Anggi Ika Satri

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GEOMETRY TRANSFORMATION

1

HALF ROTATION

Definition:

A half round on a point A is a equivalent SA that defined for every point on field as

follows:

1) If P ≠ A then SP(A) = A’ so that P midpoint segment AA’

2) SP(P) = P

Half rotation is a transformation

Proof:

Wll be proven 𝑆𝐴 bijective.

To proving 𝑆𝐴 bijective then must be proved first 𝑆𝐴 surjective and injective.

(1) Will be proved 𝑆𝐴 surjective

To showing 𝑆𝐴 surjective, will be shown ∃𝑃′ ∈ 𝑉 ∋ 𝑆𝐴(𝑃) = 𝑃′

Take any 𝑃′ ∈ 𝑉

𝑃′ ∈ 𝑉 ∋ 𝑃′ = 𝑆𝐴(𝑃)

𝑖𝑓 𝑃 = 𝐴, 𝑡ℎ𝑒𝑛 𝑆𝐴(𝐴) = 𝐴′ = 𝐴

So, ∀ 𝑃′ ∈ 𝑉 ∃ 𝑃′ = 𝑃 = 𝑆𝐴(𝑃)

If 𝑃 ≠ 𝐴 then A be axis line segment′ , Its means 𝑆𝐴(𝑃) = 𝑃′

So, 𝑆𝐴 surjective

(2) Will be proved 𝑆𝐴 injective

Suppose 𝐵1 ≠ 𝐵2

Case I

𝐵1 = 𝐵2 = 𝐴

To 𝐵1 = 𝐴 then 𝑆𝐴(𝐵1) = 𝐵1 = 𝐵1′………………..1*)

To 𝐵2 = 𝐴 then 𝑆𝐴(𝐵2) = 𝐵2 = 𝐵2′…………………2*)

From 1*) and 2*) then obtained 𝑆𝐴(𝐵1) ≠ 𝑆𝐴(𝐵2)

A A’ =

Sp(A)

180°

P

fulcru

m

Anggi Ika Satri

1201125107

GEOMETRY TRANSFORMATION

2

Case II

𝐵1 ≠ 𝐵2 ≠ 𝐴

Take any 𝐵1, 𝐵2 ∈ 𝑉 𝑤𝑖𝑡ℎ 𝐵1 ≠ 𝐵2

𝐵1 ≠ 𝐴, 𝐵2 ≠ 𝐴, 𝐵2, 𝐵2, 𝐴 𝑐𝑜𝑙𝑖𝑛𝑒𝑎𝑟

So that 𝑆𝐴(𝐵1) = 𝐵1′ and 𝑆𝐴(𝐵2) = 𝐵2′

Suppose that 𝑆𝐴(𝐵1) = 𝑆𝐴(𝐵2)

Because 𝑆𝐴(𝐵1) = 𝑆𝐴(𝐵2)

Then 𝐵1′ = 𝑆𝐴(𝐵1) = 𝑆𝐴(𝐵2) = 𝐵2′

So that obtained 𝐵1′ = 𝐵2′ and 𝐵1 = 𝐵2

Based on theorem, “Through two point scan only be made one line”

Its contradiction with the statement that 𝐵1 ≠ 𝐵2

Supposition 𝐵1 ≠ 𝐵2 𝑡ℎ𝑒𝑛 𝑆𝐴(𝐵1) = 𝑆𝐴(𝐵2) must be canceled.

So, 𝑆𝐴(𝐵1) ≠ 𝑆𝐴(𝐵2) then,𝑆𝐴 injective

From (1) and (2) then obtained 𝑆𝐴 surjective and 𝑆𝐴 injective

Because 𝑆𝐴 surjective and 𝑆𝐴 injective, then 𝑆𝐴 bijective

Because 𝑆𝐴 bijective, then 𝑆𝐴 is a transformation.

So, proven that a half round is a transformation.

Example :

1. Known three points A, B, P were colinear and different. Draw SASB(P)!

Answer:

SASB(P) = SA(SB(P))

Draw SB(P) first

B

P’ = SB(P)

P

A

Anggi Ika Satri

1201125107

GEOMETRY TRANSFORMATION

3

Transformationed P’ to point A

2. Known coordinates of the point A and B is (1,1) and (2,3). If SB(A) = A’,

determine coordinates of the point A’!

Answer :

Point B as a fulcrum

Point B is the middle point of the line segment

AA’

To determine the coordinates A’ using

the pattern looking for a midpoint.

(x,y) = (𝑥1+𝑥2

2,

𝑦1+𝑦2

2)

(2,3) = (1+𝑥2

2,

1+𝑦2

2)

x = 𝑥1+𝑥2

2 y =

𝑦1+𝑦2

2

2 = 1+𝑥2

2 3 =

1+𝑥2

2

4 = 1 + 𝑥2 6 = 1 + 𝑦2

𝑥2 = 4 − 1 𝑦2 = 6 − 1

𝑥2 = 3 𝑦2 = 5

So, the coordinate point A’ is (3,5).

B

P’ = SB(P)

P

A

P’’ = SASB(P)

1

2

3

4

2

5

Y

3 4

A

’

B

A

1 X

Anggi Ika Satri

1201125107

GEOMETRY TRANSFORMATION

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Theorem 1.1

Suppose 𝑨 a point, 𝒈 and 𝒉 two perpendicular lines that intersect in 𝑨. Then 𝑺𝑨 =

𝑴𝒈𝑴𝒉.

Proof :

Known 𝐴 a point, 𝑔 and ℎ two perpendicular lines that intersect in 𝐴.

a) Case I : 𝑃 ≠ 𝐴

Because 𝑔 ⊥ ℎ then can be formed an axis system orthogonal with 𝑔 as X axis and

ℎ as Y axis. 𝐴 as central point.

Take point 𝑃 ∈ 𝑉

Look at picture 7.2

Show that to every 𝑃 obtain 𝑆𝐴(𝑃) = 𝑀𝑔𝑀ℎ(𝑃)

Suppose 𝑃(𝑥, 𝑦) ≠ 𝐴 and 𝑆𝐴(𝑃) = 𝑃′′(𝑥1, 𝑦1)

Because 𝑆𝐴(𝑃) = 𝑃′′ then 𝐴 midpoint 𝑃𝑃′ so that

(0,0) = (𝑥1 + 𝑥

2,𝑦1 + 𝑦

2)

Obtained 𝑥1 + 𝑥 = 0 ⟺ 𝑥1 = −𝑥 and 𝑦1 + 𝑦 = 0 ⟺ 𝑦1 = −𝑦

It means 𝑆𝐴(𝑃) = (−𝑥, −𝑦)………………………………………………(1)

Reflection composition

𝑀𝑔𝑀ℎ(𝑃) = 𝑀𝑔[𝑀ℎ(𝑃)]

= 𝑀𝑔(−𝑥, 𝑦)

= (−𝑥, −𝑦)

It means 𝑀𝑔𝑀ℎ(𝑃) = (−𝑥, −𝑦)……………………………………………(2)

𝑃′′(−𝑥, −𝑦)

ℎ

𝑔 𝐴

P(x,y)

𝑋

𝑃′(−𝑥, 𝑦)

Anggi Ika Satri

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GEOMETRY TRANSFORMATION

5

From equation (1) and (2) obtained 𝑆𝐴(𝑃) = 𝑀𝑔𝑀ℎ(𝑃).

So , 𝑆𝐴 = 𝑀𝑔𝑀ℎ

b) Case II : 𝑃 = 𝐴

Based on definition, 𝑆𝐴(𝐴) = 𝐴……………………………………………(1*)

𝑀𝑔𝑀ℎ(𝐴) = 𝑀𝑔(𝐴) = 𝐴……………………………………………….(2*)

From equation (1*) and (2*) obtained 𝑆𝐴(𝐴) = 𝑀𝑔𝑀ℎ(𝐴).

So , 𝑆𝐴 = 𝑀𝑔𝑀ℎ.

Theorem 1.2

If 𝒈 and 𝒉 two lines that perpendicular then 𝑴𝒈𝑴𝒉 = 𝑴𝒉𝑴𝒈

Proof :

a) Case I : 𝑃 ≠ 𝐴

Because 𝑃 ≠ 𝐴, then 𝑀𝑔𝑀ℎ(𝑃) = 𝑆𝐴(𝑃).

𝑀ℎ𝑀𝑔(𝑃) = 𝑀ℎ (𝑀𝑔(𝑃)) = 𝑀ℎ((𝑥, −𝑦)) = (−𝐷, −𝑦) = 𝑆𝐴(𝑃).

Obtained 𝑀𝑔𝑀ℎ(𝑃) = 𝑆𝐴(𝑃) = 𝑀ℎ𝑀𝑔(𝑃)

So, 𝑀𝑔𝑀ℎ = 𝑀ℎ𝑀𝑔

b) Case II : 𝑃 = 𝐴

Because 𝑃 = 𝐴, then 𝑀𝑔𝑀ℎ(𝐴) = 𝑀𝑔(𝐴) = 𝐴

𝑀ℎ𝑀𝑔(𝐴) = 𝑀ℎ(𝐴) = 𝐴

So that obtained 𝑀𝑔𝑀ℎ(𝐴) = 𝑀ℎ𝑀𝑔(𝐴).

So, 𝑀𝑔𝑀ℎ = 𝑀ℎ𝑀𝑔.

Theorem 1.3

If 𝑺𝑨 half round, then 𝑺−𝟏𝑨 = 𝑺𝑨.

Proof :

Suppose 𝑔 and ℎ two lines that perpendicular then 𝑀𝑔𝑀ℎ = 𝑆𝐴 with 𝐴 intersection

between 𝑔 and ℎ.

(𝑀𝑔𝑀ℎ)−1 = 𝑀−1ℎ𝑀−1

𝑔 = 𝑆−1𝐴.

Because 𝑀−1ℎ = 𝑀ℎ and 𝑀−1

𝑔 = 𝑀𝑔 then 𝑀ℎ𝑀𝑔 = 𝑆−1𝐴.

Because 𝑔 ⊥ ℎ, then based theorem 7.2, 𝑀𝑔𝑀ℎ = 𝑀ℎ𝑀𝑔.

Anggi Ika Satri

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GEOMETRY TRANSFORMATION

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So that obtained 𝑆−1𝐴 = 𝑀ℎ𝑀𝑔 = 𝑀𝑔𝑀ℎ = 𝑆𝐴.

So, 𝑆−1𝐴 = 𝑆𝐴.

Theorem 1.4

If 𝑨 = (𝒂, 𝒃) and 𝑷 = (𝒙, 𝒚) then 𝑺𝑨(𝑷) = (𝟐𝒂 − 𝒙, 𝟐𝒃 − 𝒚).

Proof

a) Case I : 𝑃 ≠ 𝐴

Suppose 𝑃" = (𝑥1, 𝑦1) and 𝑆𝐴(𝑃) = 𝑃" then 𝐴 midpoint 𝑃𝑃" so that obtained

(𝑎, 𝑏) = ((𝑥1 + 𝑥

2) , (

𝑦1 + 𝑦

2))

Then 𝑥1+𝑥

2= 𝑎 and

𝑦1+𝑦

2= 𝑏 so that obtained

𝑥1+𝑥

2= 𝑎 ⟺ 𝑥1 + 𝑥 = 2𝑎 ⟺ 𝑥1 = 2𝑎 − 𝑥……………………………..(1*)

𝑦1+𝑦

2= 𝑏 ⟺ 𝑦1 + 𝑦 = 2𝑏 ⟺ 𝑦1 = 2𝑏 − 𝑦………………………………(2*)

From equation (1*) and (2*) then (𝑥1, 𝑦1) = ( 2𝑎 − 𝑥), (2𝑏 − 𝑦)

Because 𝑆𝐴(𝑃) = 𝑃", then 𝑆𝐴(𝑃) = (𝑥1, 𝑦1) = ( 2𝑎 − 𝑥), (2𝑏 − 𝑦)

So, 𝑆𝐴(𝑃) = (2𝑎 − 𝑥, 2𝑏 − 𝑦).

b) Case II : 𝑃 = 𝐴

Because 𝑃 = 𝐴, then(𝑥, 𝑦) = (𝑎, 𝑏) it means 𝑎 = 𝑥 and 𝑏 = 𝑦.

𝑆𝐴(𝐴) = 𝐴 = (𝑎, 𝑏)

(𝑎, 𝑏) = ((2𝑎 − 𝑎), (2𝑏 − 𝑏))

= ((2𝑎 − 𝑥), (2𝑏 − 𝑦))

So, 𝑆𝐴(𝑃) = (2𝑎 − 𝑥, 2𝑏 − 𝑦).

Task:

Anggi Ika Satri

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GEOMETRY TRANSFORMATION

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1. Known : line 𝑔 and point 𝐴, 𝐴 ∉ 𝑔

Asked :

a) Draw line 𝑔1 = 𝑆𝐴(𝑔) and why 𝑔 is a line?

b) Prove that 𝑔′//𝑔.

Answer :

a. 𝑔′ = 𝑆𝐴(𝑔)

Because 𝑔 is a line, then 𝑆𝐴(𝑔) is a line too. (isometry).

b. 𝑔′ ∕∕ 𝑔

proof :

𝑃 ∈ 𝑔, 𝑄 ∈ 𝑔

Because 𝑃 ∈ 𝑔 then A midpoint 𝑃𝑃′ with 𝑃′ = 𝑆𝐴(𝑃)

Because 𝑄 ∈ 𝑔 then A midpoint 𝑄𝑄′ with 𝑄′ = 𝑆𝐴(𝑄)

Look at ∆𝐴𝑃𝑄′ 𝑎𝑛𝑑 ∆𝐴𝑄𝑃′

To prove that 𝑔′ ∕∕ 𝑔 then must be shown that ∆𝐴𝑃𝑄′ 𝑎𝑛𝑑 ∆𝐴𝑄𝑃′ are

congruent.

𝑚(< 𝑃𝐴𝑄′) = 𝑚(< 𝑄𝐴𝑃′) (opposite angel)

𝑃𝐴 = 𝐴𝑃′ (because A midpoint 𝑃𝑃′ )

𝑄′𝐴 = 𝐴𝑄 (because A midpoint 𝑄𝑄′ )

According to the definition of congruence (Side Angle Side)

So ∆𝐴𝑃𝑄′ ≅ ∆𝐴𝑄𝑃′

Because ∆𝐴𝑃𝑄′ ≅ ∆𝐴𝑄𝑃′ then 𝑃𝑄′ = 𝑄𝑃′

Because 𝑃𝑄′ = 𝑄𝑃′ then 𝑔′ ∕∕ 𝑔

2. Known : points A, B, C colinear

Draw :

P Q

𝑆𝐴(𝑃) = 𝑃′

𝑔′ = 𝑆𝐴(𝑔)

A

𝑆𝐴(𝑄) = 𝑄′

𝑔

Anggi Ika Satri

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GEOMETRY TRANSFORMATION

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a) Line 𝑔 and ℎ such as 𝑀𝑔(𝐵) = 𝐵 and 𝑆𝐴 = 𝑀𝑔𝑀ℎ

b) Line 𝑘 and 𝑚 such as 𝑀−1𝑘(𝐶) = 𝐶 and 𝑆𝐴 = 𝑀𝑘𝑀𝑚

Drawing :

a) 𝑀𝑔(𝐵) = 𝐵 and 𝑆𝐴 = 𝑀𝑔𝑀ℎ

b) 𝑀−1𝑘(𝐶) = 𝐶 and 𝑆𝐴 = 𝑀𝑘𝑀𝑚

3. Known: A = (2,3)

Asked:

a. SA( C ) if C = (2,3)

b. SA( D ) if D = (-2,7)

c. SA( E ) if E= (4,-1)

d. SA( P ) if P = (x,y)

Answer:

a. C = (2,3)

SA( C ) = (2.2 - 2, 2.3 - 3)

= (2,3)

b. D = (-2,7)

SA( D ) = (2.2-(-2), 2.3-7)

= (6,-1)

c. E = (4,-1)

SA( E ) = (2.2-4, 2.3-(-1))

= (0,7)

d. P = (x,y)

SA( P ) = (2.2-x, 2.3-y)

= (4-x, 6-y)

4. Known : B = (1, -3)

𝑔

𝐴 ℎ

𝐵

Anggi Ika Satri

1201125107

GEOMETRY TRANSFORMATION

9

Determine :

a. SB(D) if D (-3, 4)

b. E if SB(E) = (-2, 5)

c. SB(P) if P= (x, y)

Answer:

a. D (-3, 4)

SB(D) = (2.1-(-3), 2.(-3)-4)

= (5, -10)

b. SB(E) = (-2, 5)

Suppose E = (x, y)

Then, 2.1- x = -2 2.(-3)-y = 5

⇔2 – x = -2 ⇔-6-y = 5

⇔x = 4 ⇔y = -11

so, E= (4, -11)

c. P = (x, y)

SB(P) = (2.1-x, 2.(-3)-y)

= (2-x, -6 - y)

5. Known : D = (0, -3) and B = (2, 6)

a. SB(B) = (2.2-2, 2.6-6)

= (2, 6)

SDSB(B) = SD(2,6)

= (2.0-2, 2.(-3) – 6)

= (-2, -12)

b. K = (1, -4)

SB(K) = (2.2-1, 2.6-(-4)

= (3, 16)

SDSB(K) = SD(3,16)

=(2.0- 3, 2.(-3)-16)

= (-3, -22)

c. SD(K) = (2.0-1, 2.(-3)-(-4))

= (-1, -2)

Anggi Ika Satri

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GEOMETRY TRANSFORMATION

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SBSD(K) = SB(-1, -2)

= (2.2-(-1),2.6-(-2))

= (5, 14)

d. According to the theorem 1.3

if SA is a half round then S-1A = SA

then, SD-1 (K) = SD(K) = (-1,-2)

and, SB-1(K) = SB(K)

so, (SDSB)-1 (K) = SB-1SD

-1 (K)

= SB-1(-1, -2)

= SB(-1, -2)

= (2.2 - (-1), 2.6 - (-2))

= (5, 14)

e. P = (x, y)

SB(P) = (2.2 – x, 2.6 – y)

= (4 – x, 12 – y)

SDSB(P) = SD(4 – x, 12 – y)

= (2.0 – (4 – x), 2.(-3) – (12 – y))

= ( - 4 + x, - 6 – 12 + y)

= (x - 4, y - 18)