Exercise Transformation of Geometry

Exercise Transformation of Geometry

By : Aditya Prihandhika

1. Known line and line as can be seen in the picture. By using a compass and ruler,

draw the line of with is a reflection on the line.

Answer:

2. Known lines of and points as can be seen in the picture below. is an

isometry with and . If , draw .

Answer:

g’

h

g

o

o

Known: and ,

Because then Because and T is an

isometry, then

.

So, to draw t’ make t’ line

through B that is perpendicular

with u.

A

B

u

s

t

3. Known line, circle with the center is and as in the picture.

a) Draw

b) What is the relation between and ?

c) Draw

Answer:

a)

b) Look at and

Because and

We get

Based on the theorem,(angle, side, angle) then .

c)

B

C

A

t

B’

A’

C’ P

O

D’

D

4. Known line.

a) Draw so that (it means: by , and the result

of reflection of t is coincide).

b) Draw a triangle that is coincide with the co-domain by .

c) Draw a quadrangle that is coincide with the co-domain by

Answer:

a)

b)

c)

To draw that is coincide with , then

is must a isosceles triangle with as a symmetry axis, is

coincide with , so that .

So,

l=l’

O=O’

t

To draw l circle that is coincide with (l), then the center of

l circle is must be on the reflection axis of , so that

t

To draw a quadrangle that is coincide with the co-domain by

, then it is must the reflection of t is coincide with the

quadrilateral axis of symmetry.

A=A’

B=C’ C=B’ t

O

5. Known lines of and .

Write an equation line of .

Answer:

Because is a reflection on , then it is an isometry.

So, based on the theorem ”an isometry map a line become a line”, and ,

then is a line.

Point is an intersection between line and axis.

Point is an intersection between line dan line.

So, and .

Because then

So, will through point , and will through .

Coordinate of point

–

substitute to the line equation , obtained:

–

So,

g

Y

X A(1,0)

B(0, 2

1)

C

g’

h:x = -1

A’(-3,0) D

Coordinate of

Point is an intersection of with axis.

–

Becase it is an isometry, then

So,

Suppose, point

The abscissa of point is

Obtained

So,

So, through points and

Equation of line: 10

1

12

1

12

1

y

xx

xx

yy

yy

)1(3

)1(

x

1

1

y=

2

1

x

1 y = 2

1x

y = 12

1

2

1x

y = 2

3

2

1x

032 yx

So,

6. Known line and .

Write the line equation of .

Answer:

Because is a reflection on , then it is an isometry.

So, based on the theorem ”an isometry map a line become a line”, and ,

then is a line.

Point is an intersection between line and axis.

Point C is an intersection between line and .

So and .

Because then

So will through point , and will through

Coordinate of point

substitute to the line equation , obtained:

– 3

2

So 3

2

X B(

3

4 ,0)

A’(0,0)

A(0,4)

C D

Y g

h

Coordinate of

Point is an intersection of with sumbu .

– = 4

Because its is an isometry, then

So,

Suppose point

Ordinate of point is 4

Obtained and

So,

So, through point 3

2 and

Line equation of : 20

2

12

1

12

1

y

xx

xx

yy

yy

)3

2(0

)3

2(

x

2

2

y=

3

23

2x

2 y = -2 )12

3( x

y = -3x -2 +2

y = -3x

03 yx

So, 03 yx

7. Known lines of , and

.

Write the equation of lines:

a). b).

c). d).

Answer:

a).

Because Mg is a reflection on g then it is an isometry.

Based on the theorem,“ An isometry map a line become a”, and Mg(h) = h’, then h’ is

a line.

Point O(0,0) is a intersection between lines of g and h.

So, Og and Oh.

Because Og then Mg(O) = O

So h’ will through point O(0,0)

Take any point on h, suppose A(1,1), then h’ also will through A’ = Mg(A).

A(x,y) gM

A’(x,-y) , g = {(x,y) | y = 0}

So, A(1,1) gM

A’(1,-1)

So, h’ line through point O(0,0) and A’(1,-1)

Line equation of h’:

01

0

12

1

12

1

y

xx

xx

yy

yy

01

0

xxy

1

g: y=0 h’: y=-x

h: y=x

Y

X

A(1,1)

A’(1,-1)

So, h’ = {(x,y) | y = -x}.

b).

Beacuse Mh is a reflection on h, then it is an isometry.

Based on the theore, “ An isometry map a line become a line”, and Mh(g) = g’, then g’

is a line.

point O(0,0) is an intersection between lines of g and h.

So, Og and Oh.

Because Oh then Mh(O) = O

So g’ will through point O(0,0)

Take any point on g, suppose C(1,0), then g’ will also through C’ = Mh(g).

C(x,y) gM

C’(y,x)

So, C(1,0) gM

C’(0,1)

So, g’ line through point O(0,0) and C’(0,1)

Line equation of g’:

01

0

12

1

12

1

y

xx

xx

yy

yy

00

0

x 0 x

So, g’ = {(x,y) | x = 0}.

Y

X = g:y=0

h: y=x

C’(0,1)

C(1,0)

g’: x=0

c).

Because Mg is a reflection on g, then it is an isometry.

Based on the theore, “An isometry map a line become a line”, and Mg(k) = k’, then k’

is a line.

Point P(2,0) is an intersection between lines of g and k.

So, Pg and Pk.

Because Pg then Mg(P) = P, then k’ will through point P(2,0)

Take any point on k, suppose B(2,2

1), then k’ will also through B’ = Mg(B).

B(x,y) gM

B’(x’,y’) = B’(x,-y)

So, B(2,2

1)

gM

B’(2,-2

1)

So, k’ line through point P(2,0) and B’(2,-2

1)

So, k’ = k = {(x,y) | x = 2}.

d).

O g:y=0

B(2,

)

P(2,0)

k : x=2 Y

X

B’(2,-

)

k'

’

k’: y=2

Y

X B(2,0)

A(2,2) B’(0,2)

k: x=2 h: y=x

Because Mh is a reflection on h, then it is an isometry.

Based on the theore, “An isometry map a line become a line”, and Mh(k) = k’, then k’

is a line.

Point A(2,2) is an intersection between lines of h and k.

So, Ah and Ak.

Because Ah then Mh(A) = A

So k’ will through point A(2,2)

Take any point on k, suppose B(2,0), because h: y = x then Mh(B) = (0,2) = B’.

So k’ through A and B’

Line equation of k’:

22

2

12

1

12

1

y

xx

xx

yy

yy

02

0

x2 y

So, g’ = {(x,y) | y=2}.

8. If g = {(x,y) | y = x} and h = {(x,y) |y = 3 – 2x}, determine the line equation of Mg(h).

Answer:

Because Mg is a reflection on h, then it is an isometry.

Menurut teorema, Based on the theore, “An isometry map a line become a line”, and

Mg(h) = h’, then h’ is a line.

Point A is an intersection between lines of g and h.

Y

X

g: y=x

C’(0,2

3)

B(0,3)

C(2

3,0)

B’(3,0) A

So, Ag and Ah.

Because Ag then Mg(A) = A

So h’ will through point A

Take point B(0,3) and C(2

3,0) because g: y = x then Mg(B) = B’ and Mg(C)=C’.

So h’ through B’ and C’


02

3

0

12

1

12

1

y

xx

xx

yy

yy

30

3

x

2

9

2

33 xy

936 xy

0963 yx

So, h’ = {(x,y) | 0963 yx }.

9. If g = {(x,y) | y = -x} and h = {(x,y) |3y = x + 3}, investigate, is A(-2,-4) located on

the line of h’ = Mg(h).

Answer:


Y

X B(-3,0) C’

C(0,1)

B’(0,3)

D

g: y=-x

h: 3y=x+3



Point D is an intersection between lines of g and h

So, Dg and Dh.

Because Dg then Mg(D) = D

So h’ will through point D

Take point B(-3,0) and C(0,1) because g: y = - x then Mg(B) = B’ and Mg(C)=C’.

So h’ through B’ and C’


03

0

12

1

12

1

y

xx

xx

yy

yy

)1(0

)1(

x3)1( xy 33 xy

So, h’ = {(x,y) | 33 xy }

Will be investigated, is A(-2,-4) located on line of h’ = Mg(h)

Substitute A(-2,-4) on h’: y = 3x + 3

Then h’ : -4 = 3(-2) + 3

-4 = -3 ( the wrong statement)

Obtained A(-2,-4) doesn’t meet the equation of h’: y = 3x + 3, it means A(-2,-4) isn’t

located on line of h’ = Mg(h)

10. Known circle of l= 432:,22 yxyx

T is an isometry which is map point A(2,3) on A’(1,-7). Determine the equation of set

T(l). Is the co-domain l also a circle?

Answer:

l = 432:,22 yxyx

A’=T(A) with A(2,3) and A’(1,-7).

L is acircle with center (2,3) and radius=2.

Because A is a center of circle l, then A’=(1,-7) is a center of circle l’=T(l).

So that T(l)=l’= 471:,22 yxyx

Co-domain l is l’ is a circle because isometry T preserve the magnitude of the angle, it

is 360o.

11. Known five lines g, g’, h, h’, and k so that g’=Mk(g), and h’=Mk(h). If g’//h’ prove

that g//h.

Answer:

It has g’//h’.

And g//h

If g isn’t parallel with h, then based on the theorem that isometry Mk preserve the

alignment of two lines, obtained g’ is not parallel with h.

Whereas it has g’//h’, then the assumption must be canceled.

It means, g//h.

12. Known lines g, h, and h’ so that h’=Mg(h). Are the expressions below true?

a) If h’//h, then h//g.

b) If h’=h then h=g.

c) If h’ h={A}, then A g.

Answer:

a) True

b) True

c) True

g h’ h

h

h’

g

h

A

h' g

13. Prove that this property: If then Mh(g)=g. Is this means that if P g then

Mh(P)=P?

Answer:

It has

And Mh(g)=g.

Because Mh preserve the magnitude of the angle between the two angles g and h at

90°, then the angle between g’ and h is also 90o. So that g’ a straightener of g.

So, g’ is coincide with g so that Mh(g)=g.

Case I. P g, P h then Mh(P)=P.

Case II. P g, Ph. Because Mh is an isometry then OP=OP’. Obtained P=P’. So,

Mh(P) P.

15. If g = {(x,y) | y = 2x + 3} and h = {(x,y) |y = 2x + 1}, determine the line equation of

h’ = Mg(h).

Answer:


P

h

g P’

P

h

g P’

Y

X

D(0,3)

C

,1)

A

B(0,1)

h: y=2x+1

g: y=2x+3 h’

E

F



Point A(- 2

1,0 ) is an intersection between line and axis.

Point B(0,1) is an intersection between line and axis.

Point C(- 2

3,0 ) is an intersection between line and axis.

Point D(0,3) is an intersection between line and axis.

So that AC =1, BD =1

Obtained h’ intersect X axis at point F(-2

5,0)

h’ intersect Y axis at point E(0,5)

Line equation of h’ through F and E so that the equation of g’:

05

0

12

1

12

1

y

xx

xx

yy

yy

)2

5(0

)2

5(

x

)2

5(5

2

5 xy 25105 xy

052 xy

So, h’ = {(x,y) | 052 xy }

16. A transformation T is determined by T(P)=(x+1,2y) for all P(x,y).

a) If A(0,3) and B(1,-1) determine A’=T(A) and B’=T(B). And also determine the

equation AB and ''BA .

b) If C(c,d) AB investigate, is C’=T(C) AB

c) If D’(e,f) AB investigate, is D AB with D’=T(D).

d) Based on the theorem, stated that if the transformation T an isometry then map a

line is a line. Is the opposite true?

Answer:

T(P)=(x+1,2y) P(x,y)

a) A(0,3), B(1,-1)

A’=T(A)=(0+1,2x3)=(1,6)

B’=T(B)=(1+1,2x(-1))=(2,-2)

034

441

1

1

4

1

10

1

)1(3

)1(

12

1

12

1

xy

xy

xy

xy

xx

xx

yy

yyAB

0148

1682

1

2

8

2

21

2

)2(6

)2(

''12

1

12

1

xy

xy

xy

xy

xx

xx

yy

yyBA

b) C(c,d) AB

It will be investigated C’=T(C) ''BA

Because A’=T(A), B’=T(B), then ''BA is co-domain of AB .

So that if C AB then C’=T(C) ''BA

c) D’(e,f) AB is investigated, is D AB with D’=T(D).

Because ''BA is co-domain AB then if D’ AB definitely D AB .

d) It has h’ is a line.

It will be shown h is a line with h’=T(h).

If h is not a line then h’=T(h) is not a line.

Whereas h’ is a line.

Then the assumption must be canceled. It means, a line h.

So, if h’ is a line then h is also a line with h’=T(H).

18. How many reflections line with the following properties:

a) An isosceles triangle reflected on itself?

b) A rectangular reflected on itself?

c) An regular quadrilateral reflected on itself?

Answer:

a) 1 reflection

b) 2 reflections

c) 4 reflections

Tasks: (Page 45)

1. On the picture 4.10, there are three collinear points, they are and are

isometries with while

. Which groups including the T and S?

Answer:

P

Q

R

P’

R’

Q’

R’’

Q’

’

P’’

P

Q

R

P’

R’

R’’

Q’

’

P’’

So:

an opponent isometry and is a direct isometry.

2. Isometry map at ; at and C at Z. If is an opponent isometry, determine the

point of .

3. An isometry map at , at and at . If S is a direct isometry, determine U.

Answer:

4. Known a point and two transformations and which are defined as follows:

, . If , and . is a midpoint of line

segment AP while is a midpoint of ''PP . Which groups including the

transformation and ?

Answer:

Illustration:

From the picture obatained is an opponent isometry because APPA "

And is a direct isometry because APPA '

D Z

W

F

E

P A P’ P”

B

A

C

Y

X

Z

5. Determine the coordinates of point on the axis so that BPXAPO .

Known that and .

Answer:

and .

Suppose

β α

6-x x P

B(6,5)

A(0,3)

In order to BPXAPO then,

4

9

8

18

5318

6

53

tantan

x

xx

xx

So, in other to BPXAPO then

6

5

Y

X

6. A ray emanating from point an be directed to point on the mirror

which is drawn as a line of . There is a line of

. The reflected ray intersect line on the point . Determine the

coordinates of point .

Answer:

7. Known lines of and and points and .

Known if P’=Mg(P), P”=Mh(P’), R’=Mg(R), and R”=Mh(R).

a. Draw P’ and R”

b. Compare the distance PR and P”R”

Y

X

h: x=-1

g; y=x

A(6,4)

P(2,2)

A’

Z

Coordinate of A’(4,6)

The ray equation of

0442

164122

)4(4)6(2

42

4

62

6

12

1

12

1

xy

xy

xy

xy

xx

xx

yy

yy

If then

So, y = -4

So, the coordinate is (-1,-4)

Answer:

a.

b. Because PR = P’R

’ (isometry preserving the distance)

Then distance of with h = distance of P’’ with h

Distance of R’ with h = distance of R

’’ with h

So distance of P’R

’ = distance of P

’’R

’’

Because distance of PR = distance of P’R

’ and distance of P

’R

’ = distance of P

’’R

’’,

then distance of PR = distance of P’’R

’’.

8. Known that T and S are equivalent so that for all points P occur T(P) = P’ and S(P’) =

P’’.

W is a function which is defined for all P as W(P) = P’’.

Is W a transformation?

Answer:

W sis a function so that points P P”S W(P) = P”.

Shown W id surjective

Think any point A(x,y)

It is clear A(x,y) T

A’(x’,y’) S

A”(x”,y”), or

A(x,y) W

A”(x”,y”)

P

R

h

g

R’

P”

P’

R”

So, point A A”S W(P) = P”.

So, W is surjective.

Shown W is injective

Think any point B(x,y) and C with B≠C.

It is clear B W

B” = W(B)

C W

C” = W(C) , with W(B) ≠ W(C)

So, point B and C with B ≠ C occur W(B) ≠ W(C).

So, W is injective.

So, because W is surjective and injective then W is a transformation.

9. R is a transformation ehich is defined for all points P as R =

a) Investigate, is R an isometry?

b) Fi R is an isometry, is R direct or opponent isometry?

Answer:

R is a transformation

P , R =

a) Is R an isometry

Take P1 , P2

R = =

R = =

It will be shown =

=√

√

= √( )

= √

√

√

Oibtained =

So, R preserve the distance, so that R is an isometry.

b) Is R direct or opponent isometry

Take any points P, Q, S sre collinear.

Suppose P , Q , and S .

Then , and with R = , R =

, and R =

10. Known a g line and point A, A’ and B so that Mg(A) = A’ and line

AB // g. By using a

ruler, determine point B’ = Mg(B)

Answer:

g

B’

B A

A’

X

Y

Q

P

S

Exercise Transformation of Geometry

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