Cambridge International Mathematics

Transformation geometry (Chapter 20) 403

If P(x,y) istranslatedhunits in thex-direction andkunits

in they-direction to become P^0 (x^0 ,y^0 ), then x^0 =x+hand

y^0 =y+k:

We write P(x,y)

¡h

k

¢

¡¡!

P^0 (x+h,y+k)

where P^0 is called theimageof the object P and

¡h

k

¢

is called thetranslation vector.

x^0 =x+h

y^0 =y+k

)

are called thetransformation equations.

Example 1 Self Tutor

Triangle OAB with vertices O(0,0),A(2,3)and B(¡ 1 ,2)is translated

¡ 3

2

¢

.

Find the image vertices and illustrate the object and image.

O( 0 , 0 )

¡ 3

2

¢

¡¡!

O^0 ( 3 , 2 )

A( 2 , 3 )

¡ 3

2

¢

¡¡!

A^0 ( 5 , 5 )

B(¡ 1 , 2 )

¡ 3

2

¢

¡¡!

B^0 ( 2 , 4 )

Example 2 Self Tutor

On a set of axes draw the line with equation y=^12 x+1.

Find the equation of the image when the line is translated through

¡ 3

¡ 1

¢

:

Under a translation, the image of a line is a parallel

line, and so will have the same gradient.

) the image of y=^12 x+1has the form

y=^12 x+c.

The object contains the point(0,1) since the

y-intercept is 1.

Since (0,1)

¡ 3

¡ 1

¢

¡¡¡!

(3,0),(3,0)lies on the image.

) 0=^12 (3) +c

) c=¡^32

) the equation of the image is y=^12 x¡^32.

y

x

5

5

A

B

O

O'

A'

B'

y

x

yk¡+¡

y

xxh¡+¡

P,()xy¡

P'()xy'',¡

h

k

³

h

k

́

O

O

y

x

& 1 *

3

• 
  

object

image

()0 ¡1,

()3 ¡0,

2

1

2

y=^1 x- 1

2

y=^1 x+ 1

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y:\HAESE\IGCSE01\IG01_20\403IGCSE01_20.CDR Friday, 3 October 2008 3:33:14 PM PETER