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## Introduction

### Discussion : Discuss these Research Topics

Research one of the following geometrical ideas and describe it to your group:

1. taxicab geometry,
2. spherical geometry,
3. fractals,
4. the Koch snowflake.

## Circle Geometry

### Terminology

The following is a recap of terms that are regularly used when referring to circles.

• arc: An arc is a part of the circumference of a circle.
• chord: A chord is defined as a straight line joining the ends of an arc.
• radius: The radius, rr, is the distance from the centre of the circle to any point on the circumference.
• diameter: The diameter is a special chord that passes through the centre of the circle. The diameter is the straight line from a point on the circumference to another point on the circumference, that passes through the centre of the circle.
• segment: A segment is the part of the circle that is cut off by a chord. A chord divides a circle into two segments.
• tangent: A tangent is a line that makes contact with a circle at one point on the circumference. (ABAB is a tangent to the circle at point PP).

### Axioms

An axiom is an established or accepted principle. For this section, the following are accepted as axioms.

1. The Theorem of Pythagoras, which states that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides. In ABCABC, this means that (AB)2+(BC)2=(AC)2(AB)2+(BC)2=(AC)2
2. A tangent is perpendicular to the radius, drawn at the point of contact with the circle.

### Theorems of the Geometry of Circles

A theorem is a general proposition that is not self-evident but is proved by reasoning (these proofs need not be learned for examination purposes).

Theorem 1 The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord.

Proof:

Consider a circle, with centre OO. Draw a chord ABAB and draw a perpendicular line from the centre of the circle to intersect the chord at point PP. The aim is to prove that APAP = BPBP

1. OAPOAP and OBPOBP are right-angled triangles.
2. OA=OBOA=OB as both of these are radii and OPOP is common to both triangles.

Apply the Theorem of Pythagoras to each triangle, to get:

O A 2 = O P 2 + A P 2 O B 2 = O P 2 + B P 2 O A 2 = O P 2 + A P 2 O B 2 = O P 2 + B P 2
(1)

However, OA=OBOA=OB. So,

O P 2 + A P 2 = O P 2 + B P 2 A P 2 = B P 2 and AP = B P O P 2 + A P 2 = O P 2 + B P 2 A P 2 = B P 2 and AP = B P
(2)

This means that OPOP bisects ABAB.

Theorem 2 The line drawn from the centre of a circle, that bisects a chord, is perpendicular to the chord.

Proof:

Consider a circle, with centre OO. Draw a chord ABAB and draw a line from the centre of the circle to bisect the chord at point PP. The aim is to prove that OPABOPAB In OAPOAP and OBPOBP,

1. AP=PBAP=PB (given)
3. OPOP is common to both triangles.

OAPOBPOAPOBP (SSS).

O P A ^ = O P B ^ O P A ^ + O P B ^ = 180 ( APB is a str. line ) O P A ^ = O P B ^ = 90 O P A B O P A ^ = O P B ^ O P A ^ + O P B ^ = 180 ( APB is a str. line ) O P A ^ = O P B ^ = 90 O P A B
(3)

Theorem 3 The perpendicular bisector of a chord passes through the centre of the circle.

Proof:

Consider a circle. Draw a chord ABAB. Draw a line PQPQ perpendicular to ABAB such that PQPQ bisects ABAB at point PP. Draw lines AQAQ and BQBQ. The aim is to prove that QQ is the centre of the circle, by showing that AQ=BQAQ=BQ. In OAPOAP and OBPOBP,

1. AP=PBAP=PB (given)
2. QPA=QPBQPA=QPB (QPABQPAB)
3. QPQP is common to both triangles.

QAPQBPQAPQBP (SAS). From this, QA=QBQA=QB. Since the centre of a circle is the only point inside a circle that has points on the circumference at an equal distance from it, QQ must be the centre of the circle.

#### Circles I

1. Find the value of xx:

Theorem 4 The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circumference of the circle.

Proof:

Consider a circle, with centre OO and with AA and BB on the circumference. Draw a chord ABAB. Draw radii OAOA and OBOB. Select any point PP on the circumference of the circle. Draw lines PAPA and PBPB. Draw POPO and extend to RR. The aim is to prove that AOB^=2·APB^AOB^=2·APB^.AOR^=PAO^+APO^AOR^=PAO^+APO^ (exterior angle = sum of interior opp. angles) But, PAO^=APO^PAO^=APO^ (AOPAOP is an isosceles ) AOR^=2APO^AOR^=2APO^ Similarly, BOR^=2BPO^BOR^=2BPO^. So,

A O B ^ = A O R ^ + B O R ^ = 2 A P O ^ + 2 B P O ^ = 2 ( A P O ^ + B P O ^ ) = 2 ( A P B ^ ) A O B ^ = A O R ^ + B O R ^ = 2 A P O ^ + 2 B P O ^ = 2 ( A P O ^ + B P O ^ ) = 2 ( A P B ^ )
(4)

#### Circles II

1. Find the angles (aa to ff) indicated in each diagram:

Theorem 5 The angles subtended by a chord at the circumference of a circle on the same side of the chord are equal.

Proof:

Consider a circle, with centre OO. Draw a chord ABAB. Select any points PP and QQ on the circumference of the circle, such that both PP and QQ are on the same side of the chord. Draw lines PAPA, PBPB, QAQA and QBQB. The aim is to prove that AQB^=APB^AQB^=APB^.

A O B ^ = 2 A Q B ^ at centre = twice at circumference and AOB ^ = 2 A P B ^ at centre = twice at circumference 2 A Q B ^ = 2 A P B ^ A Q B ^ = A P B ^ A O B ^ = 2 A Q B ^ at centre = twice at circumference and AOB ^ = 2 A P B ^ at centre = twice at circumference 2 A Q B ^ = 2 A P B ^ A Q B ^ = A P B ^
(5)

Theorem 6 (Converse of Theorem (Reference)) If a line segment subtends equal angles at two other points on the same side of the line, then these four points lie on a circle.

Proof:

Consider a line segment ABAB, that subtends equal angles at points PP and QQ on the same side of ABAB. The aim is to prove that points AA, BB, PP and QQ lie on the circumference of a circle. By contradiction. Assume that point PP does not lie on a circle drawn through points AA, BB and QQ. Let the circle cut APAP (or APAP extended) at point RR.

A Q B ^ = A R B ^ 's on same side of chord but AQB ^ = A P B ^ ( given ) A R B ^ = A P B ^ but this cannot be true since ARB ^ = A P B ^ + R B P ^ ( ext. of ) A Q B ^ = A R B ^ 's on same side of chord but AQB ^ = A P B ^ ( given ) A R B ^ = A P B ^ but this cannot be true since ARB ^ = A P B ^ + R B P ^ ( ext. of )
(6)

the assumption that the circle does not pass through PP, must be false, and AA, BB, PP and QQ lie on the circumference of a circle.

#### Circles III

1. Find the values of the unknown letters.

Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. The vertices of a cyclic quadrilateral are said to be concyclic.

Theorem 7 The opposite angles of a cyclic quadrilateral are supplementary.

Proof:

Consider a circle, with centre OO. Draw a cyclic quadrilateral ABPQABPQ. Draw AOAO and POPO. The aim is to prove that ABP^+AQP^=180ABP^+AQP^=180 and QAB^+QPB^=180QAB^+QPB^=180.

O 1 ^ = 2 A B P ^ 's at centre O 2 ^ = 2 A Q P ^ 's at centre But, O 1 ^ + O 2 ^ = 360 2 A B P ^ + 2 A Q P ^ = 360 A B P ^ + A Q P ^ = 180 Similarly , QAB ^ + QPB ^ = 180 O 1 ^ = 2 A B P ^ 's at centre O 2 ^ = 2 A Q P ^ 's at centre But, O 1 ^ + O 2 ^ = 360 2 A B P ^ + 2 A Q P ^ = 360 A B P ^ + A Q P ^ = 180 Similarly , QAB ^ + QPB ^ = 180
(7)

Theorem 8 (Converse of Theorem (Reference)) If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Proof:

Consider a quadrilateral ABPQABPQ, such that ABP^+AQP^=180ABP^+AQP^=180 and QAB^+QPB^=180QAB^+QPB^=180. The aim is to prove that points AA, BB, PP and QQ lie on the circumference of a circle. By contradiction. Assume that point PP does not lie on a circle drawn through points AA, BB and QQ. Let the circle cut APAP (or APAP extended) at point RR. Draw BRBR.

Q A B ^ + Q R B ^ = 180 opp. 's of cyclic quad but QAB ^ + QPB ^ = 180 ( given ) Q R B ^ = Q P B ^ but this cannot be true since QRB ^ = Q P B ^ + R B P ^ ( ext . of ) Q A B ^ + Q R B ^ = 180 opp. 's of cyclic quad but QAB ^ + QPB ^ = 180 ( given ) Q R B ^ = Q P B ^ but this cannot be true since QRB ^ = Q P B ^ + R B P ^ ( ext . of )
(8)

the assumption that the circle does not pass through PP, must be false, and AA, BB, PP and QQ lie on the circumference of a circle and ABPQABPQ is a cyclic quadrilateral.

##### Circles IV
1. Find the values of the unknown letters.

Theorem 9 Two tangents drawn to a circle from the same point outside the circle are equal in length.

Proof:

Consider a circle, with centre OO. Choose a point PP outside the circle. Draw two tangents to the circle from point PP, that meet the circle at AA and BB. Draw lines OAOA, OBOB and OPOP. The aim is to prove that AP=BPAP=BP. In OAPOAP and OBPOBP,

2. OAP=OPB=90OAP=OPB=90 (OAAPOAAP and OBBPOBBP)
3. OPOP is common to both triangles.

OAPOBPOAPOBP (right angle, hypotenuse, side) AP=BPAP=BP

##### Circles V
1. Find the value of the unknown lengths.

Theorem 10 The angle between a tangent and a chord, drawn at the point of contact of the chord, is equal to the angle which the chord subtends in the alternate segment.

Proof:

Consider a circle, with centre OO. Draw a chord ABAB and a tangent SRSR to the circle at point BB. Chord ABAB subtends angles at points PP and QQ on the minor and major arcs, respectively. Draw a diameter BTBT and join AA to TT. The aim is to prove that APB^=ABR^APB^=ABR^ and AQB^=ABS^AQB^=ABS^. First prove that AQB^=ABS^AQB^=ABS^ as this result is needed to prove that APB^=ABR^APB^=ABR^.

A B S ^ + A B T ^ = 90 ( TB SR ) B A T ^ = 90 ( 's at centre ) A B T ^ + A T B ^ = 90 ( sum of angles in BAT ) A B S ^ = A B T ^ However, AQB ^ = A T B ^ ( angles subtended by same chord AB ) A Q B ^ = A B S ^ S B Q ^ + Q B R ^ = 180 ( SBT is a str. line ) A P B ^ + A Q B ^ = 180 ( ABPQ is a cyclic quad ) S B Q ^ + Q B R ^ = A P B ^ + A Q B ^ AQB ^ = A B S ^ A P B ^ = A B R ^ A B S ^ + A B T ^ = 90 ( TB SR ) B A T ^ = 90 ( 's at centre ) A B T ^ + A T B ^ = 90 ( sum of angles in BAT ) A B S ^ = A B T ^ However, AQB ^ = A T B ^ ( angles subtended by same chord AB ) A Q B ^ = A B S ^ S B Q ^ + Q B R ^ = 180 ( SBT is a str. line ) A P B ^ + A Q B ^ = 180 ( ABPQ is a cyclic quad ) S B Q ^ + Q B R ^ = A P B ^ + A Q B ^ AQB ^ = A B S ^ A P B ^ = A B R ^
(9)
##### Circles VI
1. Find the values of the unknown letters.

Theorem 11 (Converse of (Reference)) If the angle formed between a line, that is drawn through the end point of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle.

Proof:

Consider a circle, with centre OO and chord ABAB. Let line SRSR pass through point BB. Chord ABAB subtends an angle at point QQ such that ABS^=AQB^ABS^=AQB^. The aim is to prove that SBRSBR is a tangent to the circle. By contradiction. Assume that SBRSBR is not a tangent to the circle and draw XBYXBY such that XBYXBY is a tangent to the circle.

A B X ^ = A Q B ^ ( tan - chord theorem ) However , ABS ^ = A Q B ^ ( given ) A B X ^ = A B S ^ But , ABX ^ = A B S ^ + X B S ^ can only be true if , XBS ^ = 0 A B X ^ = A Q B ^ ( tan - chord theorem ) However , ABS ^ = A Q B ^ ( given ) A B X ^ = A B S ^ But , ABX ^ = A B S ^ + X B S ^ can only be true if , XBS ^ = 0
(10)

If XBS^XBS^ is zero, then both XBYXBY and SBRSBR coincide and SBRSBR is a tangent to the circle.

##### Applying Theorem (Reference)
1. Show that Theorem(Reference) also applies to the following two cases:
##### Exercise 1: Circle Geometry I

BDBD is a tangent to the circle with centre OO. BOADBOAD. Prove that:
1. CFOECFOE is a cyclic quadrilateral
2. FB=BCFB=BC
3. COE///CBFCOE///CBF
5. OEBC=CDCOOEBC=CDCO

##### Exercise 2: Circle Geometry II

FDFD is drawn parallel to the tangent CBCB Prove that:
2. AFE|||CBDAFE|||CBD
3. FC.AGGH=DC.FEBDFC.AGGH=DC.FEBD

## Co-ordinate Geometry

### Equation of a Circle

We know that every point on the circumference of a circle is the same distance away from the centre of the circle. Consider a point (x1,y1)(x1,y1) on the circumference of a circle of radius rr with centre at (x0,y0)(x0,y0).

In Figure 24, OPQOPQ is a right-angled triangle. Therefore, from the Theorem of Pythagoras, we know that:

O P 2 = P Q 2 + O Q 2 O P 2 = P Q 2 + O Q 2
(29)

But,

P Q = y 1 - y 0 O Q = x 1 - x 0 O P = r r 2 = ( y 1 - y 0 ) 2 + ( x 1 - x 0 ) 2 P Q = y 1 - y 0 O Q = x 1 - x 0 O P = r r 2 = ( y 1 - y 0 ) 2 + ( x 1 - x 0 ) 2
(30)

But, this same relation holds for any point PP on the circumference. In fact, the relation holds for all points PP on the circumference. Therefore, we can write:

( x - x 0 ) 2 + ( y - y 0 ) 2 = r 2 ( x - x 0 ) 2 + ( y - y 0 ) 2 = r 2
(31)

for a circle with centre at (x0,y0)(x0,y0) and radius rr.

For example, the equation of a circle with centre (0,0)(0,0) and radius 4 is:

( y - y 0 ) 2 + ( x - x 0 ) 2 = r 2 ( y - 0 ) 2 + ( x - 0 ) 2 = 4 2 y 2 + x 2 = 16 ( y - y 0 ) 2 + ( x - x 0 ) 2 = r 2 ( y - 0 ) 2 + ( x - 0 ) 2 = 4 2 y 2 + x 2 = 16
(32)

Figure 25
Khan academy video on circles - 1

#### Exercise 3: Equation of a Circle I

Find the equation of a circle (centre OO) with a diameter between two points, PP at (-5,5)(-5,5) and QQ at (5,-5)(5,-5).

#### Exercise 4: Equation of a Circle II

Find the center and radius of the circle

x2-14x+y2+4y=-28x2-14x+y2+4y=-28.

### Equation of a Tangent to a Circle at a Point on the Circle

We are given that a tangent to a circle is drawn through a point PP with co-ordinates (x1,y1)(x1,y1). In this section, we find out how to determine the equation of that tangent.

We start by making a list of what we know:

1. We know that the equation of the circle with centre (x0,y0)(x0,y0) and radius rr is (x-x0)2+(y-y0)2=r2(x-x0)2+(y-y0)2=r2.
2. We know that a tangent is perpendicular to the radius, drawn at the point of contact with the circle.

As we have seen in earlier grades, there are two steps to determining the equation of a straight line:

1. Step 1. Calculate the gradient of the line, mm.
2. Step 2. Calculate the yy-intercept of the line, cc.

The same method is used to determine the equation of the tangent. First we need to find the gradient of the tangent. We do this by finding the gradient of the line that passes through the centre of the circle and point PP (line ff in (Reference)), because this line is a radius line and the tangent is perpendicular to it.

m f = y 1 - y 0 x 1 - x 0 m f = y 1 - y 0 x 1 - x 0
(40)

The tangent (line gg) is perpendicular to this line. Therefore,

m f × m g = - 1 m f × m g = - 1
(41)

So,

m g = - 1 m f m g = - 1 m f
(42)

Now, we know that the tangent passes through (x1,y1)(x1,y1) so the equation is given by:

y - y 1 = m ( x - x 1 ) y - y 1 = - 1 m f ( x - x 1 ) y - y 1 = - 1 y 1 - y 0 x 1 - x 0 ( x - x 1 ) y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 ) y - y 1 = m ( x - x 1 ) y - y 1 = - 1 m f ( x - x 1 ) y - y 1 = - 1 y 1 - y 0 x 1 - x 0 ( x - x 1 ) y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 )
(43)

For example, find the equation of the tangent to the circle at point (1,1)(1,1). The centre of the circle is at (0,0)(0,0). The equation of the circle is x2+y2=2x2+y2=2.

Use

y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 ) y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 )
(44)

with (x0,y0)=(0,0)(x0,y0)=(0,0) and (x1,y1)=(1,1)(x1,y1)=(1,1).

y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 ) y - 1 = - 1 - 0 1 - 0 ( x - 1 ) y - 1 = - 1 1 ( x - 1 ) y = - ( x - 1 ) + 1 y = - x + 1 + 1 y = - x + 2 y - y 1 = - x 1 - x 0 y 1 - y 0 ( x - x 1 ) y - 1 = - 1 - 0 1 - 0 ( x - 1 ) y - 1 = - 1 1 ( x - 1 ) y = - ( x - 1 ) + 1 y = - x + 1 + 1 y = - x + 2
(45)

#### Co-ordinate Geometry

1. Find the equation of the cicle:
1. with centre (0;5)(0;5) and radius 5
2. with centre (2;0)(2;0) and radius 4
3. with centre (5;7)(5;7) and radius 18
4. with centre (-2;0)(-2;0) and radius 6
5. with centre (-5;-3)(-5;-3) and radius 33
1. Find the equation of the circle with centre (2;1)(2;1) which passes through (4;1)(4;1).
2. Where does it cut the line y=x+1y=x+1?
1. Find the equation of the circle with center (-3;-2)(-3;-2) which passes through (1;-4)(1;-4).
2. Find the equation of the circle with center (3;1)(3;1) which passes through (2;5)(2;5).
3. Find the point where these two circles cut each other.
2. Find the center and radius of the following circles:
1. (x-9)2+(y-6)2=36(x-9)2+(y-6)2=36
2. (x-2)2+(y-9)2=1(x-2)2+(y-9)2=1
3. (x+5)2+(y+7)2=12(x+5)2+(y+7)2=12
4. (x+4)2+(y+4)2=23(x+4)2+(y+4)2=23
5. 3(x-2)2+3(y+3)2=123(x-2)2+3(y+3)2=12
6. x2-3x+9=y2+5y+25=17x2-3x+9=y2+5y+25=17
3. Find the x-x- and y-y- intercepts of the following graphs and draw a sketch to illustrate your answer:
1. (x+7)2+(y-2)2=8(x+7)2+(y-2)2=8
2. x2+(y-6)2=100x2+(y-6)2=100
3. (x+4)2+y2=16(x+4)2+y2=16
4. (x-5)2+(y+1)2=25(x-5)2+(y+1)2=25
4. Find the center and radius of the following circles:
1. x2+6x+y2-12y=-20x2+6x+y2-12y=-20
2. x2+4x+y2-8y=0x2+4x+y2-8y=0
3. x2+y2+8y=7x2+y2+8y=7
4. x2-6x+y2=16x2-6x+y2=16
5. x2-5x+y2+3y=-34x2-5x+y2+3y=-34
6. x2-6nx+y2+10ny=9n2x2-6nx+y2+10ny=9n2
5. Find the equations to the tangent to the circle:
1. x2+y2=17x2+y2=17 at the point (1;4)(1;4)
2. x2+y2=25x2+y2=25 at the point (3;4)(3;4)
3. (x+1)2+(y-2)2=25(x+1)2+(y-2)2=25 at the point (3;5)(3;5)
4. (x-2)2+(y-1)2=13(x-2)2+(y-1)2=13 at the point (5;3)(5;3)

## Transformations

### Rotation of a Point about an angle θθ

First we will find a formula for the co-ordinates of PP after a rotation of θθ.

We need to know something about polar co-ordinates and compound angles before we start.

#### Polar co-ordinates

 Notice that : sinα=yrsinα=yr∴y=rsinα∴y=rsinα and cosα=xrcosα=xr∴x=rcosα∴x=rcosα
so PP can be expressed in two ways:
1. P(x;y)P(x;y) rectangular co-ordinates
2. or: P(rcosα;rsinα)P(rcosα;rsinα) polar co-ordinates.

#### Compound angles

(See derivation of formulae in Ch. 12)

sin ( α + β ) = sin α cos β + sin β cos α cos ( α + β ) = cos α cos β - sin α sin β sin ( α + β ) = sin α cos β + sin β cos α cos ( α + β ) = cos α cos β - sin α sin β
(46)

#### Now consider P'P' after a rotation of θθ

P(x;y)=P(rcosα;rsinα)P'(rcos(α+θ);rsin(α+θ))P(x;y)=P(rcosα;rsinα)P'(rcos(α+θ);rsin(α+θ))
(47)
Expand the co-ordinates of P'P'
x-co-ordinate ofP'=rcos(α+θ)=rcosαcosθ-sinαsinθ=rcosαcosθ-rsinαsinθ=xcosθ-ysinθx-co-ordinate ofP'=rcos(α+θ)=rcosαcosθ-sinαsinθ=rcosαcosθ-rsinαsinθ=xcosθ-ysinθ
(48)
y-co-ordinate of P'=rsin(α+θ)=rsinαcosθ+sinθcosα=rsinαcosθ+rcosαsinθ=ycosθ+xsinθy-co-ordinate of P'=rsin(α+θ)=rsinαcosθ+sinθcosα=rsinαcosθ+rcosαsinθ=ycosθ+xsinθ
(49)

which gives the formula P'=(xcosθ-ysinθ;ycosθ+xsinθ)P'=(xcosθ-ysinθ;ycosθ+xsinθ).

So to find the co-ordinates of P(1;3)P(1;3) after a rotation of 45, we arrive at:

P ' = ( x cos θ - y sin θ ) ; ( y cos θ + x sin θ ) = ( 1 cos 45 - 3 sin 45 ) ; ( 3 cos 45 + 1 sin 45 ) = 1 2 - 3 2 ; 3 2 + 1 2 = 1 - 3 2 ; 3 + 1 2 P ' = ( x cos θ - y sin θ ) ; ( y cos θ + x sin θ ) = ( 1 cos 45 - 3 sin 45 ) ; ( 3 cos 45 + 1 sin 45 ) = 1 2 - 3 2 ; 3 2 + 1 2 = 1 - 3 2 ; 3 + 1 2
(50)
##### Rotations

Any line OPOP is drawn (not necessarily in the first quadrant), making an angle of θθ degrees with the xx-axis. Using the co-ordinates of PP and the angle αα, calculate the co-ordinates of P'P', if the line OPOP is rotated about the origin through αα degrees.

 PP αα 1. (2, 6) 60∘∘ 2. (4, 2) 30∘∘ 3. (5, -1) 45∘∘ 4. (-3, 2) 120∘∘ 5. (-4, -1) 225∘∘ 6. (2, 5) -150∘∘

### Characteristics of Transformations

Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and size, and that enlargement preserves shape but not size.

#### Geometric Transformations:

Draw a large 15××15 grid and plot ABCABC with A(2;6)A(2;6), B(5;6)B(5;6) and C(5;1)C(5;1). Fill in the lines y=xy=x and y=-xy=-x. Complete the table below , by drawing the images of ABCABC under the given transformations. The first one has been done for you.

 Description Transformation (translation, reflection, Co-ordinates Lengths Angles rotation, enlargement) A ' ( 2 ; - 6 ) A ' ( 2 ; - 6 ) A ' B ' = 3 A ' B ' = 3 B ^ ' = 90 ∘ B ^ ' = 90 ∘ ( x ; y ) → ( x ; - y ) ( x ; y ) → ( x ; - y ) reflection about the xx-axis B ' ( 5 ; - 6 ) B ' ( 5 ; - 6 ) B ' C ' = 4 B ' C ' = 4 tan A ^ = 4 / 3 tan A ^ = 4 / 3 C ' ( 5 ; - 2 ) C ' ( 5 ; - 2 ) A ' C ' = 5 A ' C ' = 5 ∴ A ^ = 53 ∘ , C ^ = 37 ∘ ∴ A ^ = 53 ∘ , C ^ = 37 ∘ ( x ; y ) → ( x + 1 ; y - 2 ) ( x ; y ) → ( x + 1 ; y - 2 ) ( x ; y ) → ( - x ; y ) ( x ; y ) → ( - x ; y ) ( x ; y ) → ( - y ; x ) ( x ; y ) → ( - y ; x ) ( x ; y ) → ( - x ; - y ) ( x ; y ) → ( - x ; - y ) ( x ; y ) → ( 2 x ; 2 y ) ( x ; y ) → ( 2 x ; 2 y ) ( x ; y ) → ( y ; x ) ( x ; y ) → ( y ; x ) ( x ; y ) → ( y ; x + 1 ) ( x ; y ) → ( y ; x + 1 )

A transformation that leaves lengths and angles unchanged is called a rigid transformation. Which of the above transformations are rigid?

## Exercises

1. ΔABCΔABC undergoes several transformations forming ΔA'B'C'ΔA'B'C'. Describe the relationship between the angles and sides of ΔABCΔABC and ΔA'B'C'ΔA'B'C' (e.g., they are twice as large, the same, etc.)
 Transformation Sides Angles Area Reflect Reduce by a scale factor of 3 Rotate by 90∘∘ Translate 4 units right Enlarge by a scale factor of 2
2. ΔDEFΔDEF has E^=30E^=30, DE=4 cm DE=4 cm , EF=5 cm EF=5 cm . ΔDEFΔDEF is enlarged by a scale factor of 6 to form ΔD'E'F'ΔD'E'F'.
1. Solve ΔDEFΔDEF
2. Hence, solve ΔD'E'F'ΔD'E'F'
3. ΔXYZΔXYZ has an area of 6 cm 26 cm 2. Find the area of ΔX'Y'Z'ΔX'Y'Z' if the points have been transformed as follows:
1. (x,y)(x+2;y+3)(x,y)(x+2;y+3)
2. (x,y)(y;x)(x,y)(y;x)
3. (x,y)(4x;y)(x,y)(4x;y)
4. (x,y)(3x;y+2)(x,y)(3x;y+2)
5. (x,y)(-x;-y)(x,y)(-x;-y)
6. (x,y)(x;-y+3)(x,y)(x;-y+3)
7. (x,y)(4x;4y)(x,y)(4x;4y)
8. (x,y)(-3x;4y)(x,y)(-3x;4y)

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