of the curve gives the rate at which a quantity is changing.
For example, the gradient of a distance-time curve gives the rate of change of distance with respect to time, which actually gives us the velocity.
You may have seen tangents whilst working with circles.
Here is an example of two tangents to points on a circle.
\(BA\) and \(BC\) each touch the circle at just one point.
1, Chapter 5, Task 1, Example 1,
Example
Below we have drawn the graph of \(y = x^{3}\) for \(-2 \leq x \leq 2\).
On the graph we have drawn the tangents to the curve at \(x = -1\) and \(x = 1\).
We can find the gradients of the tangents at these points from the graph.
By drawing the triangles as shown under each tangent, we can see that the gradients of both tangents are \(3\).
0, Chapter 6, Summary,
Summary
Simultaneous (linear) equations
can be solved by finding the point of intersection of the two straight lines.
They take the form
\(ax+by=c\) \(dx+ey=f\) (\(a,b,c,d,e\) and \(f\) constants)
Quadratic functions
are parabolic in shape. For example,
There are of the form \(f\left(x\right) = ax^{2} + bx + c\) \(\left(a \neq 0\right)\) For example, \(f\left(x\right) = 2x^{2} - x+1\)
Quadratic equations
are of the form \(ax^{2} + bx + c = 0\) \(\left(a \neq 0\right)\) For example, \(2x^{2} - x + 1= 0\)
Cubic functions
They are of the form \(f\left(x\right) = ax^{3} + bx^{2} + cx + d\) (\(a \neq 0\))
For example, \(f\left(x\right) =x^{3}\), \(f\left(x\right) =2x^{3}-x+1\)
Cubic equations
are of the form \(ax^{3} +bx^{2}+cx+d=0\) \(\left(a \neq 0\right)\)
For example, \(2x^{3}-x+1=0\)
Reciprocal functions
are of the form
They are the form \(f\left(x\right) = \frac{k}{x}\)
For example, \(f\left(x\right)=\frac{1}{x}, f\left(x\right)=-\frac{2}{x}\)
Roots of a quadratic
equation there can be \(2\) or \(1\) or \(0\) roots for a quadratic equation. For example,
Roots of a cubic equation
there can be 3 or 2 or 1 root of a quadratic equation. For example,
Tangent
a line that touches a curve at one pointonly, as shown opposite.
Then consider the triangle \(OAC\). As \(OA\) and \(OC\) are both radii of the circle, triangle \(OAC\) is an isosceles triangle with \(b=c\). Recall also that \(a=70^{\circ}\)
Angles subtended at the circumference by a chord (on the same side of the chord) are equal; that is, in the diagram
\(a = b\)
Proof
Because angle \(𝑎\) is on the circumference, the angle at the centre is \(2𝑎\), according to Result 5. But if you consider the angle \(𝑏\), which is also at the circumference, the angle at the centre is \(2𝑏\), according to Result 5.
Thus \(2𝑎 = 2𝑏\); divide both sides to obtain \(𝑎 = 𝑏\), as required.
1, Chapter 3.2, Cyclic quadrilaterals,
Cyclic quadrilaterals
In cyclic quadrilaterals (quadrilaterals where all \(4\) vertices lie on a circle), opposite angles sum to \(180^{\circ}\):
\(a + c = 180^{\circ}\)
and
\(b + d = 180^{\circ}\)
Proof
Construct the diagonals \(AC\) and \(BD\), as shown.
Then label the angles subtended by \(AB\) as w; that is \(angle\;ADB = angle\;ACB\) (= w)
Similarly for the other chords, the angles being marked \(x, 𝑦\) and \(z\) as shown.
Now that we have labelled the angles, as shown on the right, we can prove the result
In triangle \(ABD\), the sum of the angles is \(180^{\circ}\) , so
In the diagram the chords \(AB\) and \(CD\) are parallel. Prove that the triangles \(ABE\) and \(DEC\) are isosceles.
Solution:
The angle \(a\) and \(angle BDC\) are angles in the same segment, so \(angle BDC = a\)
Since \(AB\) and \(DC\) are parallel, angle \(a\) and \(angle ACD\) are equal alternate angles
\(angle ACD = a = angle BDC\)
Hence in triangle \(DEC\), the base angles at \(C\) and \(D\) are equal, so the triangle is isosceles.
The angle at \(B\), \(angle ABD\), equals the angle at \(C\), \(angle ACD\), because they are angles in the same segment hence
\(angle ABD = angle ACD = a\)
Hence triangle \(ABE\) is isosceles, since the angles at \(A\) and \(B\) are equal.
0, Chapter 4, Circles and Tangents,
Circles and Tangents
Three further important results are now presented.
Result 8
If two tangents are drawn from a point \(T\) to a circle with centre \(O\), and \(P\) and \(R\) are the points of contact of the tangents with the circle, then, using symmetry,
\(PT = RT\)
Triangles \(TPO\) and \(TRO\) are congruent*
* Important note - Congruent shapes are identical; that is, corresponding angles are equal and corresponding sides are equal
1, Chapter 4.1, Alternate Segment,
Alternate Segment
Result 9
The angle between a tangent and a chord equals an angle at the circumference subtended by the same chord;
e.g. \(𝑎 = 𝑏\) in the diagram.
This is known as the alternate segment theorem and needs a proof, as it is not obviously true!
Proof
We need to show that \(angle\;RPT = angle\;PQR\)
Construct the diameter \(POS\), as shown.
We know that angle \(SRP = 90^{\circ}\) since \(PS\) is a diameter.
\(\begin{align*} AP \times PC&= BP \times PD\\ 2.5x&=1 \times 4\\ x&=1.6\;cm \end{align*}\)
0, Chapter 5, Summary,
Summary
A bearing gives the direction or position of something, or the direction of movement, relative to a fixed point. Bearings are of the form of angles, expressed in degrees as three-digit numbers; they are measured from north in a clockwise direction.
For example, \(060^{\circ}\;210^{\circ}\)
A tangent is a line that touches only one point on the circumference of a circle. A tangent is always perpendicular to the radius of the circle.
The point of tangency is the point where a tangent touches the circumference of a circle.
A chord is a line joining any two points on the circle. The perpendicular bisector is a second line that divides the first line in half and is at right angles to it. The perpendicular bisector of a chord always passes through the centre of the circle.
An angle subtended by a chord in a circle is shown in the diagram opposite. Angles subtended at the circumference on the same side of a circle by a chord are equal.
An angle subtended at the centre of a circle by an arc is twice the angle subtended at a point on the circumference (alternate segment theorem).
Interactive Exercises:
Angles, Circles and Tangents Interactive Exercises, https://www.cimt.org.uk/sif/geometry/g2/interactive.htm
Angles and Circles 1, https://www.cimt.org.uk/sif/geometry/g2/interactive/s1.html
Angles and Circles 2, https://www.cimt.org.uk/sif/geometry/g2/interactive/s2.html
Circles and Tangents, https://www.cimt.org.uk/sif/geometry/g2/interactive/s3.html