To solve the problem of rotating the vertices [tex]\(G(-4,0)\)[/tex], [tex]\(H(5,2)\)[/tex], and [tex]\(I(-2,-3)\)[/tex] counterclockwise by [tex]\(270^\circ\)[/tex] about the origin, we need to transform each vertex using the properties of rotation.
### Step-by-Step Solution
Rotation matrix for [tex]\(270^\circ\)[/tex] counterclockwise:
When rotating a point [tex]\((x, y)\)[/tex] counterclockwise by [tex]\(270^\circ\)[/tex], the rotation transformation is equivalent to rotating [tex]\(90^\circ\)[/tex] clockwise. The transformation for [tex]\(90^\circ\)[/tex] clockwise is given by:
[tex]\[ (x', y') = (y, -x) \][/tex]
Using this formula:
1. Rotate [tex]\(G(-4, 0)\)[/tex]:
[tex]\[
G' = (0, -(-4)) = (0, 4)
\][/tex]
2. Rotate [tex]\(H(5, 2)\)[/tex]:
[tex]\[
H' = (2, -5)
\][/tex]
3. Rotate [tex]\(I(-2, -3)\)[/tex]:
[tex]\[
I' = (-3, 2)
\][/tex]
### Result
Therefore, the coordinates of the vertices of [tex]\(\triangle G'H'I'\)[/tex] after a [tex]\(270^\circ\)[/tex] counterclockwise rotation are:
- [tex]\(G' = (0, 4)\)[/tex]
- [tex]\(H' = (2, -5)\)[/tex]
- [tex]\(I' = (-3, 2)\)[/tex]
### Answer
From the given multiple-choice options, the correct one is:
C. [tex]\(G'(0, 4), H'(2, -5), I'(-3, 2)\)[/tex]
