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Randy draws triangle [tex]$ABC$[/tex] on the coordinate plane with vertices [tex]$A (7,-4)$[/tex], [tex]$B (10,3)$[/tex], and [tex]$C (6,1)$[/tex]. He then translates the figure so the coordinates of the image are [tex]$A^{\prime}(5,1)$[/tex], [tex]$B^{\prime}(8,8)$[/tex], and [tex]$C^{\prime}(4,6)$[/tex]. What rule did he use to draw the image?

A. [tex]$T_{-5,2}(x, y)$[/tex]
B. [tex]$T_{-2,5}(x, y)$[/tex]
C. [tex]$T_{2,-5}(x, y)$[/tex]
D. [tex]$T_{5,-2}(x, y)$[/tex]


Sagot :

To determine which translation rule Randy used to draw the image of triangle [tex]\( ABC \)[/tex], follow these steps:

1. Identify the coordinates before and after translation:
- Before translation: [tex]\( A(7, -4) \)[/tex], [tex]\( B(10, 3) \)[/tex], and [tex]\( C(6, 1) \)[/tex].
- After translation: [tex]\( A'(5, 1) \)[/tex], [tex]\( B'(8, 8) \)[/tex], and [tex]\( C'(4, 6) \)[/tex].

2. Calculate the translation vector using point [tex]\( A \)[/tex] and [tex]\( A' \)[/tex]:
[tex]\[ \text{Translation vector} = (x' - x, y' - y) \][/tex]
For point [tex]\( A \)[/tex] to [tex]\( A' \)[/tex]:
[tex]\[ \text{Translation vector} = (5 - 7, 1 - (-4)) = (-2, 5) \][/tex]

3. Verify the translation vector with the other points [tex]\( B \)[/tex] and [tex]\( B' \)[/tex], and [tex]\( C \)[/tex] and [tex]\( C' \)[/tex]:
For point [tex]\( B \)[/tex] to [tex]\( B' \)[/tex]:
[tex]\[ \text{Translation vector} = (8 - 10, 8 - 3) = (-2, 5) \][/tex]
For point [tex]\( C \)[/tex] to [tex]\( C' \)[/tex]:
[tex]\[ \text{Translation vector} = (4 - 6, 6 - 1) = (-2, 5) \][/tex]

Since the translation vector [tex]\((-2, 5)\)[/tex] is the same for all three points, the translation is consistent.

4. Determine the translation rule:
The translation vector [tex]\((-2, 5)\)[/tex] tells us that each point on the figure is translated 2 units to the left and 5 units up. Therefore, the rule used by Randy is:
[tex]\[ T_{-2,5}(x, y) \][/tex]

So, the correct translation rule Randy used is:
[tex]\[ \boxed{T_{-2,5}(x, y)} \][/tex]
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