- The first step is already shown, so begin with the coordinates of \(A'\). - A reflection across the \(x\)-axis keeps one coordinate unchanged. - Check that the final point is the same distance from the \(x\)-axis as \(A'\).

1. From the graph, the translation takes \(A=(-3,2)\) to \(A'=(1,2)\). 2. Reflection across the \(x\)-axis sends \((x,y)\) to \((x,-y)\). 3. Therefore, \(A''=(1,-2)\).

\(A''=(1,-2)\)

- A reflection across the y-axis leaves one coordinate unchanged. Does the first pair fit that pattern? - A translation keeps the direction and size of a move constant. Describe the change from \(P'\) to \(P''\).

1. From \(P\) to \(P'\), only the sign of the x-coordinate changes, so the first move is reflection across the y-axis. 2. From \(P'\) to \(P''\), the x-coordinate stays fixed while the y-coordinate increases by \(3\), so the second move is translation \(3\) units up.

First reflect across the y-axis; then translate \(3\) units up.

- Compare corresponding vertices in the original and intermediate triangles first. - Then compare the intermediate triangle with the final triangle. - Look for a change that is identical at every vertex in each step.

1. Comparing \(ABC\) with \(A'B'C'\), each vertex moves \(4\) units to the right and keeps the same \(y\)-coordinate. The first transformation is a translation \(4\) units right. 2. Comparing \(A'B'C'\) with \(A''B''C''\), each \(x\)-coordinate stays the same while each \(y\)-coordinate changes sign. The second transformation is reflection across the \(x\)-axis.

First translate \(4\) units to the right; then reflect across the \(x\)-axis.

- Focus on what a reflection and a translation preserve. - Think about what it means for one figure to be carried exactly onto another by rigid motions. - Do not replace the requested transformation argument with a triangle-congruence criterion.

1. Reflection across \(PQ\) maps \(\triangle ABC\) to \(\triangle A'B'C'\). 2. The stated translation maps \(\triangle A'B'C'\) to \(\triangle DEF\). 3. A reflection and a translation are both rigid motions, so each preserves distances and angle measures. 4. Their composition is therefore a sequence of rigid motions that maps every vertex and side of \(\triangle ABC\) onto the corresponding vertex and side of \(\triangle DEF\). 5. By the rigid-motion definition of congruence, \(\triangle ABC\cong\triangle DEF\).

The reflection followed by the translation is a sequence of rigid motions mapping \(\triangle ABC\) exactly onto \(\triangle DEF\). Therefore, \(\triangle ABC\cong\triangle DEF\) by the rigid-motion definition of congruence.

- Think about what happens to a point's signed perpendicular distance after the first reflection and then the second. - The net motion is perpendicular to both parallel mirror lines. - Compare the translation distance with the separation of the two mirrors.

1. The composition of reflections across two parallel lines is a translation perpendicular to the lines. 2. The translation distance is twice the distance between the reflection lines: \(2\cdot3=6\) units. 3. Because the reflection across \(g\) is followed by the reflection across \(h\), the translation is from \(g\) toward \(h\), which is upward in the diagram.

A translation \(6\) units upward, perpendicular to \(g\) and \(h\).

- A horizontal move does not change a point's height. How can you use that to check your first point? - A point and its image across the x-axis are equally far from the axis. Which coordinate records that distance?

1. Read \(P=(-2,1)\) on the grid. 2. Moving three units right gives \(P'=(1,1)\). 3. Reflecting this point across the x-axis gives \(P''=(1,-1)\).

\(P'=(1,1)\) and \(P''=(1,-1)\).

- How would you recognize two points the same distance from the y-axis on opposite sides? - Mark the intermediate point first so you do not apply both rules to the original point.

1. Reflection across the y-axis changes the sign of the x-coordinate and leaves the y-coordinate unchanged. 2. The intermediate point is \(P'=(-2,1)\). 3. Moving this image three units right gives \(P''=(1,1)\).

\(P'=(-2,1)\) and \(P''=(1,1)\).

- Read the original vertices from the grid and change their x-coordinate signs first. - Use each reflected point, not the original point, when applying the translation. - Think about whether either move changes the triangle's size or shape.

1. Read \(A=(1,1)\), \(B=(3,1)\), and \(C=(2,3)\). 2. After the reflection, \(A'=(-1,1)\), \(B'=(-3,1)\), and \(C'=(-2,3)\). 3. Adding \(4\) to each x-coordinate and subtracting \(1\) from each y-coordinate gives \(A''=(3,0)\), \(B''=(1,0)\), and \(C''=(2,2)\). 4. Both a reflection and a translation preserve side lengths and angles, so the triangles are congruent.

\(A''=(3,0)\), \(B''=(1,0)\), \(C''=(2,2)\). Yes; both moves are rigid motions.

- Work out the position after the first move in each order. - Reflecting across the x-axis changes the sign of the vertical coordinate. - Moving north adds to the vertical coordinate.

1. In the stated order, reflection gives \((1,-2)\), then moving north gives \((1,1)\). 2. In the student's order, moving north gives \((1,5)\), then reflection gives \((1,-5)\). 3. The two positions differ, so the order matters for these transformations.

Stated order: \((1,1)\). Student's order: \((1,-5)\). Yes, the order matters.

- Apply the two reflections in order and track which coordinate sign changes each time. - Compare the original equation with the final equation. - Which single transformation changes both \(x\) and \(y\) to their opposites?

1. Reflecting across the y-axis replaces \(x\) with \(-x\), so \(t(x) = -\frac{2}{3}x + 5\). 2. Reflecting across the x-axis multiplies all outputs by \(-1\), so \(u(x) = \frac{2}{3}x - 5\). 3. Reflecting across both coordinate axes is equivalent to a \(180^\circ\) rotation about the origin. That rotation maps \(s\) directly to \(u\).

1. \(t(x) = -\frac{2}{3}x + 5\) 2. \(u(x) = \frac{2}{3}x - 5\) 3. A \(180^\circ\) rotation about the origin

- Read each original vertex and its corresponding final image vertex carefully. - Compare what happens to the \(x\)-coordinates and the \(y\)-coordinates separately. - Interpret a sign change in one coordinate as a reflection. - Interpret adding or subtracting a constant from one coordinate as a translation.

1. The vertices are \(A(1, 2)\), \(B(4, 1)\), \(C(5, 4)\), and \(D(2, 5)\). The final image vertices are \(A''(-1, -1)\), \(B''(-4, -2)\), \(C''(-5, 1)\), and \(D''(-2, 2)\). 2. For each corresponding pair, the \(x\)-coordinate changes sign and the \(y\)-coordinate decreases by \(3\). Therefore, Rule II is correct: \((x, y)\to(-x, y-3)\). 3. Reflect \(ABCD\) across the \(y\)-axis, then translate the result \(3\) units down.

a) \(A(1, 2)\), \(B(4, 1)\), \(C(5, 4)\), \(D(2, 5)\); \(A''(-1, -1)\), \(B''(-4, -2)\), \(C''(-5, 1)\), \(D''(-2, 2)\). b) Rule II: \((x, y)\to(-x, y-3)\). c) Reflect across the \(y\)-axis, then translate \(3\) units down.

- Use the vertex whose final image is at the origin to determine what the translation must accomplish. - Apply the same translation to the other vertices before deciding on the rotation. - Compare the translated rays from the origin with the corresponding final rays.

1. Vertex \(A\) must first move to the origin because its final image \(A''\) is the origin, and a rotation about the origin leaves the origin fixed. 2. Since \(A=(-3,-1)\), translating by \(\langle 3,1\rangle\) sends \(A\) to \((0,0)\). The same translation sends \(B\) to \((2,0)\) and \(C\) to \((0,2)\). 3. A \(90^\circ\) counterclockwise rotation about the origin sends \((2,0)\) to \((0,2)\) and \((0,2)\) to \((-2,0)\), matching \(B''\) and \(C''\). 4. Therefore, the required sequence is translation by \(\langle 3,1\rangle\), followed by a \(90^\circ\) counterclockwise rotation about the origin.

Translate by \(\langle 3,1\rangle\), then rotate \(90^\circ\) counterclockwise about the origin.

- Undo the second transformation before trying to identify the first one. - A reflection can be undone by applying the same reflection again. - After recovering the intermediate image, compare the displacement of corresponding vertices.

1. Reflection across the \(y\)-axis is its own inverse, so reflect the final image back across the \(y\)-axis. 2. This gives the intermediate vertices \(A'=(-2,2)\), \(B'=(0,2)\), and \(C'=(-1,4)\). 3. Compare corresponding original and intermediate vertices. From \(A=(-4,-1)\) to \(A'=(-2,2)\), the change is \(\langle 2,3\rangle\). 4. The same change takes \(B\) to \(B'\) and \(C\) to \(C'\), so the first transformation is translation by \(\langle 2,3\rangle\).

The translation vector is \(\langle 2,3\rangle\).

- The two mirror lines share one fixed point. What must happen to that point under both reflections? - Compare the directed angle from the first mirror line to the second. - The equivalent rotation uses twice that directed angle.

1. The composition of reflections across intersecting lines is a rotation about their intersection. 2. The rotation angle is twice the directed angle from the first mirror line to the second mirror line. 3. The marked angle from \(g\) to \(h\) is \(40^\circ\) counterclockwise, so the equivalent rotation is \(80^\circ\) counterclockwise about \(O\).

A rotation of \(80^\circ\) counterclockwise about \(O\).

- Read \(T\circ R\) as applying \(R\) first and then \(T\). - Work through the two sequences separately rather than combining their rules immediately. - Compare the two final coordinates.

1. From the graph, \(P=(1,2)\). 2. For \(T\circ R\), first reflect: \((1,2)\mapsto(-1,2)\). Then translate: \((-1,2)\mapsto(2,2)\). 3. For \(R\circ T\), first translate: \((1,2)\mapsto(4,2)\). Then reflect: \((4,2)\mapsto(-4,2)\). 4. The final images differ, so changing the order changes the composition.

\(T\circ R\) sends \(P\) to \((2,2)\), while \(R\circ T\) sends \(P\) to \((-4,2)\). The order matters.

- Compare one side of the original triangle with its matching side to work out how much it has turned. - For a single rotation, the unknown center must stay the same distance from a point and its image. - For the second description, see how far the rotated image of one vertex is from its target; then check the other two. - What happens to the direction of a side or to clockwise/counterclockwise order under each type of move?

1. Read \(A=(1,1)\), \(B=(3,1)\), \(C=(1,2)\), \(D=(-3,2)\), \(E=(-3,4)\), and \(F=(-4,2)\). The vector \(AB=\langle2,0\rangle\) becomes \(DE=\langle0,2\rangle\), so the rotation is \(90^\circ\) counterclockwise. 2. If its center is \(S=(h,k)\), rotating \(A\) about \(S\) gives \((h-(1-k),\;k+(1-h))\). Set this equal to \(D=(-3,2)\). Then \(h+k=-2\) and \(k-h=1\), so \(S=(-\frac{3}{2},-\frac{1}{2})\). 3. Relative to \(S\), the vectors to \(B\) and \(C\) are \(\langle\frac{9}{2},\frac{3}{2}\rangle\) and \(\langle\frac{5}{2},\frac{5}{2}\rangle\). Quarter-turns give \(E=(-3,4)\) and \(F=(-4,2)\), so the single rotation works for all vertices. 4. For the second description, rotate about the origin by \(90^\circ\) counterclockwise: \(A'=(-1,1)\), \(B'=(-1,3)\), and \(C'=(-2,1)\). The common translation vector to \(D,E,F\) is \(\langle-2,1\rangle\). 5. A translation keeps \(AB\) horizontal, while \(DE\) is vertical. A reflection reverses orientation, but the ordered triangles \(ABC\) and \(DEF\) both run counterclockwise. Neither single move works.

a) Rotate \(90^\circ\) counterclockwise about \(S=(-\frac{3}{2},-\frac{1}{2})\). It sends \(A\to D\), \(B\to E\), and \(C\to F\). b) Rotate \(90^\circ\) counterclockwise about the origin, then translate by \(\langle-2,1\rangle\). c) A translation cannot turn a horizontal side vertical. A reflection would reverse the triangle's orientation, but the matching triangles have the same orientation.

- Follow the horizontal coordinate through each mirror line. The vertical coordinate stays the same. - Does the final movement depend on the locations of both lines or on their separation? - Consider what changes when you swap the order of the two mirror lines. - For the last part, start with \((x,y)\) instead of a particular point.

1. Reflection across \(x=a\) sends \((x,y)\) to \((2a-x,y)\). Reflection of that image across \(x=b\) gives \((2b-(2a-x),y)=(x+2(b-a),y)\). 2. Since \(P\) moves from \(x=-2\) to \(x=6\), \(2(b-a)=8\) and \(b-a=4\). For example, \((a,b)=(0,4)\) or \((1,5)\); both work. 3. Every point moves \(8\) units right, so \(Q''=(9,4)\). 4. Reversing the order gives the horizontal change \(2(a-b)=-8\), sending \(P(-2,1)\) to \((-10,1)\). 5. The expression \((x+2(b-a),y)\) shows that the change depends only on the separation and order of the mirror lines, not on the starting point. Thus the combined motion is a translation.

a) \(b-a=4\); for example, \((a,b)=(0,4)\) and \((1,5)\). b) \(Q''=(9,4)\). c) \((-10,1)\). d) The two reflections send any \((x,y)\) to \((x+2(b-a),y)=(x+8,y)\), an \(8\)-unit translation to the right.