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Triangle [tex][tex]$A B C$[/tex][/tex] has vertices [tex][tex]$A(-3,1), B(-3,4)$[/tex][/tex], and [tex][tex]$C(-7,1)$[/tex][/tex].

1. Part A: If [tex][tex]$\triangle A B C$[/tex][/tex] is translated according to the rule [tex][tex]$(x, y) \rightarrow (x+4, y-3)$[/tex][/tex] to form [tex][tex]$\triangle A^{\prime} B^{\prime} C^{\prime}$[/tex][/tex], how is the translation described in words? (3 points)

2. Part B: Where are the vertices of [tex][tex]$\triangle A^{\prime} B^{\prime} C^{\prime}$[/tex][/tex] located? Show your work or explain your steps. (4 points)

3. Part C: Triangle [tex][tex]$A^{\prime} B^{\prime} C^{\prime}$[/tex][/tex] is rotated [tex][tex]$90^{\circ}$[/tex][/tex] counterclockwise about the origin to form [tex][tex]$\triangle A^{\circ} B^{\circ} C^{\circ}$[/tex][/tex]. Is [tex][tex]$\triangle A B C$[/tex][/tex] congruent to [tex][tex]$\triangle A^{\prime} B^{\circ} C^{\prime \prime}$[/tex][/tex]? Give details to support your answer. (3 points)



Answer :

GinnyAnswer
Certainly! Let's address each part of the problem step-by-step.

### Part A
The translation rule given is [tex]\((x, y) \rightarrow (x+4, y-3)\)[/tex]. Translating each point 4 units to the right and 3 units down affects the coordinates of each vertex of the triangle.

Solution:
The translation can be described in words as:
"Translate each point 4 units to the right and 3 units down."

### Part B
Now let's apply the translation rule to each vertex of triangle [tex]\(ABC\)[/tex].

- Original vertices:
- [tex]\(A(-3, 1)\)[/tex]
- [tex]\(B(-3, 4)\)[/tex]
- [tex]\(C(-7, 1)\)[/tex]

- Applying the translation [tex]\((x, y) \rightarrow (x + 4, y - 3)\)[/tex]:

For Vertex [tex]\(A\)[/tex]:
- [tex]\(A(-3, 1) \rightarrow A'(-3 + 4, 1 - 3)\)[/tex]
- [tex]\(A' = (1, -2)\)[/tex]

For Vertex [tex]\(B\)[/tex]:
- [tex]\(B(-3, 4) \rightarrow B'(-3 + 4, 4 - 3)\)[/tex]
- [tex]\(B' = (1, 1)\)[/tex]

For Vertex [tex]\(C\)[/tex]:
- [tex]\(C(-7, 1) \rightarrow C'(-7 + 4, 1 - 3)\)[/tex]
- [tex]\(C' = (-3, -2)\)[/tex]

Solution:
The coordinates of the vertices of [tex]\(\triangle A'B'C'\)[/tex] are:
- [tex]\(A'(1, -2)\)[/tex]
- [tex]\(B'(1, 1)\)[/tex]
- [tex]\(C'(-3, -2)\)[/tex]

### Part C
We need to rotate [tex]\(\triangle A'B'C'\)[/tex] 90 degrees counterclockwise about the origin:

The rule for rotating a point [tex]\((x, y)\)[/tex] 90 degrees counterclockwise about the origin is [tex]\((x, y) \rightarrow (-y, x)\)[/tex].

Using this rule:

For Vertex [tex]\(A'\)[/tex]:
- [tex]\(A'(1, -2) \rightarrow A''(-(-2), 1)\)[/tex]
- [tex]\(A'' = (2, 1)\)[/tex]

For Vertex [tex]\(B'\)[/tex]:
- [tex]\(B'(1, 1) \rightarrow B''(-1, 1)\)[/tex]
- [tex]\(B'' = (-1, 1)\)[/tex]

For Vertex [tex]\(C'\)[/tex]:
- [tex]\(C'(-3, -2) \rightarrow C''(-(-2), -3)\)[/tex]
- [tex]\(C'' = (2, -3)\)[/tex]

Solution:
The coordinates of the vertices of [tex]\(\triangle A''B''C''\)[/tex] after the 90 degrees counterclockwise rotation are:
- [tex]\(A''(2, 1)\)[/tex]
- [tex]\(B''(-1, 1)\)[/tex]
- [tex]\(C''(2, -3)\)[/tex]

### Congruence
Translation and rotation are both rigid transformations, meaning they preserve the size and shape of the triangles. Therefore, any triangle subjected to these transformations remains congruent to its original shape.

Conclusion:
Yes, [tex]\(\triangle ABC\)[/tex] is congruent to [tex]\(\triangle A''B''C''\)[/tex] because the transformations performed (translation followed by a 90-degree counterclockwise rotation) preserve the lengths of sides and angles. Therefore, the triangles maintain congruency.

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