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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)
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.
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.
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.
Solution:
a) False
b) True
c) True
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