QUESTION IMAGE
Question
from unit 1, lesson 6 5 the point (-2, -3) is rotated 90° counterclockwise using center (0, 0). what are the coordinates of the image? a (-3, -2) b (-3, 2) c (3, -2) d (3, 2)
Step1: Recall 90° CCW rotation rule
For a point \((x, y)\) rotated 90° counterclockwise about the origin, the new coordinates are \((-y, x)\).
Step2: Apply the rule to \((-2, -3)\)
Here, \(x = -2\) and \(y = -3\). Substitute into the rule: new \(x\) is \(-y = -(-3)=3\), new \(y\) is \(x = -2\)? Wait, no, wait. Wait, the correct rule is: 90° counterclockwise rotation about origin: \((x,y)\to(-y,x)\). Wait, let's check again. Wait, original point is \((-2, -3)\). So \(x=-2\), \(y = -3\). Then, applying \((x,y)\to(-y, x)\): \(-y = -(-3)=3\), \(x=-2\)? Wait, no, that can't be. Wait, maybe I mixed up. Wait, no, the correct formula for 90° counterclockwise rotation about the origin is \((x, y) \mapsto (-y, x)\). Wait, let's take a simple point, say (1,0). Rotated 90° CCW, it becomes (0,1). Wait, no: (1,0) rotated 90° CCW is (0,1)? Wait, no, (1,0) is on the x-axis. Rotating 90° CCW around origin, it should go to (0,1)? Wait, no, (1,0) rotated 90° CCW is (0,1)? Wait, no, actually, the rotation matrix for 90° CCW is \(
\). So multiplying by \(
\) gives \(
\). So for (1,0): \(
=(0,1)\), which is correct. For (0,1): \(
=(-1,0)\), correct. For (-2, -3): applying the matrix, we get \(
=(3, -2)\)? Wait, no: wait, the matrix is \(
\), so \(
=
=
\). So the new coordinates are (3, -2), which is option C.
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C. (3, -2)