Question

question
determine a series of transformations that would map polygon abcde onto polygon abcde

Explanation:

Step1: Analyze Translation

First, we can consider translating the original polygon \(ABCDE\) to align the corresponding vertices. Let's find the translation vector. For point \(A(-1, -1)\) to \(A'(6, 6)\), the horizontal translation is \(6 - (-1)=7\) and vertical translation is \(6 - (-1)=7\)? Wait, no, maybe first rotation. Wait, the original polygon and the image seem to be rotated. Let's check the orientation. Alternatively, first rotate the polygon \(ABCDE\) 90 degrees counterclockwise (or clockwise) and then translate. Wait, let's take a vertex, say \(A(-1, -1)\), \(A'(6, 6)\). Let's first rotate \(ABCDE\) 90 degrees counterclockwise about the origin. The rotation rule for 90 degrees counterclockwise is \((x,y)\to(-y,x)\). For \(A(-1, -1)\), after rotation: \((-(-1), -1)=(1, -1)\). Not matching \(A'(6,6)\). Wait, maybe 90 degrees clockwise? Rule \((x,y)\to(y, -x)\). For \(A(-1, -1)\): \((-1, 1)\). No. Alternatively, rotate about a point. Wait, maybe first translate, then rotate, then translate again. Wait, let's look at the coordinates. Original polygon \(ABCDE\) has vertices: Let's assume \(A(-1, -1)\), \(B(-2, -3)\)? Wait, no, the graph: \(A\) is at \((-1, -1)\), \(B\) at \((-2, -3)\)? Wait, no, the y-axis is horizontal? Wait, the axes: the horizontal axis is x, vertical is y? Wait, the graph has x-axis (vertical) and y-axis (horizontal)? Wait, maybe the axes are swapped. Let's reorient: Let's consider the horizontal axis as y (right is positive y) and vertical axis as x (up is positive x). So \(A\) is at (x=-1, y=-1), \(A'\) is at (x=6, y=6). So to map \(A(-1, -1)\) to \(A'(6, 6)\), let's first rotate 90 degrees counterclockwise. The rotation matrix for 90 degrees counterclockwise is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). So applying to \(A(-1, -1)\): \(x' = 0*(-1)+(-1)*(-1)=1\), \(y' = 1*(-1)+0*(-1)=-1\). No. Wait, maybe the first step is to rotate the polygon \(ABCDE\) 90 degrees counterclockwise about the origin, then translate. Wait, maybe I made a mistake in axis orientation. Let's assume the standard coordinate system: x-axis horizontal (right positive), y-axis vertical (up positive). Then the original polygon \(ABCDE\) is in the third quadrant (x negative, y negative), and \(A'B'C'D'E'\) is in the first quadrant. Let's take vertex \(A(-1, -1)\), \(A'(6, 6)\). Let's first rotate \(ABCDE\) 90 degrees counterclockwise about the origin. The rotation of a point \((x,y)\) 90 degrees counterclockwise is \((-y, x)\). So \(A(-1, -1)\) becomes \((1, -1)\). Then translate 5 units right and 7 units up: \((1 + 5, -1 + 7)=(6, 6)\), which matches \(A'\). Let's check another vertex, say \(B\). Let's assume \(B(-2, -3)\) (from the graph). Rotate 90 degrees counterclockwise: \((-(-3), -2)=(3, -2)\). Translate 5 right and 7 up: \((3 + 5, -2 + 7)=(8, 5)\). Wait, but \(B'\) should be... Wait, maybe my vertex coordinates are wrong. Alternatively, the correct sequence: First, rotate polygon \(ABCDE\) 90 degrees counterclockwise about the origin. Then translate 7 units right and 7 units up? Wait, no. Alternatively, the steps are: 1. Rotate the polygon \(ABCDE\) 90 degrees counterclockwise about the origin. 2. Translate the rotated polygon 7 units to the right and 7 units up. Let's verify with \(A(-1, -1)\): Rotate 90 CCW: \((1, -1)\). Translate (7,7): \((1 + 7, -1 + 7)=(8,6)\). No, \(A'\) is (6,6). Wait, maybe rotate 90 degrees clockwise. Rotation rule for 90 degrees clockwise: \((x,y)\to(y, -x)\). For \(A(-1, -1)\): \((-1, 1)\). Translate (7,5): \((-1 + 7, 1 + 5)=(6,6)\). Yes! Let's check: \(A(-1, -1)\) rotated 90 degrees clockwise: \((-1, 1)\). Then translate 7 units right (x+7) and 5 units up (y+5): \((-1 + 7, 1 + 5)=(6,6)\), which matches \(A'\). Let's check another vertex, say \(E\). Assume \(E(-4, -2)\). Rotate 90 degrees clockwise: \((-2, 4)\). Translate 7 right and 5 up: \((-2 + 7, 4 + 5)=(5,9)\). Wait, but \(E'\) should be... Maybe my initial vertex coordinates are incorrect. Alternatively, the correct sequence is: First, rotate the polygon \(ABCDE\) 90 degrees clockwise about the origin, then translate it 7 units to the right and 5 units up. Alternatively, another approach: The transformation from \(ABCDE\) to \(A'B'C'D'E'\) involves a rotation (to change the orientation) and translations (to move the position). A common sequence is: 1. Rotate the polygon \(ABCDE\) 90 degrees counterclockwise (or clockwise) about the origin. 2. Translate the rotated polygon by a vector to align with \(A'B'C'D'E'\). After analyzing, a possible series is: Rotate polygon \(ABCDE\) 90 degrees clockwise about the origin, then translate 7 units to the right and 5 units up. Or, first translate, then rotate, then translate. But a more accurate sequence: Let's take vertex \(A(-1, -1)\) to \(A'(6, 6)\). The vector from \(A\) to \(A'\) is \((7,7)\), but the orientation is changed, so rotation is involved. So first, rotate \(ABCDE\) 90 degrees counterclockwise (to match the orientation), then translate. Wait, maybe the correct steps are: 1. Rotate the polygon \(ABCDE\) 90 degrees counterclockwise about the origin. 2. Translate the rotated polygon 7 units to the right and 7 units up. But earlier calculation didn't match. Alternatively, the problem allows for a series of transformations, such as a rotation followed by a translation, or translation followed by rotation, etc. A possible correct series: Rotate polygon \(ABCDE\) 90 degrees counterclockwise about the origin, then translate 8 units to the right and 7 units up. Wait, maybe I need to look at the graph again. The original polygon is in the third quadrant (x negative, y negative), and the image is in the first quadrant, rotated. So a 90-degree counterclockwise rotation (which changes (x,y) to (-y,x)) and then translation. For example, take point \(A(-1, -1)\): after 90-degree CCW rotation: (1, -1). Then translate (5,7): (1+5, -1+7)=(6,6), which is \(A'\). So rotation 90 degrees CCW about the origin, then translation 5 units right and 7 units up. Let's confirm with another point, say \(B\). Assume \(B(-2, -3)\): after 90-degree CCW rotation: (3, -2). Translate 5 right and 7 up: (3+5, -2+7)=(8,5). If \(B'\) is (8,5), that works. So the series of transformations is: Rotate the polygon \(ABCDE\) 90 degrees counterclockwise about the origin, then translate it 5 units to the right and 7 units up. Alternatively, another possible sequence: Translate the polygon \(ABCDE\) 1 unit right and 1 unit up (to center at origin), rotate 90 degrees counterclockwise, then translate 6 units right and 6 units up. But the simplest series is likely a rotation (90 degrees counterclockwise or clockwise) followed by a translation. So a possible answer is: Rotate polygon \(ABCDE\) 90 degrees counterclockwise about the origin, then translate it 7 units to the right and 7 units up (or other translation vector that aligns the vertices). Wait, maybe the correct steps are: 1. Rotate the polygon \(ABCDE\) 90° clockwise about the origin. 2. Translate the rotated polygon 7 units to the right and 5 units up. But regardless, the key is to identify the rotation (to change orientation) and translation (to move position). So the series of transformations could be: First, rotate the polygon \(ABCDE\) 90 degrees counterclockwise about the origin, and then translate it 7 units horizontally (right) and 7 units vertically (up). Or, alternatively, translate first, then rotate. But the main transformations are rotation and translation.

Step2: Confirm the Transformations

Let's formalize: Let's take vertex \(A(-1, -1)\).
- Rotation 90° counterclockwise about the origin: The rule is \((x, y) \to (-y, x)\). So \(A(-1, -1) \to (1, -1)\).
- Translation: We need to move \((1, -1)\) to \(A'(6, 6)\). The translation vector is \((6 - 1, 6 - (-1))=(5, 7)\). So translate 5 units right and 7 units up.

Checking another vertex, say \(B\) (assume \(B(-2, -3)\)):
- Rotation 90° counterclockwise: \((3, -2)\).
- Translation 5 right, 7 up: \((3 + 5, -2 + 7)=(8, 5)\). If \(B'\) is \((8, 5)\), this matches.

Thus, the series of transformations is: Rotate polygon \(ABCDE\) 90 degrees counterclockwise about the origin, then translate it 5 units to the right and 7 units up. (Alternatively, other valid sequences like rotate clockwise then translate, but this is a valid one.)

Answer:

A series of transformations that maps polygon \(ABCDE\) onto \(A'B'C'D'E'\) is: Rotate \(ABCDE\) 90° counterclockwise about the origin, then translate the rotated polygon 5 units to the right and 7 units up. (Other valid sequences, such as different rotations or translations, may also apply.)