Question

Triangle XYZ has vertices X(–1, –3), Y(0, 0), and Z(1, –3). Malik rotated the triangle 90° clockwise about the origin. What is the correct set of image points for triangle X’Y’Z’?
X’(3, –1), Y’(0, 0), Z’(3, 1)
X’(1, 3), Y’(0, 0), Z’(–1, 3)
X’(1, –3),Y’(0, 0), Z’(–1, –3)
X’(–3, 1), Y’(0, 0), Z’(–3, –1)

Answer 1

To rotate a point (x, y) 90 degrees clockwise about the origin, you can use the transformation formula:

\[
(x, y) \rightarrow (y, -x)
\]

Now let's apply this transformation to the vertices of triangle XYZ:

1. For point X(–1, –3):
\[
X' = (y, -x) = (-3, 1)
\]

2. For point Y(0, 0):
\[
Y' = (0, 0) \quad \text{(remains the same)}
\]

3. For point Z(1, –3):
\[
Z' = (y, -x) = (-3, -1)
\]

Putting these points together, we have:
- \(X'(-3, 1)\)
- \(Y'(0, 0)\)
- \(Z'(-3, -1)\)

Looking at the options provided, the correct set of image points for triangle \(X'Y'Z'\) is:

**None of the provided options exactly matches the calculated coordinates.**

However, if we summarize:
- The transformed coordinates are \(X'(-3, 1)\), \(Y'(0, 0)\), and \(Z'(-3, -1)\).

Please double-check the options you have; if the question does allow for any errors or variations, the closest match might be among them, but the exact answer does not align.