Exercise Transformation of Geometry

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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

Transcript of 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’

Line equation of h’:

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:

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

Y

X B(-3,0) C’

C(0,1)

B’(0,3)

D

g: y=-x

h: 3y=x+3

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

Mg(h) = h’, then h’ is a line.

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’

Line equation of h’:

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:

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

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

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(- 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