Note on Translation

  • Note
  • Things to remember
  • Exercise

In a translation, every point of the object must be moved in the same direction and for the same distance.When you are performing a translation, the initial object is called the pre-image and the object after the translation is called the image. The original object and its translation have the same shape and size, and they face in the same direction. For example, ΔABC is translated by 3 units to the right.

A general rule:

T (x, y)\(\rightarrow\) (x + a, y + b)

In the above graph,

T (-1, 1) \(\rightarrow\) A'(-1 + 3, 1 + 0) = A' (2,1)

T(-2, -1) \(\rightarrow\) B'(-2 + 3, -1 + 0) = B'(1, -1)

T(0, -2) \(\rightarrow\) C' (0 + 3, -2 + 0) = C' (3, -2)

Example

Find the image of the points P (4, 3) under the translation(x, y) → (x+3, y-3).

Solution:

The translation is (x, y) → (x+3, y-3)

The translation tells you to add 3 to the x-value and subtract 3 from the y- value.

So, (4, 3) → (4+3, 3-3) =(7, 0)

  • Translation is a transformation that "slides" an object at a fixed distance and in a given direction. 
  • The original object and its translation have the same shape and size, and they face in the same direction. 
  • An example of translation in architecture would be the stadium setting. The seats are the same size and shape and face in the same direction.
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Very Short Questions

Solution:

The translation is (x, y) → (x+2, y-2)

The translation tells you to add 2 to the x-value and subtract 2 from the y- value.

So, the image of (3, 2) → (3+2, 2-2) = (5, 0)

Solution:

To translate then the above triangle, we proceed as follows:

(x, y) → (x+9, y-7)

(-6, 6) → (-6 + 9, 6 -7) = (3, 1)

(-6, 1) → (-6 + 9, 1 - 7) = (3, -6)

(-1, 1) → (-1 + 9, 1-7) = (8, -6)

Solution:

The vertices of \(\triangle\) are A(0, 6), B(3, -2) and C(4, 0).

Translation vector (T) =\(\begin{pmatrix} 0\\5 \end{pmatrix}\)

When , T =\(\begin{pmatrix} a\\b \end{pmatrix}\), P(x, y) \(\rightarrow\) P'(x+a, y+a)

\(\therefore\) A(0, 6)\(\rightarrow\) A'(0+0, 6+5) = A'(0, 11)

B(3, -2)\(\rightarrow\) B'(3+0, -2+5) = B'(3, 3)

C(4, 0) \(\rightarrow\) C'(4+0, 0+5) = C'(4, 5)

\(\triangle\)ABC and \(\triangle\)A'B'C' are shown on the graph.

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Solution:

According to the question,

Presenting the points A(4, -5) in the graph, under the translation of 3 units right and 4 units up.

A(4, -5) \(\rightarrow\) A'(4+3, -5+4) = A'(7, -1)

Here,

A(4, -5) \(\rightarrow\)A' (7, -1)

Showing in the graph.

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Solution:

The translation tells us to add 5 to the x-value and add 4 units to the y-axis.

According to the Question,

\(\triangle\)ABC \(\triangle\)A'B'C'

A(4,5)\(\rightarrow\) A'(9,9)

B(1,3)\(\rightarrow\) B'(6,7)

C(4,3)\(\rightarrow\) C'(9,7)

Presenting in the graph.

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