Exercise Transformation of Geometry
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