Transformations Answers

7/31/2019 Transformations Answers

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Home Study

Modules KS4Foundation Level

Transformations

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Home Study Modules Foundation Level: TRANSFORMATIONS

Mathematics4U www. gcsemathematics4u.co.uk

Answers

1. a) Translation: The vector describing the translation.

This is all that is necessary since the upper number in the vector gives thedistance travelled in the x-direction and the lower number the distancetravelled in the y-direction. The sign of these numbers shows whether thetranslation is left/right or up/down.Dont

write vectors like fractions with the upper number underlined!

b) Reflection: The mirror line.

The mirror line is all that is required. This may be described in several wayssuch as The line AB, the x-axis, the line x = 0 or the line y = x.

Dont forget a mathematical mirror is two way, unlike a normal physicalmirror. In other words, in a mathematical mirror you can reflect from eitherside.

c) Rotation: The centre of rotation, the angle of rotation and the direction of rotation(clockwise or anticlockwise).

If the angle of rotation is 180o

(a half turn), we do not have to say whether itis clockwise or anticlockwise since they give the same result. For all otherangles, the direction must

be specified because a rotation of 90

oclockwise

is not the same as a rotation of 90oanticlockwise, for example.

It is also very important to specify the centre of rotation as this determinesthe position of the image (the rotated shape).

d) Enlargement: The centre of enlargement and the scale factor.

Again, the centre of enlargement is important since that determines the finalposition of the image.The scale factor will be greater than 1 if the shape has become larger,between 0 and 1 if the shape has shrunk and negative if the image is on theother side of the centre of enlargement to the original shape.

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2.

a) A to D b) A to K

c) A to B d) C to H

e) K to B f) B to K

C

K

J

I

H

F

G

D

13)(

B A

65( ) E

65( )

13( )

As you will see from the first two examples which are illustrated above,the upper number in the vector gives the distance moved to the left orright (positive being to the right and negative to the left) and the lowernumber gives the distance moved up or down (positive being up andnegative being down).The easiest way to see how far the shape has moved is to track just onepoint as we have done here. The other points will obviously move by the

same amount. This technique of tracking one point is also very useful forthe other transformations as we shall soon see.

139( )

40( )

25( )

25( )

K to B and B to K illustrate the fact that moving in the opposite directionchanges the signs of the numbers in the vectors. (Obvious really, butworth emphasising!)

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g) C to K h) H to I0

8( )41( )

i) J to D j) G to F39( )

22( )3.

Z

A

B

63(

43(

c) The vector from A to B is 10

6

)

)

)

(

As I am sure you will appreciate by now, it is much easier to discussideas if they have names and this applies to transformations as muchas anything else. So make sure you know the two terms object andimage.

Object is the shape you start with before the transformation (anytransformation) and the image is the new shape after thetransformation.

We shall be using these terms in this module.

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4.M1 M1

M2

M2 M1

M1

M2M2

The first couple of examples are easy to reflect and some people arenaturally good at reflecting even the strangest shapes. But what if youfind reflecting more complicated shapes difficult? All you need to do isreflect one point at a time until you build up the complete shape.

Lets look at the shape with a circle as an example.

M1

M2

Take the point A. We need to reflect it across the mirror line, sothat a ray joining A and its reflection, A, crosses the mirror lineat right angles. All we need to do is count the squares (or in thiscase, the diagonals of squares) from A to the mirror.

A

This is one and a half diagonals, so count one and a halfdiagonals on the other side of the mirror line to get the point A.

Do this for as many points as you need to build up the reflectedshape.A

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5.

x = 0 y = 1

Shaded shape Dotted shapeis image Is object

(reflection). (original shape)

2

1

01

2

3

4

3

4

32 1 1 2 3 4

4

2

1

01

2

3

4

3

4

32 1 1 2 3 4

4

y

y

xx

x = 0 is, of course, the y-axis, so the first shape must bereflected in this.When a shape crosses the mirror line, just treat it as two shapesseparated by the mirror and reflect one across the mirror oneway and the other across the mirror the other way.This demonstrates what we mean when we say a mathematical

mirror is two-way.

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6. a)x = 1

b) The transformation that maps shape C directly onto shape B is a translation

c) The transformation that maps shape C directly onto shape A is a rotation of 180o,centre (4, 1). {More on how to find the centre of rotation later}

2

1

01

2

3

4

4

3 2 1 1 2 3 44

5

6

3

5

6

5 66 5

y

x

y = x

A

B C

D E

F

x = 4

X

y =

1

The line y =x is, of course, the line going through the

points (1,1), (2,

2), (3,

3) etc. If you feel you need to

revise this, study the module on Graphs.

By the way, this flag shape is a great shape to use if you want toexperiment with transformations because it only has four points andit is very easy to see whether it has been reflected, rotated ortranslated.If you try your experiments with a square or equilateral triangleinstead, how are you going to know if the shape has been rotatedor reflected?

10 0( )

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7.6

y

5

The reflection of the point (2, 2) in the line y = x + 3 is (1, 5)

8.

The reflection of the point (4, 6) in the line x + y = 2 is (4, 2)

2

1

01234

4

3 2 1 1 2 3 44

3

y = x + 3

6 5 5 6

x

56

Draw the line y = x + 3 (this can be done by plotting just two or three pointssuch as (0, 3), (3, 6) and (

6,

3) or by using y = mx + c.

Plot the point (2, 2) and draw a ray towards the mirror line as we have done

before. Count the diagonals (one and a half) and plot the image the samedistance on the other side of the mirror line.

2

1

01234

4

3 2 1 1 2 3 44

5

6y

Plot the line y =-x 2 and thepoint (

4,

6). Count the

diagonals of the squares fromthe point to the mirror line andcontinue the same distance onthe other side of the mirror.

y =-x 2 3

56

5 66 5

This brings you to (4, 2).

x

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9.

6y

5

A1 is the point (4, 1), B1 is (1, 2) and C1 is (1, 5)

10.

Here we think of the shape as being made up of three points(the vertices) and we simply reflect these in the mirror line.

2

1

0123

4

4

3 2 1 1 2 3 44

3

56

5 66 5

x

y = xA

B

Centre of Rotation

Remember this is find the centre of rotation when theangle of rotation is 180

oand is quite straightforward.

The technique is to join corresponding points between theobject and the image. It is really only necessary to jointwo pairs to find their meeting point, but we have drawfour pairs to show that it doesnt matter which ones youchoose, they always meet in the same place.

Image

Object

A1 B1 C

C1

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P

Page 10

11. Q

W R

O

SV

U T

The important thing to remember here is that of corresponding letters.

a) Pentagon STUVW to pentagon SRQPW is a reflection in the line SW.

b) Kite QRSO to kite WPQO is an anticlockwise rotation of 90o, centre O.

When we say, for example, pentagon STUVW to pentagon SRQPW, we

mean point S goes to point S, point T goes to point R, point U goes topoint Q etc. It is important to know this rule. If you did not know this, youmight say there are two ways of going from STUVW to SRQPW areflection in SW or a rotation of 180

oabout O.

c) Triangle OVU to triangle OUT is an anticlockwise rotation of 45o about O.

d) Octagon PQRSTUVW to Octagon UVWPQRST is an anticlockwise rotation of 135o

about O. {You may have said a clockwise rotation of 225o about O. Given a choice, we normally give therotation with the smaller angle}

Dont forget that with a rotation you must give the angle, directionand centre of rotation (and dont forget to say it is a rotation!)

Now, heres a case where you can get easy marks (or lose some easy marks if youare not careful!). When the examination paper is set, the examiners have to allocatethe correct number of marks to that paper, so they make a pot of coffee, get the

chocky biscuits out and sit down round a table, distributing the marks throughout allthe parts of all the questions.Do you think they will get to the end of the paper and have allocated every singleavailable mark? Not a chance!There will either not be enough marks, in which case they will need to trim theallocation a little or (and this is where you can score) there will be marks left over sothey have to give a few more out. One easy place to add extra marks is fortransformations one for the word rotation, one for the angle, one for the directionand one for the centre, if youre lucky, so make sure you mention all these things you never know your luck!


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12.

Image

Object

P

a) The transformation is an enlargement, scale factor 2, centre P.

It is very easy to find the centre of enlargement. Simply joincorresponding points (dotted lines) and backtrack them untilthey meet. The meeting point is the centre of enlargement.

b) If the object and image were swapped, the transformation would be anenlargement, scale factor , centre P

We always, of course, use the word enlargement for this type oftransformation, even when the shape becomes smaller. We simply use afraction to indicate the reduction in size.The reason we do this is that we may want to multiply the scale factorstogether. Say a shape is enlarged with S.F 4 and the image enlargedwith S.F , multiplying these gives 4 = 2. In other words the finalimage is twice as big as the original shape.We can string as many of these together as we wish to find the overall

enlargement, but you dont need to worry about this at Foundation Level.

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Here is a very important point.When you enlarge a shape using rays as we did above, you find the distance fromthe centre of enlargement to each point on the object in turn, multiply by the scalefactor and use the new distance to find the position of the new point on the image.Sometimes you do this on a grid in which case you can count squares to help you

do this. At other times there is no grid and you need to measure with a ruler.

C

Whichever way you do it, you must measure both the original distance and theenlarged distance from the centre of enlargement.

Lets look at the enlargement from point C above as an example.

X

Y

First you measure the distance CX and multiply this by the scale factor (in thiscase, 5).Then you find the point Y by measuring the enlarged distance from C, notfrom X.If you measure from X, the scale factor of the enlargement will be 1 greater thanit should be (in this case, 6).You then repeat this for all points on the shape, every time measuring from C,as in the diagram above. The diagram below shows you how NOT to do it.

C

X

Y

WRONG

RIGHT5CX

5CX

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14. Find the area using the formula base height with XY as the base(i.e. the triangle is upside down).

And dont forget you must draw a diagram for a question like this!(And dont start a sentence with the word And.)

a) Area of triangle OXY = base height

= (8 3) 6

= 5 6 = 15 units2

b) Area of triangle OX1Y1 = base height

= (4 1.5) 3 {Thinking of X1Y1 as the base now}

= 2.5 3 = 3.75 units2

O

P X (3, 6) Y (8, 6)6

5

4

Y1X13

2

1

101 3 52 4 96 7 8

You could work this out by finding the area of triangle OPY and subtracting the areaof triangle OPX and you would, of course, end up with the same answer.

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Page 15

15.12y

The transformation that is equivalent to both the given transformations isa rotation of 180o about the point (7, 6).

9

8

6

5

4

3

11

10

2

1

7

0 1 2 3 4 5 6 7 8 9 10 11 120

x

L

Centre of Rotation

y = 6

x = 7

Another great module completed Well done!

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