The **Translation** is a transformation in which
each point of the given object is displaced through definite distance and
direction. The displacement is defined by a translation vector (a, b).

The following are the properties of translation:

1.
The object and the image
under the translation are congruent.

2.
The lines joining any point
of the object with its corresponding image are parallel and equal.

*Example 1:** Find the image of ΔABC under the translation through translation
vector PQ**.*

*Solution:*

*To get the image of ΔABC under the translation vector PQ draw
the lines through A, B, and C which are parallel and equal to PQ in the same
direction to get the images of A, B, and C and join them to get ΔA’B’C’.*

**Translation Using Co-ordinates**

The image of the geometrical figures under the translation through certain translation vector can be obtained with the help of co-ordinates.

**Translation Through Translation Vector T = (a, b)**

Let A(2, 1) be a point and T = (2, 3) be the translation
vector. To translate A(2, 1) through translation vector T = (2, 3), it has to
move 2 units right and then 3 units up which is the point (4, 4)

Hence, A’(4, 4) is the image of the point A(2, 1) under the translation vector T = AA’ = (2, 3).

Let us see the following table of some other points on the same graph and their corresponding images under the translation through translation vector T = (2, 3).

From the above table, we can see that the image of any
point under the translation through translation vector (a, b) is obtained by
adding a to x-coordinate and b to y-coordinate of the given point.

**Worked Out Examples**

*Example 2:** If A(-4, -4), B(-2, -1) and C(-1, -5) are the vertices of a
triangle ABC. Find the co-ordinates of the image of ΔABC under the translation
T =(6, 5). Draw ΔABC and its image on the same graph paper.*

*Solution:*

*As A(-4, -4), B(-2, -1) and C(-1, -5) are the vertices of ΔABC, the co-ordinates of the vertices of image of ΔABC can be obtained by using the formula as below:*

*Drawing ΔABC and ΔA’B’C’ on the same graph paper we have figure
as shown.*

* *

*Example 3:** A(-4, 2), B(0, 1) and C(-2, -3) are the vertices of ΔABC. If A’(-1,
6) be the image under the translation of vertex A. Find the images B’ and
C’ of B and C under the same translation.*

*Solution:*

*Here, as the image of A(-4, 2) under the translation is A’(-1,
6),*

*Now, under the same translation, the image of B(0, 1) is B’(0+3,
1+4) i.e. B’(3, 5) and the image of C(-2, -3) is C’(-2+3, -3+4) i.e. C’(1, 1).*

*Example 4:** Translate a point A(2, 7) under translation T _{1} = (4, -3)
to the point A’. Translate A’ to A’’ under another translation T_{2} =(2,
4). Find the translation vector which translates A to A’’.*

*Solution:*

*Now, let T be the translation vector which translates A(2, 7) to
A’’(8, 8).*

*Example 5:** Translate the point M(4, -5) by the translation vector T = (-3, 4)
and find the image point M’ and write down the translation vector which maps M’
to M.*

*Solution:*

*Let T’ = (a, b) be the translation vector which maps M’(1, -1)
back to M(4, -5). Then,*

*Therefore,*

*1+a = 4*

*or, a = 4 – 1 *

*or, a = 3*

*And,*

*-1+b = -5*

*or, b = -5 + 1*

*or, b = -4*

*∴** The required translation vector is (3, -4) which is called the inverse of T and written as T ^{-1}.*

*i.e. If T = (a, b) then
T^{-1} = (-a, -b).*

* *

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