Answer :
Sure, let's solve this problem step-by-step.
### Part A: Translation Description
The translation rule given is [tex]\((x, y) \rightarrow (x + 4, y - 3)\)[/tex]. This means that we need to translate each point 4 units to the right and 3 units downwards. Hence, the translation can be described in words as:
Translate 4 units to the right and 3 units downwards.
### Part B: Vertices of [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex]
To find the vertices of the translated triangle [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex], we apply the translation rule to each of the original vertices [tex]\(A(-3, 1)\)[/tex], [tex]\(B(-3, 4)\)[/tex], and [tex]\(C(-7, 1)\)[/tex].
1. For vertex [tex]\(A(-3, 1)\)[/tex]:
[tex]\[ A' = (x + 4, y - 3) = (-3 + 4, 1 - 3) = (1, -2) \][/tex]
2. For vertex [tex]\(B(-3, 4)\)[/tex]:
[tex]\[ B' = (x + 4, y - 3) = (-3 + 4, 4 - 3) = (1, 1) \][/tex]
3. For vertex [tex]\(C(-7, 1)\)[/tex]:
[tex]\[ C' = (x + 4, y - 3) = (-7 + 4, 1 - 3) = (-3, -2) \][/tex]
Therefore, the vertices are:
[tex]\[ A' = (1, -2), \quad B' = (1, 1), \quad C' = (-3, -2) \][/tex]
### Part C: Rotation of [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex]
Now, we rotate [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex] [tex]\(90^\circ\)[/tex] counterclockwise about the origin. The rotation rule for [tex]\(90^\circ\)[/tex] counterclockwise is [tex]\((x, y) \rightarrow (-y, x)\)[/tex]. We apply this to each of the vertices [tex]\(A'(1, -2)\)[/tex], [tex]\(B'(1, 1)\)[/tex], and [tex]\(C'(-3, -2)\)[/tex].
1. For vertex [tex]\(A'(1, -2)\)[/tex]:
[tex]\[ A'' = (-(-2), 1) = (2, 1) \][/tex]
2. For vertex [tex]\(B'(1, 1)\)[/tex]:
[tex]\[ B'' = (-(1), 1) = (-1, 1) \][/tex]
3. For vertex [tex]\(C'(-3, -2)\)[/tex]:
[tex]\[ C'' = (-(-2), -3) = (2, -3) \][/tex]
The vertices of the rotated [tex]\(\triangle A'' B'' C''\)[/tex] are:
[tex]\[ A'' = (2, 1), \quad B'' = (-1, 1), \quad C'' = (2, -3) \][/tex]
### Congruency Check
To determine if [tex]\(\triangle A B C\)[/tex] is congruent to [tex]\(\triangle A'' B'' C''\)[/tex], we compare the side lengths of both triangles.
1. Original triangle [tex]\(\triangle A B C\)[/tex]:
[tex]\[ \text{Side } AB = \|A - B\| = \sqrt{(-3 - (-3))^2 + (1 - 4)^2} = \sqrt{0^2 + (-3)^2} = 3 \][/tex]
[tex]\[ \text{Side } BC = \|B - C\| = \sqrt{(-3 - (-7))^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 \][/tex]
[tex]\[ \text{Side } CA = \|C - A\| = \sqrt{(-7 - (-3))^2 + (1 - 1)^2} = \sqrt{(-4)^2 + 0^2} = 4 \][/tex]
2. Rotated triangle [tex]\(\triangle A'' B'' C''\)[/tex]:
[tex]\[ \text{Side } A''B'' = \|A'' - B''\| = \sqrt{(2 - (-1))^2 + (1 - 1)^2} = \sqrt{3^2 + 0^2} = 3 \][/tex]
[tex]\[ \text{Side } B''C'' = \|B'' - C''\| = \sqrt{(-1 - 2)^2 + (1 - (-3))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5 \][/tex]
[tex]\[ \text{Side } C''A'' = \|C - A\| = \sqrt{(2 - 2)^2 + (-3 - 1)^2} = \sqrt{0^2 + (-4)^2} = 4 \][/tex]
Since the side lengths of [tex]\(\triangle A B C\)[/tex] and [tex]\(\triangle A'' B'' C''\)[/tex] match, the triangles are congruent.
### Conclusion:
Yes, [tex]\(\triangle ABC\)[/tex] is congruent to [tex]\(\triangle A'' B'' C''\)[/tex].
Thus, we have provided complete and detailed steps to answer each part of the problem correctly.
### Part A: Translation Description
The translation rule given is [tex]\((x, y) \rightarrow (x + 4, y - 3)\)[/tex]. This means that we need to translate each point 4 units to the right and 3 units downwards. Hence, the translation can be described in words as:
Translate 4 units to the right and 3 units downwards.
### Part B: Vertices of [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex]
To find the vertices of the translated triangle [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex], we apply the translation rule to each of the original vertices [tex]\(A(-3, 1)\)[/tex], [tex]\(B(-3, 4)\)[/tex], and [tex]\(C(-7, 1)\)[/tex].
1. For vertex [tex]\(A(-3, 1)\)[/tex]:
[tex]\[ A' = (x + 4, y - 3) = (-3 + 4, 1 - 3) = (1, -2) \][/tex]
2. For vertex [tex]\(B(-3, 4)\)[/tex]:
[tex]\[ B' = (x + 4, y - 3) = (-3 + 4, 4 - 3) = (1, 1) \][/tex]
3. For vertex [tex]\(C(-7, 1)\)[/tex]:
[tex]\[ C' = (x + 4, y - 3) = (-7 + 4, 1 - 3) = (-3, -2) \][/tex]
Therefore, the vertices are:
[tex]\[ A' = (1, -2), \quad B' = (1, 1), \quad C' = (-3, -2) \][/tex]
### Part C: Rotation of [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex]
Now, we rotate [tex]\(\triangle A^{\prime} B^{\prime} C^{\prime}\)[/tex] [tex]\(90^\circ\)[/tex] counterclockwise about the origin. The rotation rule for [tex]\(90^\circ\)[/tex] counterclockwise is [tex]\((x, y) \rightarrow (-y, x)\)[/tex]. We apply this to each of the vertices [tex]\(A'(1, -2)\)[/tex], [tex]\(B'(1, 1)\)[/tex], and [tex]\(C'(-3, -2)\)[/tex].
1. For vertex [tex]\(A'(1, -2)\)[/tex]:
[tex]\[ A'' = (-(-2), 1) = (2, 1) \][/tex]
2. For vertex [tex]\(B'(1, 1)\)[/tex]:
[tex]\[ B'' = (-(1), 1) = (-1, 1) \][/tex]
3. For vertex [tex]\(C'(-3, -2)\)[/tex]:
[tex]\[ C'' = (-(-2), -3) = (2, -3) \][/tex]
The vertices of the rotated [tex]\(\triangle A'' B'' C''\)[/tex] are:
[tex]\[ A'' = (2, 1), \quad B'' = (-1, 1), \quad C'' = (2, -3) \][/tex]
### Congruency Check
To determine if [tex]\(\triangle A B C\)[/tex] is congruent to [tex]\(\triangle A'' B'' C''\)[/tex], we compare the side lengths of both triangles.
1. Original triangle [tex]\(\triangle A B C\)[/tex]:
[tex]\[ \text{Side } AB = \|A - B\| = \sqrt{(-3 - (-3))^2 + (1 - 4)^2} = \sqrt{0^2 + (-3)^2} = 3 \][/tex]
[tex]\[ \text{Side } BC = \|B - C\| = \sqrt{(-3 - (-7))^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 \][/tex]
[tex]\[ \text{Side } CA = \|C - A\| = \sqrt{(-7 - (-3))^2 + (1 - 1)^2} = \sqrt{(-4)^2 + 0^2} = 4 \][/tex]
2. Rotated triangle [tex]\(\triangle A'' B'' C''\)[/tex]:
[tex]\[ \text{Side } A''B'' = \|A'' - B''\| = \sqrt{(2 - (-1))^2 + (1 - 1)^2} = \sqrt{3^2 + 0^2} = 3 \][/tex]
[tex]\[ \text{Side } B''C'' = \|B'' - C''\| = \sqrt{(-1 - 2)^2 + (1 - (-3))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5 \][/tex]
[tex]\[ \text{Side } C''A'' = \|C - A\| = \sqrt{(2 - 2)^2 + (-3 - 1)^2} = \sqrt{0^2 + (-4)^2} = 4 \][/tex]
Since the side lengths of [tex]\(\triangle A B C\)[/tex] and [tex]\(\triangle A'' B'' C''\)[/tex] match, the triangles are congruent.
### Conclusion:
Yes, [tex]\(\triangle ABC\)[/tex] is congruent to [tex]\(\triangle A'' B'' C''\)[/tex].
Thus, we have provided complete and detailed steps to answer each part of the problem correctly.
