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[tex]$\triangle CAR$[/tex] has coordinates [tex]$C(2,4)$[/tex], [tex]$A(1,1)$[/tex], and [tex]$R(3,0)$[/tex]. A translation maps point [tex]$C$[/tex] to [tex]$C^{\prime}(3,2)$[/tex]. Find the coordinates of [tex]$A^{\prime}$[/tex] and [tex]$R^{\prime}$[/tex] under this translation.

A. [tex]$A^{\prime}(-2,2)$[/tex]; [tex]$R^{\prime}(2,-2)$[/tex]
B. [tex]$A^{\prime}(4,-2)$[/tex]; [tex]$R^{\prime}(2,-1)$[/tex]
C. [tex]$A^{\prime}(2,-1)$[/tex]; [tex]$R^{\prime}(4,-2)$[/tex]
D. [tex]$A^{\prime}(-1,0)$[/tex]; [tex]$R^{\prime}(-2,2)$[/tex]


Sagot :

Let's go through the solution step-by-step:

1. Identify the translation vector:
- The original coordinates for point [tex]\( C \)[/tex] are [tex]\( (2, 4) \)[/tex].
- The new coordinates for point [tex]\( C \)[/tex] after the translation are [tex]\( (3, 2) \)[/tex].

So, the translation vector [tex]\(\vec{T}\)[/tex] is calculated by:
[tex]\[ \vec{T} = (C' - C) = (3 - 2, 2 - 4) = (1, -2) \][/tex]

2. Apply the translation vector to point [tex]\( A \)[/tex] (original coordinates [tex]\( (1, 1) \)[/tex]):
- Using the translation vector [tex]\(\vec{T}\)[/tex]:
[tex]\[ A' = A + \vec{T} = (1, 1) + (1, -2) = (1 + 1, 1 - 2) = (2, -1) \][/tex]

3. Apply the translation vector to point [tex]\( R \)[/tex] (original coordinates [tex]\( (3, 0) \)[/tex]):
- Again, using the translation vector [tex]\(\vec{T}\)[/tex]:
[tex]\[ R' = R + \vec{T} = (3, 0) + (1, -2) = (3 + 1, 0 - 2) = (4, -2) \][/tex]

4. Verification with given options:
- Comparing the new coordinates:
- For [tex]\( A' \)[/tex]: [tex]\( (2, -1) \)[/tex]
- For [tex]\( R' \)[/tex]: [tex]\( (4, -2) \)[/tex]

These coordinates match the third given option:
[tex]\[ A' = (2, -1) \quad \text{and} \quad R' = (4, -2). \][/tex]

Therefore, the correct option is:
[tex]\[ \boxed{3} \][/tex]