Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
