Exercise Of Translation

By : Aditya Prihandhika

Task 1 1. Given points A, B, and C were colinear

a. Draw GAB(A) and GAB(B)

b. Draw GAB(C)

c. Draw the lines g and h with gA and GAB=MhMg

d. Draw lines g and h so gC then GAB=MhMg

A B

C

A B=GAB(A) A’=GAB(B)

A B

C C’=GAB(C)

h g

A B

C

GAB(A) =B

MhMg(A)=B } GAB=MhMg

A B

g h

A

g k

B

m

A

m’

B

2. Diketahui : Titik-titik A, B, dan garis g sehingga g AB.

a. Lukislah garis h sehingga MhMg= GAB

b. Lukislah garis k sehingga MgMk= GAB

c. line m so m’ = GAB(m)

h g

A B

GAB(A)= B

MhMg = Mh(Mg(A))=Mh(B)=B } MhMg=GAB

GAB(A)= B

MgMk = Mg(Mk(A))=Mg(A)=B } MgMk=GAB

GAB (m) = B

m’ = B

d. The point of C so GBA(C) = B

GAB(C) = B

3. Known : Lines g//h and the A point not on the lines..

a. Draw the point B so MhMg= GAB

Jelas GAB(A)= MhMg(A)= Mh(A’)=B

b. Draw the point C so MgMh= GAC

It’s clear GAC(A)= MgMh(A)= Mg(A’)=C

m’ = GAB(m)

g h

A Mg(A)=A’ B= Mh(A’)

g h

C= Mg(A’ ) A Mh(A)=A’

A B C

A B

P

C

D

P

P’

P”

P’

P”

P

4. Known the points A, B, C, D and line g

Draw !

a) GCD GAB (P)

GAB (P) = P’ where PP’ = AB

GCD (P) = P” where P’P” = CD

b) GCD GBA (P)

GBA (P) = P’ Where PP’ = BA

GCD (PP) = P” where P’P” = CD

h’ = GDC (h)

h

g = GABGDC (h)

P

P’

P”

P”’ = G3

AB (P)

c) Line h so GAB GCD (h) = g

d) G3AB (P)

5. Is the phrases below are true or false:

a. If GAB=MgMh then GAB=MhMg..(False)

Proof:

Known GAB=MgMh.

MgMh ≠ MhMg (the product of two reflection is not commutative)

Then GAB ≠ MhMg.

So If GAB=MgMh then GAB ≠ MhMg

b. Every translation is an involution. (False)

Proof:

Assume : GAB=MhMg.

So we get (GAB)-1= (MhMg)-1

= Mg-1Mh-1

= MgMh

≠ GAB.

So GAB is not involution.

c. GABGAB= GCD with (True)

Proof:

Take an arbitrary point P.

If GABGAB(P)=P4 and GCD(P)=P5, maka akan dibuktithen will be prove

P4=P5.

Because GAB(P)=P2 then

GAB(P2)=P4 then and

GABGAB(P)=P4 then

So , akibatnya .54 PP

So GABGAB(P)= GCD(P).

Because P is an arbitary point, then GABGAB= GCD.

d. If M is midpoint , then (True)

e. If g’ = (g), then g’//g (True)

6. If A(2,3) and B(4,-7) determine the equation of a line g and h so that

Answer :

We know g and h and the distance between g and h

The line equation

So

Assume A ∈ g then the line equation of g

The distance between g and h , A ∈ g so h through C point, so that C

midpoint AB

)

)

So C (-1,5)

The line equation h AB and through C(-1,5)

So g : y =

And h : y =

7. Known: the points A(-1,3), B(-5,-1), and C(2,4).

a. Determine ).(' CGC AB

Answer:

Because )(' CGC AB then

So that 242 22 xx and .044 22 yy

So ).0,2()(' CGC AB

b. Find the equation of lines g and h so gC and MhMg= GAB.

Answer:

It;s clear .14

4

15

31

12

12

xx

yymAB

In order that MhMg= GAB then it must be g//h and ., ABhABg

so we get

Because g//h then 1 hg mm .

suppose line h through the point D then

So we get

.1

11

1

g

g

gAB

m

m

mm

2

212

212

2

2

2

2

412

412

2

2

2

2

12

2

12412

12

2

12

2

412

21

)4()4()4()2(

)31()15()4()2(

])()[()()(

yx

yx

yyxxyyxx

ABCD

ABCD

222

2

2

2

222

2

2

2

2

12

2

12

2

12

2

12

22

)4()4()4()2(

)31()15()4()2(

)()()()(

'

'

yx

yx

yyxxyyxx

ABCC

ABCC

So 042 221

2 xx and .244 221

2 yy

The point D (0,2).

The equation of a line g through point C(2,4) with 1gm is

6

24

)2(14

)( 11

xy

xy

xy

xxmyy

And the equation of a line h through point D(0,2) with 1hm is

.2

2

)0(12

)( 11

xy

xy

xy

xxmyy

8. Known: A(2,1), B(5,-3)

a.

Suppose then

so that:

and

So C’(7,-2)

b. with

Suppose

then so that

and

So

10. Known : The points A=(2,-1), B=(3,4), and g={(x,y)\y+2x=4}.

a. Determine GAB(P) if P(x,y).

Answer :

It’s clear that BAGAB )(

).4,3()1,2(

)4,3()1,2(

ba

GAB

So that 132 aa and .541 bb

So ).5,1(),()( yxyxGPG ABAB

b. Detrmine D so that GAB(D)=(1,3).

Answer:

Suppose the point ),( 11 yxD then

).3,1()5,1(

)3,1(),(

)3,1()(

11

11

yx

yxG

DG

AB

AB

So that 011 11 xx and .235 11 yy

So the point D(0,-2).

c. Determine an equation for line h, so that ).(gGh AB

Answer:

.32

4225

4)1(25

)42()(

yx

xy

xy

xyGgGh ABAB

Task 2 1. Given directed line segmets points C and P.

a) Determine GABSC(P)

Answer:

GABSC(P)=GAB[SC(P)]

=GAB(P’) with C is midpoint

=P” with

b) Determine SCGAB(P)

Answer:

SCGAB(P)=SC[GAB(P)]

=SC(P’) with

=P” with C is midpoint

c) Find all points X so that GABSC(X)=X

Answer:

Based on theorem 10. 6 we get GABSC=SD

Take an arbitary X

GABSC(X)=SD(X)

We get SD(X)=X, it’s mean X=D’

Take an arbitary point E with and point D is midpoint it’s

mean

We get GABSC(X) = GABSC(D)

= GAB[SC(X)]

=GAB(D’), with C is midpoint D’,

it’s mean =D with =X

so the point X is a midpoint where

2. Known points A, B, C colinear

a) Determine D so that SDSC=GAB

Answer:

Based on theorem 10. 5 point C and point D is linear, where 2

b) Determine E so that SASBSC=SE

Answer:

Based on the result of the theorem 10. 6, we get point E linear with

point where,

c) Determine F so that GABSC=SF

Answer:

Based on the result of the theorem 10. 6, we get point F is a midpoint

it’s mean where

3. a) For every point P=(x,y), S defined by S(P)=(x+a,y+b). Determine S-1(P).

Answer:

Based on theorem 7. 3 S-1(P)=S(P)

=(x+a,y+b)

b) if G1 and G2 are translations, investigate whether the G1G2=G2G1.

Answer:

Take an arbitary point P

Suppose G1=GAB and G2=GCD

G1G2(P)=G1[G2(P)]

=G1(P’) with

=P” with

So , ………(1)

G2G1(P)=G2[G1(P)]

=G2(P’) with

=P” with

So , ………(2)

Based on (1) and (2) applies GABGCD=GCDGAB

G1G2=G2G1

4. G is translation defined as follows:

If P=(x,y) then G(P)=(x+2,y+3). Known C=(1,-7).

Determine the coordinate D so that SDSC=G

Answer:

SDSC(P)=G(P)

SD[(2-x,-14-y)]=(x+2,y+3)

Suppose D(a,b)

[2a-(2-x),2b-(-14-y)]=(x+2,y+3)

2a-(2-x)=x+2

2a=x+2+2-x

2a=4

a=2

2b-(-14-y)=y+3

2b=y+3-14-y

2b=-11

b=-5,5

So that D(2,-5,5)

5. If A=(1,0), B=(2,5) and C=(-3,8) are points known. Determine

coordinates point D so that GCD=SBSA.

Answer:

Suppose = then E=(1+[x+3],0+[y-8])

=(4+x,y-8)

If B is midpoint , so

x=-1

y=18

the coordinate D=(-1,18)

6. Suppose A=(a1,a2) and B=(b1,b2). Dengan using the coordinates, proof:

a) SBSA is a translation

Answer:

Take an arbitary P(x,y)

SBSA(P)=SB[SA(P)]

=SB(2a1-x,2a2-y)

=(2b1-2a1+x,2b2-2a2+y)

=[x+2(b1-a1),y+2(b2-a2)]

b) If P is a point and P’=SASB(P), then =

Answer:

Take an arbitary P(x,y)

From the result a) we get P’=[ x+2(b1-a1),y+2(b2-a2)]

=( b1–a1,b2-a2)

=[ x+2(b1-a1)-x,y+2(b2-a2)-y]

=[ 2(b1-a1),2(b2-a2)]

=2( b1–a1,b2-a2)

=2

So =

7. Known A=(2,1) and B=(-3,5)

a) If P=(x,y) determine SASB(P)

Answer:

SASB(P)=SA(2.-3-x,2.5-y)

=SA(-6-x,10-y)

=2.2-(-6-x),2.1-(10-y)

=(10+x,-8+y)

So SASB(P) =(10+x,-8+y)

b) L={(x,y)| x2+y2=4}. Find the equation of the set L’=SASB(L).

Answer:

L= x2+y2=4 it’s mean a circle with center (0,0) with radius =2

SASB(L)=SA[2.(-3)-0,2.5-0]

=SA(-6,10)

=[2.2-(-6),2.1-10]

=(10,-8)

So L’={(x,y)|(x-10)2+(y+8)2=4}