Answer :
### Part A: Describing the Translation
To describe the translation of triangle [tex]\( \triangle ABC \)[/tex] according to the rule [tex]\( (x, y) \rightarrow (x - 3, y + 4) \)[/tex]:
The translation rule [tex]\( (x, y) \rightarrow (x - 3, y + 4) \)[/tex] indicates that every point of the triangle [tex]\( \triangle ABC \)[/tex] is moved 3 units to the left and 4 units upward. Hence, the triangle [tex]\( \triangle ABC \)[/tex] is translated by moving each vertex 3 units to the left and 4 units up to form [tex]\( \triangle A'B'C' \)[/tex].
### Part B: Finding the Vertices of [tex]\( \triangle A'B'C' \)[/tex]
We start with the given vertices of [tex]\( \triangle ABC \)[/tex]:
- [tex]\( A(-3, 1) \)[/tex]
- [tex]\( B(-3, 4) \)[/tex]
- [tex]\( C(-7, 1) \)[/tex]
To find the vertices of [tex]\( \triangle A'B'C' \)[/tex] after the translation, we apply the rule [tex]\( (x, y) \rightarrow (x - 3, y + 4) \)[/tex]:
1. For vertex [tex]\( A \)[/tex]:
[tex]\[ A' = (x - 3, y + 4) = (-3 - 3, 1 + 4) = (-6, 5) \][/tex]
2. For vertex [tex]\( B \)[/tex]:
[tex]\[ B' = (x - 3, y + 4) = (-3 - 3, 4 + 4) = (-6, 8) \][/tex]
3. For vertex [tex]\( C \)[/tex]:
[tex]\[ C' = (x - 3, y + 4) = (-7 - 3, 1 + 4) = (-10, 5) \][/tex]
Thus, the vertices of [tex]\( \triangle A'B'C' \)[/tex] are:
- [tex]\( A'(-6, 5) \)[/tex]
- [tex]\( B'(-6, 8) \)[/tex]
- [tex]\( C'(-10, 5) \)[/tex]
### Part C: Rotation and Congruence
Next, we rotate [tex]\( \triangle A'B'C' \)[/tex] by [tex]\( 180^\circ \)[/tex] counterclockwise about the origin to form [tex]\( \triangle A''B''C'' \)[/tex].
The rule for a [tex]\( 180^\circ \)[/tex] rotation about the origin is [tex]\( (x, y) \rightarrow (-x, -y) \)[/tex]. Applying this to the vertices of [tex]\( \triangle A'B'C' \)[/tex]:
1. For vertex [tex]\( A' \)[/tex]:
[tex]\[ A'' = (-x, -y) = (-(-6), -(5)) = (6, -5) \][/tex]
2. For vertex [tex]\( B' \)[/tex]:
[tex]\[ B'' = (-x, -y) = (-(-6), -(8)) = (6, -8) \][/tex]
3. For vertex [tex]\( C' \)[/tex]:
[tex]\[ C'' = (-x, -y) = (-(-10), -(5)) = (10, -5) \][/tex]
Therefore, the vertices of [tex]\( \triangle A''B''C'' \)[/tex] are:
- [tex]\( A''(6, -5) \)[/tex]
- [tex]\( B''(6, -8) \)[/tex]
- [tex]\( C''(10, -5) \)[/tex]
### Congruency Statement
The triangles [tex]\( \triangle ABC \)[/tex] and [tex]\( \triangle A''B''C'' \)[/tex] are congruent. This is because both translation and rotation are rigid transformations, meaning they do not alter the distances or angles within the shapes. As such, these transformations preserve the overall size and shape of the triangle, maintaining congruency between the original triangle [tex]\( \triangle ABC \)[/tex] and the final triangle [tex]\( \triangle A''B''C'' \)[/tex]. Thus, [tex]\(\triangle ABC \cong \triangle A''B''C''\)[/tex].
To describe the translation of triangle [tex]\( \triangle ABC \)[/tex] according to the rule [tex]\( (x, y) \rightarrow (x - 3, y + 4) \)[/tex]:
The translation rule [tex]\( (x, y) \rightarrow (x - 3, y + 4) \)[/tex] indicates that every point of the triangle [tex]\( \triangle ABC \)[/tex] is moved 3 units to the left and 4 units upward. Hence, the triangle [tex]\( \triangle ABC \)[/tex] is translated by moving each vertex 3 units to the left and 4 units up to form [tex]\( \triangle A'B'C' \)[/tex].
### Part B: Finding the Vertices of [tex]\( \triangle A'B'C' \)[/tex]
We start with the given vertices of [tex]\( \triangle ABC \)[/tex]:
- [tex]\( A(-3, 1) \)[/tex]
- [tex]\( B(-3, 4) \)[/tex]
- [tex]\( C(-7, 1) \)[/tex]
To find the vertices of [tex]\( \triangle A'B'C' \)[/tex] after the translation, we apply the rule [tex]\( (x, y) \rightarrow (x - 3, y + 4) \)[/tex]:
1. For vertex [tex]\( A \)[/tex]:
[tex]\[ A' = (x - 3, y + 4) = (-3 - 3, 1 + 4) = (-6, 5) \][/tex]
2. For vertex [tex]\( B \)[/tex]:
[tex]\[ B' = (x - 3, y + 4) = (-3 - 3, 4 + 4) = (-6, 8) \][/tex]
3. For vertex [tex]\( C \)[/tex]:
[tex]\[ C' = (x - 3, y + 4) = (-7 - 3, 1 + 4) = (-10, 5) \][/tex]
Thus, the vertices of [tex]\( \triangle A'B'C' \)[/tex] are:
- [tex]\( A'(-6, 5) \)[/tex]
- [tex]\( B'(-6, 8) \)[/tex]
- [tex]\( C'(-10, 5) \)[/tex]
### Part C: Rotation and Congruence
Next, we rotate [tex]\( \triangle A'B'C' \)[/tex] by [tex]\( 180^\circ \)[/tex] counterclockwise about the origin to form [tex]\( \triangle A''B''C'' \)[/tex].
The rule for a [tex]\( 180^\circ \)[/tex] rotation about the origin is [tex]\( (x, y) \rightarrow (-x, -y) \)[/tex]. Applying this to the vertices of [tex]\( \triangle A'B'C' \)[/tex]:
1. For vertex [tex]\( A' \)[/tex]:
[tex]\[ A'' = (-x, -y) = (-(-6), -(5)) = (6, -5) \][/tex]
2. For vertex [tex]\( B' \)[/tex]:
[tex]\[ B'' = (-x, -y) = (-(-6), -(8)) = (6, -8) \][/tex]
3. For vertex [tex]\( C' \)[/tex]:
[tex]\[ C'' = (-x, -y) = (-(-10), -(5)) = (10, -5) \][/tex]
Therefore, the vertices of [tex]\( \triangle A''B''C'' \)[/tex] are:
- [tex]\( A''(6, -5) \)[/tex]
- [tex]\( B''(6, -8) \)[/tex]
- [tex]\( C''(10, -5) \)[/tex]
### Congruency Statement
The triangles [tex]\( \triangle ABC \)[/tex] and [tex]\( \triangle A''B''C'' \)[/tex] are congruent. This is because both translation and rotation are rigid transformations, meaning they do not alter the distances or angles within the shapes. As such, these transformations preserve the overall size and shape of the triangle, maintaining congruency between the original triangle [tex]\( \triangle ABC \)[/tex] and the final triangle [tex]\( \triangle A''B''C'' \)[/tex]. Thus, [tex]\(\triangle ABC \cong \triangle A''B''C''\)[/tex].
