# Point Reflection - Camden Central School

## Point Reflection

### Point Symmetry

• An image that looks the same when turned 180º is said to have point symmetry. • (the point of symmetry is the center of the shape) yes No Yes No List the upper case letters of the alphabet that have point symmetry.

### Point Reflection

– reflecting an image about a point Notation: R origin (A)→A’ In 3-9, the reflection of triangle ABC through point B is triangle DBE.

3. What is the image of A under the point reflection?

D 4 .

R B

(

C

) E  ?

5 .

R B

(

D

) A  ?

7. What is the preimage of C under the point reflection?

E 6 .

R B

(

B

) B  ?

C A B 8 .

R B

( 

CAB

) 

EDB

 ?

9 .

R B

(

CA

)  ?

ED

D E

In 10-18: Sketch the figure. Tell whether the figure has point symmetry. If it does locate the point.

12.

16.

13.

17.

14.

18.

15.

In 19-22: Sketch the figure. Carefully sketch the given figure under a reflection in point A.

B D C C A A’ C’ B’ A A’ B C’ B’ D’

### Point Reflection in the Origin

Reflect the image of AB with endpoints A(-6,3), B(3,1) through the origin. (

x

,

y

)  (?, ?)

A B

(  6 , ( 3 , 3 )  1 ) 

A

' ( 6 ,  3 )

B

' (  3 ,  1 )

A

(  6 , 3 ) (

x

### Rule for R

origin ,

y

)  ( 

x

, 

y

)

B

' (  3 ,  1 )

B

( 3 , 1 )

A

' ( 6 ,  3 )

In 2-5, find the image of the point under a reflection in the origin.

3 .

(  7 , 1 ) 

R

   ( 7 ,  1 ) 5 .

(  3 , 0 ) 

R

   ( 3 , 0 )

B

' ( 3 , 0 )

A

' ( 7 ,  1 )

7a) Using the rule (x,y) → (-x,-y), find the images of A(-5,-2), B(-1,-2), C(-1,4), and D(-3,3), namely A’, B’, C’, D’.

(

x

,

y

)  ( 

x

, 

y

)

A

(  5 ,  2 ) 

A

' ( 5 , 2 )

B

(  1 ,  2 ) 

C

(  1 , 4 ) 

B C

' ' ( ( 1 , 1 , 2  ) 4 )

D

(  3 , 3 ) 

D

' ( 3 ,  3 ) b) On one set of axes, draw quadrilateral ABCD and A’B’C’D’ You can check by extending a line from A to A’ through the origin

B

'

C

'

D

'

A

'

# Homework

### • Page 4: 2,4,6ab,8abc

In 2-5, find the image of the point under a reflection in the origin.

2 .

( 6 ,  3 ) 

R

   (  6 , 3 ) 4 .

( 0 , 0 ) 

R

   ( 0 , 0 )

A

'

B

'

A

6a) Using the rule (x,y) → (-x,-y), find the images of A(2,1), B(4,5), C(-1,3), namely A’, B’, C’.

(

x

,

y

)  ( 

x

, 

y

)

A

( 2 , 1 ) 

B

( 4 , 5 ) 

A B

' ' ( (   2 , 4 ,  1 )  5 )

C

(  1 , 3 ) 

C

' ( 1 ,  3 ) b) On one set of axes, draw triangles ABC and A’B’C’.

A

'

C

'

B

'

8a) Draw triangle RST whose vertices are R(-2,-1), S(-2,2), T(4,2), namely R’, S’, T’.

b) Find the image of each of the vertices, namely R’S’T’ (

x

,

y

)  ( 

x

, 

y

)

R

(  2 ,  1 ) 

S

(  2 , 2 ) 

T

( 4 , 2 ) 

R

' ( 2 , 1 )

S

' ( 2 ,  2 )

T

' (  4 ,  2 ) c) On one set of axes, draw triangles RST and R’S’T’.

R

'

S

'

T

'