Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
To find the coordinates of the vertices of the pre-image of a triangle after a 90° counterclockwise rotation about the origin, we need to reverse the transformation. A 90° counterclockwise rotation involves the transformation of a point [tex]\((x, y)\)[/tex] to [tex]\((-y, x)\)[/tex]. To find the pre-image, we use the inverse transformation [tex]\((x, y) \rightarrow (y, -x)\)[/tex].
Given the coordinates of the vertices of the image after rotation:
- [tex]\(A' (6, 3)\)[/tex]
- [tex]\(B' (-2, 1)\)[/tex]
- [tex]\(C' (1, 7)\)[/tex]
We will now apply the inverse transformation [tex]\((x, y) \rightarrow (y, -x)\)[/tex] to each of these vertices.
### Step 1: Finding the pre-image of [tex]\(A'\)[/tex]
For [tex]\(A' (6, 3)\)[/tex]:
- [tex]\(x = 6\)[/tex]
- [tex]\(y = 3\)[/tex]
Applying the inverse transformation:
[tex]\[ (6, 3) \rightarrow (3, -6) \][/tex]
So, the coordinates of [tex]\(A\)[/tex] are [tex]\((3, -6)\)[/tex].
### Step 2: Finding the pre-image of [tex]\(B'\)[/tex]
For [tex]\(B' (-2, 1)\)[/tex]:
- [tex]\(x = -2\)[/tex]
- [tex]\(y = 1\)[/tex]
Applying the inverse transformation:
[tex]\[ (-2, 1) \rightarrow (1, 2) \][/tex]
So, the coordinates of [tex]\(B\)[/tex] are [tex]\((1, 2)\)[/tex].
### Step 3: Finding the pre-image of [tex]\(C'\)[/tex]
For [tex]\(C' (1, 7)\)[/tex]:
- [tex]\(x = 1\)[/tex]
- [tex]\(y = 7\)[/tex]
Applying the inverse transformation:
[tex]\[ (1, 7) \rightarrow (7, -1) \][/tex]
So, the coordinates of [tex]\(C\)[/tex] are [tex]\((7, -1)\)[/tex].
### Conclusion
The coordinates of the vertices of the pre-image of the triangle are:
- [tex]\(A (3, -6)\)[/tex]
- [tex]\(B (1, 2)\)[/tex]
- [tex]\(C (7, -1)\)[/tex]
Thus, after reversing the 90° counterclockwise rotation, the original coordinates of the vertices of the pre-image are:
[tex]\[ \boxed{(3, -6), (1, 2), (7, -1)} \][/tex]
Given the coordinates of the vertices of the image after rotation:
- [tex]\(A' (6, 3)\)[/tex]
- [tex]\(B' (-2, 1)\)[/tex]
- [tex]\(C' (1, 7)\)[/tex]
We will now apply the inverse transformation [tex]\((x, y) \rightarrow (y, -x)\)[/tex] to each of these vertices.
### Step 1: Finding the pre-image of [tex]\(A'\)[/tex]
For [tex]\(A' (6, 3)\)[/tex]:
- [tex]\(x = 6\)[/tex]
- [tex]\(y = 3\)[/tex]
Applying the inverse transformation:
[tex]\[ (6, 3) \rightarrow (3, -6) \][/tex]
So, the coordinates of [tex]\(A\)[/tex] are [tex]\((3, -6)\)[/tex].
### Step 2: Finding the pre-image of [tex]\(B'\)[/tex]
For [tex]\(B' (-2, 1)\)[/tex]:
- [tex]\(x = -2\)[/tex]
- [tex]\(y = 1\)[/tex]
Applying the inverse transformation:
[tex]\[ (-2, 1) \rightarrow (1, 2) \][/tex]
So, the coordinates of [tex]\(B\)[/tex] are [tex]\((1, 2)\)[/tex].
### Step 3: Finding the pre-image of [tex]\(C'\)[/tex]
For [tex]\(C' (1, 7)\)[/tex]:
- [tex]\(x = 1\)[/tex]
- [tex]\(y = 7\)[/tex]
Applying the inverse transformation:
[tex]\[ (1, 7) \rightarrow (7, -1) \][/tex]
So, the coordinates of [tex]\(C\)[/tex] are [tex]\((7, -1)\)[/tex].
### Conclusion
The coordinates of the vertices of the pre-image of the triangle are:
- [tex]\(A (3, -6)\)[/tex]
- [tex]\(B (1, 2)\)[/tex]
- [tex]\(C (7, -1)\)[/tex]
Thus, after reversing the 90° counterclockwise rotation, the original coordinates of the vertices of the pre-image are:
[tex]\[ \boxed{(3, -6), (1, 2), (7, -1)} \][/tex]
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.