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Sequences of rigid motions

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55104110
Point \(A\) has been translated \(4\) units to the right to point \(A'\), as shown. Complete the two-step sequence by reflecting \(A'\) across the \(x\)-axis. Give the coordinates of the final point \(A''\).
Figure for problem 551041

Hints

- 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 \(A'\).

Solution

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

Answer

\(A''=(1,-2)\)
55104210
The coordinate grid shows triangle \(ABC\), an intermediate image \(A'B'C'\), and the final image \(A''B''C''\). Identify the two transformations, in order, that map \(ABC\) to \(A''B''C''\).
Figure for problem 551042

Hints

- 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.

Solution

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.

Answer

First translate \(4\) units to the right; then reflect across the \(x\)-axis.
55104410
The diagram shows triangle \(ABC\), its reflection \(A'B'C'\) across the dashed line \(PQ\), and triangle \(DEF\), obtained by translating \(A'B'C'\) so that \(A'\) maps to \(D\), \(B'\) maps to \(E\), and \(C'\) maps to \(F\). Without using SSS, SAS, ASA, or AAS, explain why \(\triangle ABC\cong\triangle DEF\). Your explanation must use the definition of congruence in terms of rigid motions.
Figure for problem 551044

Hints

- 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.

Solution

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, \(\triangle ABC\cong\triangle DEF\).

Answer

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.
55501710
Parallel lines \(g\) and \(h\) are \(3\) units apart, with \(h\) above \(g\), as shown. A figure is reflected across \(g\) and then across \(h\). What single rigid motion is equivalent to these two reflections? State its distance and direction.
Figure for problem 555017

Hints

- 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.

Solution

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.

Answer

A translation \(6\) units upward, perpendicular to \(g\) and \(h\).
51319110
The line \(s\) is given by \(s(x) = \frac{2}{3}x + 5\). 1. Reflect \(s\) across the y-axis. Call the image \(t\) and write its equation. 2. Reflect \(t\) across the x-axis. Call the new image \(u\) and write its equation. 3. Compare \(s\) and \(u\). What single rigid motion maps \(s\) directly to \(u\)?

Hints

- 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?

Solution

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 \(u\).

Answer

1. \(t(x) = -\frac{2}{3}x + 5\) 2. \(u(x) = \frac{2}{3}x - 5\) 3. A \(180^\circ\) rotation about the origin
53216810
The graph shows quadrilateral \(ABCD\) and its image \(A''B''C''D''\) after a two-step rigid-motion sequence. a) Read the coordinates of the vertices of both quadrilaterals from the graph. b) Which rule maps the coordinates \((x, y)\) of a point on \(ABCD\) to the coordinates \((x'', y'')\) of the corresponding point on \(A''B''C''D''\)? - **Rule I:** \((x, y)\to(x-5, y-3)\) - **Rule II:** \((x, y)\to(-x, y-3)\) - **Rule III:** \((x, y)\to(-x, -y)\) - **Rule IV:** \((x, y)\to(x-3, -y)\) c) Describe two basic transformations that can be performed in sequence to map \(ABCD\) onto \(A''B''C''D''\).
Figure for problem 532168

Hints

- 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.

Solution

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.

Answer

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.
55104310
The coordinate grid shows triangle \(ABC\) and its final image \(A''B''C''\). Give a sequence consisting of a translation followed by a rotation about the origin that maps \(ABC\) to \(A''B''C''\). State the translation vector and the rotation angle and direction.
Figure for problem 551043

Hints

- 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.

Solution

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.

Answer

Translate by \(\langle 3,1\rangle\), then rotate \(90^\circ\) counterclockwise about the origin.
55104510
The coordinate grid shows triangle \(ABC\) and the final image \(A''B''C''\). A two-step sequence was used: 1. Translate \(ABC\) by an unknown vector to obtain \(A'B'C'\). 2. Reflect \(A'B'C'\) across the \(y\)-axis to obtain \(A''B''C''\). Determine the translation vector used in the first step. Explain how you worked backward from the final image.
Figure for problem 551045

Hints

- 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.

Solution

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 \(\langle 2,3\rangle\).

Answer

The translation vector is \(\langle 2,3\rangle\).
55501810
Lines \(g\) and \(h\) intersect at \(O\). The directed acute angle from \(g\) to \(h\) is marked in the diagram. A figure is reflected across \(g\) and then across \(h\). What single rigid motion is equivalent to the two reflections? State its center, angle, and direction.
Figure for problem 555018

Hints

- 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.

Solution

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 \(O\).

Answer

A rotation of \(80^\circ\) counterclockwise about \(O\).
55501910
Point \(P\) is shown on the coordinate grid. Let \(R\) be reflection across the \(y\)-axis, and let \(T\) be translation by \(\langle3,0\rangle\). Find the final image of \(P\) under \(T\circ R\) and under \(R\circ T\). What do the results show about the order of these transformations?
Figure for problem 555019

Hints

- 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.

Solution

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.

Answer

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

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