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Rigid motions and congruence

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5521898
The graph shows segment \(\overline{AB}\) and its image \(\overline{A'B'}\). What single rigid motion maps \(\overline{AB}\) to \(\overline{A'B'}\)? State the motion completely.
Figure for problem 552189

Hints

- Compare each point with its corresponding primed point. - Does every point move by the same horizontal and vertical amount?

Solution

1. Point \(A(0,1)\) maps to \(A'(3,3)\), a change of \(3\) units right and \(2\) units up. 2. Point \(B(2,2)\) maps to \(B'(5,4)\), the same change. 3. Therefore, the motion is a translation \(3\) units right and \(2\) units up.

Answer

A translation \(3\) units right and \(2\) units up.
5521908
The diagram shows segment \(\overline{PQ}\) and its image \(\overline{RS}\) after a quarter-turn about \(O\). Is the rotation clockwise or counterclockwise? State the complete rotation.
Figure for problem 552190

Hints

- Track point \(P\) and its image \(R\) around the center \(O\). - Use the marked quarter-turn to decide the direction.

Solution

1. Point \(P\) rotates from the ray to the right of \(O\) to point \(R\) above \(O\). 2. That turn is \(90^\circ\) counterclockwise, and the same turn maps \(Q\) to \(S\).

Answer

A \(90^\circ\) counterclockwise rotation about \(O\).
5121488
Point \(A'\) with coordinates \((-5.2, 8)\) is the image of point \(A\) after a reflection across the \(y\)-axis. Find the coordinates of the original point \(A\). Justify your answer by describing the general rule for reflecting a point across the \(y\)-axis.

Hints

- Think about what happens when a point is flipped from one side of the vertical axis to the other. - Which coordinate controls left and right, and which coordinate controls up and down? - Apply the reflection rule in reverse to move from the image back to the original point.

Solution

1. A reflection across the \(y\)-axis changes the sign of the \(x\)-coordinate and leaves the \(y\)-coordinate unchanged: \((x, y) \to (-x, y)\). 2. Since the image has \(x\)-coordinate \(-5.2\), the original \(x\)-coordinate is \(5.2\). 3. The \(y\)-coordinate remains \(8\). 4. Therefore, the original point is \(A(5.2, 8)\).

Answer

The original point is \(A(5.2, 8)\). A reflection across the \(y\)-axis changes \((x, y)\) to \((-x, y)\).
5123858
A \(180^\circ\) rotation about \(Z(3, 3)\) maps \(A(1, 1)\) to \(C\) and \(B(4, 2)\) to \(D\). For each point, state the displacement from \(Z\), reverse that displacement for the half-turn, and find the coordinates of \(C\) and \(D\).

Hints

- Measure each point from the rotation center \(Z\), not from the origin. - What happens to a displacement vector under a half-turn? - After changing the displacement, add it back to the coordinates of \(Z\).

Solution

1. From \(Z(3, 3)\) to \(A(1, 1)\), the displacement is \((-2, -2)\). 2. A \(180^\circ\) rotation reverses that displacement to \((2, 2)\), so \(C=(3+2, 3+2)=(5, 5)\). 3. From \(Z(3, 3)\) to \(B(4, 2)\), the displacement is \((1, -1)\). 4. Reversing it gives \((-1, 1)\), so \(D=(3-1, 3+1)=(2, 4)\).

Answer

For \(A\): displacement \((-2, -2)\) becomes \((2, 2)\), so \(C(5, 5)\). For \(B\): displacement \((1, -1)\) becomes \((-1, 1)\), so \(D(2, 4)\).
5123888
A quadrilateral has \(180^\circ\) rotational symmetry about the point \(S\), where its diagonals intersect. a) What does this symmetry imply about the two parts of each diagonal? b) Which special types of quadrilaterals have this property? Name at least three examples.

Hints

- Think about what a \(180^\circ\) rotation about the midpoint of a segment does to its endpoints. - How must opposite vertices be positioned for the whole quadrilateral to map onto itself? - Recall the diagonal property that identifies parallelograms.

Solution

1. A \(180^\circ\) rotation about \(S\) maps each vertex to the opposite vertex, so \(S\) is the midpoint of both diagonals. 2. Thus \(\overline{AS}=\overline{CS}\) and \(\overline{BS}=\overline{DS}\); the diagonals bisect each other. 3. A quadrilateral whose diagonals bisect each other is a parallelogram. 4. Examples include a parallelogram, rectangle, rhombus, and square.

Answer

a) The diagonals bisect each other. b) Examples include a parallelogram, rectangle, rhombus, and square.
5123918
Match each quadrilateral with its symmetry description. Use the usual general form of each quadrilateral, and treat the kite as a kite that is not a rhombus. **Quadrilaterals:** 1. Square 2. Parallelogram 3. Kite 4. Isosceles trapezoid **Descriptions:** A: The quadrilateral has \(180^\circ\) rotational symmetry but generally has no line of symmetry. B: The quadrilateral has exactly one line of symmetry through two opposite vertices. C: The quadrilateral has four lines of symmetry and \(180^\circ\) rotational symmetry. D: The quadrilateral has exactly one line of symmetry through the midpoints of two opposite sides.

Hints

- For each quadrilateral, determine how many lines of symmetry it has. - Distinguish line symmetry from \(180^\circ\) rotational symmetry. - Imagine folding each quadrilateral along a possible symmetry line.

Solution

1. A square has four lines of symmetry and \(180^\circ\) rotational symmetry, so it matches C. 2. A general parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals but no line of symmetry, so it matches A. 3. A kite that is not a rhombus has one line of symmetry along a diagonal through two opposite vertices, so it matches B. 4. An isosceles trapezoid has one line of symmetry perpendicular to its parallel bases and through their midpoints, so it matches D.

Answer

1. C 2. A 3. B 4. D
5123944
Identify quadrilaterals from their lines of symmetry. a) Which quadrilateral has exactly two lines of symmetry, both passing through opposite vertices? b) Which quadrilateral has exactly two lines of symmetry, both passing through the midpoints of opposite sides? c) Which quadrilateral has four lines of symmetry? Describe where those lines are located.

Hints

- Imagine folding each quadrilateral so that its edges match exactly. - Picture the quadrilaterals and test possible fold lines. - Distinguish lines through vertices from lines through side midpoints. - Which quadrilateral has the greatest number of matching fold lines?

Solution

1. A rhombus that is not a square has exactly two lines of symmetry: its diagonals. 2. A rectangle that is not a square has exactly two lines of symmetry. Each passes through the midpoints of a pair of opposite sides. 3. A square has four lines of symmetry: its two diagonals and the two lines through the midpoints of opposite sides.

Answer

a) A rhombus that is not a square. b) A rectangle that is not a square. c) A square. Its lines of symmetry are the two diagonals and the two lines through the midpoints of opposite sides.
5123958
A general parallelogram that is neither a rectangle nor a rhombus and an isosceles trapezoid that is not a parallelogram have different symmetries. Explain which quadrilateral has \(180^\circ\) rotational symmetry and which has line symmetry. Then name one property of the diagonals that every isosceles trapezoid has but a general parallelogram does not.

Hints

- A figure has \(180^\circ\) rotational symmetry if a half-turn maps it onto itself. - A figure has line symmetry if reflection across a line maps it onto itself. - Picture a slanted parallelogram and an isosceles trapezoid, then compare their diagonals.

Solution

1. A general parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals, but it has no line of symmetry. 2. An isosceles trapezoid that is not a parallelogram has one line of symmetry through the midpoints of its parallel bases. 3. The diagonals of an isosceles trapezoid are congruent. The diagonals of a general parallelogram are not generally congruent; that occurs only in the special case of a rectangle.

Answer

The general parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals. The isosceles trapezoid has line symmetry through the midpoints of its parallel bases. Its diagonals are congruent, unlike those of a general parallelogram.
5123985
Use the inclusive definition that a kite is any quadrilateral with two pairs of adjacent congruent sides. a) What broad family of quadrilaterals is characterized by having at least one line of symmetry through two opposite vertices? b) How many lines of symmetry does a rectangle that is not a square have? Briefly describe their locations.

Hints

- Which quadrilaterals can be folded across a diagonal so that the two halves match? - What side-length relationships result when two vertices reflect onto each other? - Imagine folding a rectangle so that its vertices match. Where are the fold lines?

Solution

1. If a line of symmetry passes through two opposite vertices, the other two vertices are reflections of each other. 2. This creates two pairs of adjacent congruent sides, so the quadrilateral belongs to the kite family. 3. A rectangle that is not a square has exactly two lines of symmetry. 4. Each of those lines passes through the midpoints of a pair of opposite sides.

Answer

a) The kite family. b) Two lines of symmetry, each through the midpoints of a pair of opposite sides.
5124038
Determine whether each statement is true or false. Justify each answer. a) Every rectangle has \(180^\circ\) rotational symmetry about the intersection of its diagonals. b) Every rhombus has exactly two lines of symmetry. c) Every parallelogram has line symmetry. d) A square has four lines of symmetry.

Hints

- Test whether each figure maps onto itself under a reflection or a \(180^\circ\) rotation. - Check special cases. A square is also a rectangle and a rhombus. - Count the ways a square can be folded so that its halves match.

Solution

1. Statement a) is true. A rectangle is a parallelogram, and every parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals. 2. Statement b) is false. A non-square rhombus has two lines of symmetry, but a square is also a rhombus and has four. 3. Statement c) is false. A general parallelogram has no line of symmetry; only special parallelograms such as rectangles and rhombi do. 4. Statement d) is true. A square has two diagonal lines of symmetry and two lines through the midpoints of opposite sides.

Answer

a) True. b) False; a square is a rhombus with four lines of symmetry. c) False; a general parallelogram has no line of symmetry. d) True.
5174278
Point \(A(35, 42)\) is reflected across both coordinate axes in sequence. First, \(A\) is reflected across the x-axis to create \(A'\). Then \(A'\) is reflected across the y-axis to create \(A''\). Find the coordinates of \(A'\) and \(A''\).

Hints

- Which coordinate changes sign in a reflection across the x-axis? - Which coordinate changes sign in a reflection across the y-axis? - Perform the reflections one at a time, using the first image as the input for the second reflection.

Solution

1. A reflection across the x-axis keeps the x-coordinate and changes the sign of the y-coordinate. Therefore, \(A' = (35, -42)\). 2. A reflection across the y-axis keeps the y-coordinate and changes the sign of the x-coordinate. Therefore, \(A'' = (-35, -42)\).

Answer

\(A' = (35, -42)\) and \(A'' = (-35, -42)\)
5174288
For each pair, the second point is the reflection of the first point across one coordinate axis. State whether the reflection is across the x-axis or the y-axis. a) \(P(12, 5)\rightarrow P'(12, -5)\) b) \(Q(-8, 20)\rightarrow Q'(8, 20)\) c) \(R(-15, -3)\rightarrow R'(-15, 3)\)

Hints

- Compare the coordinates in each pair and identify which one stayed the same. - A reflection across the x-axis changes the sign of which coordinate? - The reflecting axis lies halfway between a point and its image.

Solution

1. For \(P\), the \(x\)-coordinate stays \(12\) and the \(y\)-coordinate changes sign, so the reflection is across the x-axis. 2. For \(Q\), the \(y\)-coordinate stays \(20\) and the \(x\)-coordinate changes sign, so the reflection is across the y-axis. 3. For \(R\), the \(x\)-coordinate stays \(-15\) and the \(y\)-coordinate changes sign, so the reflection is across the x-axis.

Answer

a) x-axis b) y-axis c) x-axis
5174298
An unknown point \(Q\) is reflected across the y-axis, producing the image \(Q'(-22, 14)\). Find the coordinates of the original point \(Q\).

Hints

- A reflection can be undone by reflecting across the same axis again. - Which coordinate changes in a reflection across the y-axis? - Keep the unchanged coordinate and reverse the sign of the other coordinate.

Solution

1. A reflection across the y-axis changes the sign of the x-coordinate and keeps the y-coordinate unchanged. 2. Since the image has x-coordinate \(-22\), the original point has x-coordinate \(22\). 3. The y-coordinate remains \(14\), so \(Q = (22, 14)\).

Answer

\(Q = (22, 14)\)
5191078
Kite \(ABCD\) is reflected across diagonal \(\overline{AC}\), and the reflection maps \(B\) to \(D\). Do not use a memorized kite-diagonal rule. Use the definition of a reflection to explain why \(\overline{AC}\) is the perpendicular bisector of \(\overline{BD}\). Then state the two resulting relationships between the diagonals.

Hints

- What geometric relationship does a reflection line have to the segment joining a point and its image? - Apply that relationship to \(B\) and \(D\). - Translate “perpendicular bisector” into two separate statements about the diagonals.

Solution

1. Since reflection across \(\overline{AC}\) maps \(B\) to \(D\), every point on the reflection line \(\overline{AC}\) is equidistant from \(B\) and \(D\). 2. Therefore, \(\overline{AC}\) is the perpendicular bisector of \(\overline{BD}\). 3. Hence \(\overline{AC}\perp\overline{BD}\), and \(\overline{AC}\) passes through the midpoint of \(\overline{BD}\).

Answer

Because reflection across \(\overline{AC}\) maps \(B\) to \(D\), \(\overline{AC}\) is the perpendicular bisector of \(\overline{BD}\). Thus the diagonals are perpendicular, and \(\overline{AC}\) bisects \(\overline{BD}\).
5241548
Points \(A(3.2, 4.5)\), \(B(1.5, 7)\), and \(C(3.2, 1.2)\) are in the coordinate plane. a) Which point is highest? Explain briefly. b) Which two points lie on the same vertical line? c) Translate point \(B\) by moving it \(1.7\) units right and \(2.5\) units down. Find the coordinates of \(B'\). Which original point is at that location?

Hints

- The y-coordinate tells how high a point is. - Points on a vertical line share an x-coordinate. - Moving right increases x, and moving down decreases y.

Solution

1. Point \(B\) is highest because its y-coordinate, \(7\), is greater than \(4.5\) and \(1.2\). 2. Points \(A\) and \(C\) share the x-coordinate \(3.2\), so they lie on the same vertical line. 3. Translating \(B(1.5, 7)\) gives \(B'(1.5+1.7, 7-2.5)=(3.2, 4.5)\), which is the location of \(A\).

Answer

a) Point \(B\) b) Points \(A\) and \(C\) c) \(B'=(3.2, 4.5)\), which is the same location as \(A\).
5241688
A quadrilateral has vertices \(A(2, 2)\), \(B(6, 2)\), \(C(6, 6)\), and \(D(2, 6)\). a) Classify the quadrilateral. Justify your answer using side lengths and the directions of its sides. b) Point \(A\) is translated \(3\) units up. Find the coordinates of \(A'\). c) The entire quadrilateral is translated \(3\) units up. Find the coordinates of all four image points.

Hints

- A square has four equal sides and four right angles. - Moving a point up changes only its y-coordinate. - Apply the same coordinate change to every vertex.

Solution

1. Each side has length \(4\) units, and adjacent sides are horizontal and vertical. Therefore, the quadrilateral is a square. 2. Moving \(A(2, 2)\) up \(3\) units gives \(A'(2, 5)\). 3. Adding \(3\) to every y-coordinate gives \(A'(2, 5)\), \(B'(6, 5)\), \(C'(6, 9)\), and \(D'(2, 9)\).

Answer

a) The quadrilateral is a square. b) \(A'=(2, 5)\) c) \(A'=(2, 5)\), \(B'=(6, 5)\), \(C'=(6, 9)\), and \(D'=(2, 9)\)
5321858
Quadrilateral \(ABCD\) has vertices \(A(1, 2)\), \(B(4, 1)\), \(C(3, 4)\), and \(D(1, 3)\). It will be rotated \(180^\circ\) about the origin \(O(0, 0)\). The image point \(A'\) is already shown. a) Find the coordinates of \(B'\), \(C'\), and \(D'\). b) Describe in words or with a rule how to find the image \(P'(x', y')\) of any point \(P(x, y)\) after a \(180^\circ\) rotation about the origin.
Figure for problem 532185

Hints

- Which point is halfway between \(A\) and its shown image \(A'\)? - Describe how \(A\) and \(A'\) are positioned relative to the origin. - Apply the same coordinate change to \(B\), \(C\), and \(D\). - Generalize the pattern for \(P(x, y)\).

Solution

1. A \(180^\circ\) rotation about the origin changes both coordinate signs, so \(B(4, 1)\to B'(-4, -1)\), \(C(3, 4)\to C'(-3, -4)\), and \(D(1, 3)\to D'(-1, -3)\). 2. In general, the rule is \((x, y)\to(-x, -y)\).

Answer

a) \(B'(-4, -1)\), \(C'(-3, -4)\), and \(D'(-1, -3)\) b) Change the sign of both coordinates: \((x, y)\to(-x, -y)\).
5331418
The graph shows triangle \(PQR\). For each vertex, multiply only the \(x\)-coordinate by \(-1\) and leave the \(y\)-coordinate unchanged. a) Give the coordinates of \(P'\), \(Q'\), and \(R'\). b) What geometric transformation does this coordinate rule represent?
Figure for problem 533141

Hints

- What happens when the sign of the \(x\)-coordinate changes but the \(y\)-coordinate stays the same? - Does each point remain the same distance from the y-axis? - Which axis acts as the line of reflection?

Solution

1. The graph gives \(P(1, 1)\), \(Q(5, 2)\), and \(R(2, 4)\). Applying \((x, y)\to(-x, y)\) gives \(P'(-1, 1)\), \(Q'(-5, 2)\), and \(R'(-2, 4)\). 2. Only the horizontal position changes, and each point remains the same distance from the y-axis. The rule is a reflection across the y-axis.

Answer

a) \(P'(-1, 1)\), \(Q'(-5, 2)\), and \(R'(-2, 4)\) b) A reflection across the y-axis.
5331428
The graph shows a figure in Quadrant II. What happens geometrically when you multiply only the \(y\)-coordinate of every vertex by \(-1\)? Explain using the coordinates of points \(A\) through \(F\).
Figure for problem 533142

Hints

- First record the coordinates of all six vertices. - How does changing a positive \(y\)-coordinate to its opposite affect vertical position? - Which axis would act as the fold line?

Solution

1. The vertices are \(A(-5, 1)\), \(B(-2, 1)\), \(C(-2, 2)\), \(D(-4, 2)\), \(E(-4, 4)\), and \(F(-5, 4)\). 2. Apply \((x, y)\to(x, -y)\): \(A'(-5, -1)\), \(B'(-2, -1)\), \(C'(-2, -2)\), \(D'(-4, -2)\), \(E'(-4, -4)\), and \(F'(-5, -4)\). 3. The \(x\)-coordinates stay the same while the vertical positions reverse. This is a reflection across the \(x\)-axis.

Answer

Multiplying only the \(y\)-coordinate by \(-1\) reflects the figure across the \(x\)-axis. The image vertices are \(A'(-5, -1)\), \(B'(-2, -1)\), \(C'(-2, -2)\), \(D'(-4, -2)\), \(E'(-4, -4)\), and \(F'(-5, -4)\).
5368388
In parallelogram \(ABCD\), diagonal \(\overline{AC}\) is drawn. Perpendicular segments from \(B\) and \(D\) meet the diagonal at \(E\) and \(F\), respectively. Use the labeled length in the diagram to find \(DF\). Justify your answer.
Figure for problem 536838

Hints

- What rotational symmetry does a parallelogram have? - Where does point \(B\) map under a \(180^\circ\) rotation about the diagonal intersection? - What happens to lengths and right angles under a rotation?

Solution

1. A parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals. 2. Under this rotation, \(B\) maps to \(D\), and diagonal \(\overline{AC}\) maps to itself. Therefore, perpendicular segment \(\overline{BE}\) maps to perpendicular segment \(\overline{DF}\). 3. Rotations preserve length, so \(DF=BE=4.5\,\text{cm}\).

Answer

\(DF=4.5\,\text{cm}\)
5368968
Determine whether translations \(u\) and \(v\) are the same. Translation \(u\) maps \(A(1, 1)\) to \(B(4, 3)\). Translation \(v\) maps \(C(3, 2)\) to \(D(6, 4)\).

Hints

- Find the horizontal and vertical change for each translation. - Compare the two translation vectors.

Solution

1. Translation \(u\) changes the coordinates by \((4-1,3-1)=(3,2)\). 2. Translation \(v\) changes the coordinates by \((6-3,4-2)=(3,2)\). 3. Both translations move every point \(3\) units right and \(2\) units up, so they are the same translation.

Answer

Yes. Both translations have vector \((3, 2)\).
5371558
Quadrilateral \(PQRS\) is a kite, and diagonal \(\overline{PR}\) is its line of reflection. The diagonals intersect at \(M\). a) Under reflection across \(\overline{PR}\), where does point \(Q\) map? b) Use that reflection and the labeled length in the diagram to find \(QS\).
Figure for problem 537155

Hints

- Which two vertices lie on opposite sides of the reflection line \(\overline{PR}\)? - What relationship does a reflection line have to the segment joining a point and its image? - Use the diagram's labeled half only after identifying the mapping.

Solution

1. Reflection across \(\overline{PR}\) maps \(Q\) to \(S\). 2. The reflection line is the perpendicular bisector of the segment joining a point and its image, so \(M\) is the midpoint of \(\overline{QS}\). 3. The diagram gives \(QM=3.5\,\text{cm}\), so \(MS=3.5\,\text{cm}\). 4. Therefore, \(QS=3.5\,\text{cm}+3.5\,\text{cm}=7\,\text{cm}\).

Answer

a) \(Q\) maps to \(S\). b) \(QS=7\,\text{cm}\).
5512338
Triangle \(ABC\) maps to triangle \(A_1B_1C_1\) by the translation shown. a) If \(AB=5\,\text{cm}\), find \(A_1B_1\). b) If \(\angle C=48^\circ\), find \(\angle C_1\). c) What relationship must \(\overline{AB}\) and \(\overline{A_1B_1}\) have besides equal length?
Figure for problem 551233

Hints

- Identify which measurements a rigid motion preserves. - Match each original vertex with its translated image. - Think about the direction of a segment before and after a translation.

Solution

1. A translation is a rigid motion, so it preserves length. Therefore, \(A_1B_1=5\,\text{cm}\). 2. A translation preserves angle measure, so \(\angle C_1=48^\circ\). 3. A nonzero translation maps a segment to a parallel segment, so \(\overline{AB}\parallel\overline{A_1B_1}\).

Answer

a) \(5\,\text{cm}\) b) \(48^\circ\) c) \(\overline{AB}\parallel\overline{A_1B_1}\)
5546078
Triangle \(ABC\) is shown on the coordinate plane. a) Rotate the original triangle \(90^\circ\) counterclockwise about the origin. Give the coordinates of \(A'\), \(B'\), and \(C'\). b) Rotate the original triangle \(270^\circ\) counterclockwise about the origin. Give the coordinates of \(A''\), \(B''\), and \(C''\). c) State the coordinate rule for each rotation.
Figure for problem 554607

Hints

- Read the original coordinates from the graph before rotating anything. - Track how a point's horizontal and vertical positions change after a quarter-turn around the origin. - Compare several original-image pairs before stating a general coordinate rule.

Solution

1. From the graph, \(A(1, 2)\), \(B(4, 2)\), and \(C(3, 5)\). 2. A \(90^\circ\) counterclockwise rotation sends \((x, y)\) to \((-y, x)\). Thus \(A'(-2, 1)\), \(B'(-2, 4)\), and \(C'(-5, 3)\). 3. A \(270^\circ\) counterclockwise rotation sends \((x, y)\) to \((y, -x)\). Thus \(A''(2, -1)\), \(B''(2, -4)\), and \(C''(5, -3)\).

Answer

a) \(A'(-2, 1)\), \(B'(-2, 4)\), \(C'(-5, 3)\) b) \(A''(2, -1)\), \(B''(2, -4)\), \(C''(5, -3)\) c) \(90^\circ\) counterclockwise: \((x, y)\to(-y, x)\); \(270^\circ\) counterclockwise: \((x, y)\to(y, -x)\)
5546088
Triangle \(ABC\) and the line \(y=x\) are shown on the coordinate plane. Reflect the triangle across \(y=x\). a) Give the coordinates of \(A'\), \(B'\), and \(C'\). b) Which vertex stays fixed, and why?
Figure for problem 554608

Hints

- Read the three original coordinates from the graph. - Compare the roles of the x-coordinate and y-coordinate when a point is mirrored across \(y=x\). - What happens to a point that already lies on the reflection line?

Solution

1. From the graph, \(A(-4, 1)\), \(B(-1, 3)\), and \(C(-2, -2)\). 2. Reflection across \(y=x\) interchanges the coordinates, so \((x, y)\to(y, x)\). 3. Therefore, \(A'(1, -4)\), \(B'(3, -1)\), and \(C'(-2, -2)\). 4. Vertex \(C\) stays fixed because \(C(-2, -2)\) lies on the line \(y=x\).

Answer

a) \(A'(1, -4)\), \(B'(3, -1)\), \(C'(-2, -2)\) b) \(C\) stays fixed because it lies on \(y=x\).
5121498
Point \(P(-12, 7)\) is reflected twice. First, it is reflected across the \(x\)-axis to create point \(P'\). Then \(P'\) is reflected across the \(y\)-axis to create point \(P''\). a) Find the coordinates of \(P'\) and \(P''\). b) Compare the coordinates of \(P\) and \(P''\). What single rigid motion maps \(P\) directly to \(P''\)?

Hints

- Work through the reflections in order. What changes when you reflect across the \(x\)-axis? What changes across the \(y\)-axis? - Compare the signs of both coordinates before and after the two reflections. - Which rigid motion about the origin changes the signs of both coordinates?

Solution

1. Reflecting \(P(-12, 7)\) across the \(x\)-axis changes the sign of the \(y\)-coordinate, so \(P'(-12, -7)\). 2. Reflecting \(P'(-12, -7)\) across the \(y\)-axis changes the sign of the \(x\)-coordinate, so \(P''(12, -7)\). 3. From \(P\) to \(P''\), both coordinate signs change. 4. The rule \((x, y)\to(-x, -y)\) is a \(180^\circ\) rotation about the origin.

Answer

a) \(P'(-12, -7)\) and \(P''(12, -7)\). b) A \(180^\circ\) rotation about the origin maps \(P\) directly to \(P''\).
5121508
Analyze these rigid motions in the coordinate plane. a) Point \(S(-6, 9)\) is rotated \(180^\circ\) about the origin. Give the coordinates of its image \(S'\). b) Point \(T\) lies on the \(y\)-axis and is reflected across the \(x\)-axis. Explain without drawing why the image \(T'\) must also lie on the \(y\)-axis. c) Which points in the coordinate plane remain fixed when they are reflected across the \(x\)-axis? Describe all such points.

Hints

- What coordinate condition describes a point on one of the axes? - When is a number equal to its opposite? - Imagine folding the coordinate plane along the line of reflection. Which points do not move?

Solution

1. A \(180^\circ\) rotation about the origin changes \((x, y)\) to \((-x, -y)\), so \(S(-6, 9)\to S'(6, -9)\). 2. Every point on the \(y\)-axis has \(x=0\). Reflection across the \(x\)-axis leaves the \(x\)-coordinate unchanged, so the image still has \(x=0\) and lies on the \(y\)-axis. 3. Reflection across the \(x\)-axis maps \((x, y)\) to \((x, -y)\). 4. A point is fixed when \(y=-y\), so \(y=0\). Thus every point on the \(x\)-axis is fixed.

Answer

a) \(S'(6, -9)\). b) A point on the \(y\)-axis has \(x=0\), and reflection across the \(x\)-axis does not change its \(x\)-coordinate. c) All points on the \(x\)-axis, or all points of the form \((x, 0)\), remain fixed.
5121528
Points \(P(2, 4)\) and \(Q(2, -2)\) are in the coordinate plane. A claim is made that the \(x\)-axis is the line of reflection that maps one point to the other. Determine whether the claim is correct. If it is not, find the equation of the actual line of reflection.

Hints

- What relationship must the line of reflection have to the segment joining a point and its image? - Find the point halfway between \(P\) and \(Q\). - Recall the equation of the \(x\)-axis.

Solution

1. A line of reflection must be the perpendicular bisector of the segment joining a point and its image. 2. Points \(P(2, 4)\) and \(Q(2, -2)\) have the same \(x\)-coordinate, so \(\overline{PQ}\) is vertical. Its perpendicular bisector is horizontal and has the form \(y=k\). 3. The midpoint has \(y\)-coordinate \(\frac{4+(-2)}{2}=1\). 4. Therefore, the line of reflection is \(y=1\), not the \(x\)-axis, whose equation is \(y=0\).

Answer

The claim is incorrect. The line of reflection is \(y=1\), the perpendicular bisector of \(\overline{PQ}\).
5121538
Rectangle \(ABCD\) has vertices \(A(1, 1)\), \(B(4, 1)\), \(C(4, 3)\), and \(D(1, 3)\). a) Reflect the rectangle across the line \(x=5\). Give the coordinates of \(A'\), \(B'\), \(C'\), and \(D'\). b) Find the area of the original rectangle and the reflected rectangle. What property of reflections does this illustrate?

Hints

- Find each vertex's horizontal distance from the line \(x=5\). - Place each image point the same distance on the other side of the line. - Use the rectangle area formula. - Consider whether a reflection changes side lengths.

Solution

1. A reflection across \(x=5\) places each image point the same horizontal distance on the other side of the line. 2. The reflected vertices are \(A'(9, 1)\), \(B'(6, 1)\), \(C'(6, 3)\), and \(D'(9, 3)\). 3. The original rectangle has width \(3\) and height \(2\), so its area is \(3\cdot2=6\) square units. 4. The image also has width \(3\) and height \(2\), so its area is \(6\) square units. A reflection is a rigid motion, so it preserves lengths and area.

Answer

a) \(A'(9, 1)\), \(B'(6, 1)\), \(C'(6, 3)\), and \(D'(9, 3)\). b) Each rectangle has area \(6\) square units. This illustrates that reflections preserve size and area.
5123308
Triangle \(ABC\) has vertices \(A(2, 1)\), \(B(6, 1)\), and \(C(4, 4)\). Reflect the triangle across the line \(x=7\), then translate the image \(3\) units up. a) Find the coordinates of the final vertices \(A_2\), \(B_2\), and \(C_2\). b) Explain why \(\triangle A_2B_2C_2\) is congruent to \(\triangle ABC\).

Hints

- For a reflection across a vertical line, compare each point's horizontal distance from the line. - What changes in an upward translation: the \(x\)-coordinate, the \(y\)-coordinate, or both? - Which properties are preserved by reflections and translations?

Solution

1. Reflection across \(x=7\) sends \((x, y)\) to \((14-x, y)\), giving \(A_1(12, 1)\), \(B_1(8, 1)\), and \(C_1(10, 4)\). 2. Translating \(3\) units up gives \(A_2(12, 4)\), \(B_2(8, 4)\), and \(C_2(10, 7)\). 3. Reflections and translations are rigid motions, so they preserve distances and angle measures. 4. Therefore, \(\triangle A_2B_2C_2\) is congruent to \(\triangle ABC\).

Answer

a) \(A_2(12, 4)\), \(B_2(8, 4)\), and \(C_2(10, 7)\) b) The triangles are congruent because the transformation sequence consists only of rigid motions.
5123878
Points \(P(2, 2)\) and \(R(6, 6)\) are opposite vertices of square \(PQRS\). The diagonals of a square bisect each other, have equal lengths, and are perpendicular. a) Find the coordinates of the point \(M\) where the diagonals intersect. b) Use a \(90^\circ\) rotation about \(M\) to find the two missing vertices. Give both possible assignments of the coordinates to \(Q\) and \(S\).

Hints

- Where is the intersection of the diagonals relative to opposite vertices of a square? - Compare the vector from \(M\) to one known vertex with the vectors from \(M\) to the missing vertices. - What coordinate change represents a quarter-turn of a vector about the origin?

Solution

1. \(M\) is the midpoint of \(\overline{PR}\), so \(M\left(\frac{2+6}{2},\frac{2+6}{2}\right)=(4, 4)\). 2. The vector from \(M\) to \(R\) is \((2, 2)\). 3. Rotating this vector \(90^\circ\) in the two possible directions gives \((-2, 2)\) and \((2, -2)\). 4. Adding these vectors to \(M\) gives the missing vertices \((2, 6)\) and \((6, 2)\). Either assignment to \(Q\) and \(S\) is valid.

Answer

a) \(M(4, 4)\). b) The missing vertices are \((6, 2)\) and \((2, 6)\). Either \(Q(6, 2), S(2, 6)\) or \(Q(2, 6), S(6, 2)\) is valid.
5123898
A kite that is not a rhombus has diagonals \(e\) and \(f\). a) Use line symmetry to explain why one diagonal bisects the other, but the two diagonals do not bisect each other. b) Which diagonal lies on the line of symmetry: the diagonal that bisects the other, or the diagonal that is bisected?

Hints

- Picture the kite and identify its line of symmetry. - What must be true of a segment whose endpoints are reflections across a line? - What additional symmetry would be needed for both diagonals to bisect each other?

Solution

1. Let diagonal \(e\) lie on the line of symmetry. 2. The endpoints of \(f\) are reflections of each other across \(e\), so \(e\) is the perpendicular bisector of \(f\). 3. Because the kite is not a rhombus, the intersection is not the midpoint of \(e\), so \(f\) does not bisect \(e\). 4. Therefore, the diagonal on the line of symmetry is the diagonal that bisects the other diagonal.

Answer

a) The diagonal on the line of symmetry is the perpendicular bisector of the other diagonal. The other diagonal does not bisect the symmetry diagonal. b) The diagonal on the line of symmetry is the one that bisects the other diagonal.
5123995
Use inclusive definitions: a kite has two pairs of adjacent congruent sides, and a trapezoid has at least one pair of parallel sides. a) A rectangle is a parallelogram with four right angles. What additional condition makes a rectangle a kite? What special quadrilateral results? b) Explain why every square can be considered a special isosceles trapezoid. Refer to its parallel sides and symmetry.

Hints

- What side-length property defines a kite? - What happens to a rectangle when adjacent sides are congruent? - What condition defines a trapezoid under the inclusive definition? - What makes a trapezoid isosceles?

Solution

1. For a rectangle to be a kite, it must have congruent adjacent sides. 2. Opposite sides of a rectangle are already congruent, so congruent adjacent sides make all four sides congruent. The rectangle is then a square. 3. A square has two pairs of parallel sides, so it meets the inclusive definition of a trapezoid. 4. For either pair chosen as bases, the other two sides are congruent and there is a line of symmetry perpendicular to the bases through their midpoints. Therefore, a square is a special isosceles trapezoid under the stated inclusive definition.

Answer

a) Adjacent sides must be congruent, making the rectangle a square. b) A square has at least one pair of parallel sides, congruent legs, and a line of symmetry perpendicular to the chosen bases. Under the inclusive definition, it is a special isosceles trapezoid.
5124008
Quadrilateral \(ABCD\) has both diagonals as lines of reflection. Reflection across \(\overline{AC}\) fixes \(A\) and \(C\) and swaps \(B\) with \(D\). Reflection across \(\overline{BD}\) fixes \(B\) and \(D\) and swaps \(A\) with \(C\). a) Use the first reflection to state two pairs of congruent sides. b) Use the second reflection to state two pairs of congruent sides. c) Combine the reflection results to classify \(ABCD\). Then state the \(180^\circ\) rotation that maps the quadrilateral onto itself.

Hints

- A reflection preserves distance from every fixed point on its reflection line. - Track the image of each endpoint of a side under each reflection. - Combine the side equalities only after deriving them from the two mappings.

Solution

1. Reflection across \(\overline{AC}\) maps \(B\) to \(D\) while fixing \(A\) and \(C\). Because reflections preserve length, \(AB=AD\) and \(CB=CD\). 2. Reflection across \(\overline{BD}\) maps \(A\) to \(C\) while fixing \(B\) and \(D\). Therefore, \(AB=CB\) and \(AD=CD\). 3. Combining these equalities shows \(AB=BC=CD=DA\), so \(ABCD\) is a rhombus. 4. Every rhombus is a parallelogram, so a \(180^\circ\) rotation about the intersection of its diagonals maps \(A\) to \(C\) and \(B\) to \(D\).

Answer

a) \(AB=AD\) and \(CB=CD\). b) \(AB=CB\) and \(AD=CD\). c) \(ABCD\) is a rhombus. A \(180^\circ\) rotation about the intersection of its diagonals maps \(A\leftrightarrow C\) and \(B\leftrightarrow D\).
5124018
Use the inclusive definition that a trapezoid has at least one pair of parallel sides. a) An isosceles trapezoid has a line of symmetry through the midpoints of its parallel sides. If it is also a parallelogram, what special type of quadrilateral must it be? b) How many lines of symmetry does a general parallelogram that is neither a rhombus nor a rectangle have? c) Name a quadrilateral that has line symmetry but does not have \(180^\circ\) rotational symmetry.

Hints

- Combine the angle properties of a parallelogram and an isosceles trapezoid. - Test whether a slanted parallelogram can be folded onto itself. - Think of a figure that can be reflected onto itself but not rotated \(180^\circ\) onto itself.

Solution

1. In a parallelogram, consecutive angles are supplementary. In an isosceles trapezoid, each pair of base angles is congruent. 2. Congruent supplementary angles are each \(90^\circ\), so a figure that is both an isosceles trapezoid and a parallelogram must be a rectangle. 3. A general parallelogram that is neither a rectangle nor a rhombus has no lines of symmetry. 4. A kite that is not a rhombus is one example of a quadrilateral with line symmetry but no \(180^\circ\) rotational symmetry. An isosceles trapezoid that is not a rectangle is another.

Answer

a) A rectangle. b) \(0\) lines of symmetry. c) One example is a kite that is not a rhombus or an isosceles trapezoid that is not a rectangle.
5124158
A rectangle and an isosceles trapezoid that is not a parallelogram each have a line of reflection that swaps one diagonal with the other. a) What does that reflection prove about the two diagonals in each figure? Name the preserved quantity that justifies your conclusion. b) A \(180^\circ\) rotation about the intersection of the diagonals maps a rectangle onto itself. Explain what this forces the intersection point to be for each diagonal. c) Suppose a quadrilateral has the reflection property from part a), the half-turn property from part b), and perpendicular diagonals. State the three resulting diagonal properties.

Hints

- In part a), focus on what happens when an entire segment is reflected onto another segment. - In part b), ask where the center of a half-turn lies relative to a point and its image. - Keep the reflection conclusion, half-turn conclusion, and given perpendicularity separate before combining them.

Solution

1. A reflection that maps one diagonal to the other preserves length, so the two diagonals are congruent in each figure. 2. In the rectangle, the \(180^\circ\) rotation swaps the endpoints of each diagonal. The rotation center must therefore be the midpoint of each diagonal, so the diagonals bisect each other. 3. Combining the two motion-based conclusions with the given perpendicularity gives three properties: the diagonals are congruent, they bisect each other, and they are perpendicular.

Answer

a) The diagonals are congruent because reflections preserve length. b) Their intersection is the midpoint of each diagonal, so the diagonals bisect each other. c) The diagonals are congruent, bisect each other, and are perpendicular.
5191088
Points \(A\) and \(C\) are the endpoints of one diagonal of a kite. Segment \(\overline{AC}\) is the kite’s line of symmetry. Describe how the other two vertices, \(B\) and \(D\), must be positioned relative to \(\overline{AC}\) and to each other.

Hints

- Think about how a line of symmetry maps one vertex to another. - Where must \(B\) and \(D\) lie so they are mirror images? - Decide where the two diagonals may intersect.

Solution

1. Choose a point \(P\) in the interior of \(\overline{AC}\). 2. Points \(B\) and \(D\) must lie on the line through \(P\) perpendicular to \(\overline{AC}\), on opposite sides of \(\overline{AC}\), with \(PB=PD\). 3. Then reflection across \(\overline{AC}\) maps \(B\) to \(D\), so \(ABCD\) is a kite.

Answer

Choose an interior point \(P\) on \(\overline{AC}\). Points \(B\) and \(D\) must lie on opposite sides of \(\overline{AC}\) along the perpendicular through \(P\), with \(PB=PD\).
5191438
Points \(P(5, 2)\), \(Q(9, 6)\), and \(R(5, 12)\) are three vertices of quadrilateral \(PQRS\). a) Find the coordinates of point \(S\) so that \(PQRS\) is a kite. b) Is the kite also a rhombus? Justify your answer by comparing the lengths of \(\overline{PQ}\) and \(\overline{QR}\).

Hints

- Look for a line through \(P\) and \(R\) that can be a line of symmetry. - Reflect \(Q\) across that line. - A rhombus has four congruent sides. - Compare the horizontal and vertical changes along \(\overline{PQ}\) and \(\overline{QR}\).

Solution

1. Points \(P\) and \(R\) lie on \(x=5\), so \(x=5\) can be the kite's line of symmetry. Point \(Q(9, 6)\) is \(4\) units right of this line, so its reflection is \(S(1, 6)\). 2. Compare the adjacent side lengths using their squares: \(PQ^2=4^2+4^2=32\) and \(QR^2=4^2+6^2=52\). 3. Since \(PQ\ne QR\), the kite does not have four congruent sides, so it is not a rhombus.

Answer

a) \(S(1, 6)\) b) No. Since \(PQ^2=32\) and \(QR^2=52\), the adjacent sides are not congruent, so the kite is not a rhombus.
5371568
Isosceles triangle \(ABC\) has base \(\overline{AB}\), so \(AC=BC\). The diagram shows the median from \(A\) and the median from \(B\). a) Name the reflection line that maps \(A\) to \(B\), and describe what it does to the midpoint of \(\overline{BC}\). b) Use that reflection to determine the length of the median from \(B\).
Figure for problem 537156

Hints

- Which symmetry of an isosceles triangle exchanges the two base vertices? - Track the entire side \(\overline{BC}\) under that reflection before tracking its midpoint. - Once the two medians are images of each other, what does rigidity say about their lengths?

Solution

1. The reflection line is the perpendicular bisector of base \(\overline{AB}\), which passes through vertex \(C\). 2. This reflection maps \(A\) to \(B\), maps \(\overline{BC}\) to \(\overline{AC}\), and therefore maps the midpoint of \(\overline{BC}\) to the midpoint of \(\overline{AC}\). 3. Thus the median from \(A\) maps to the median from \(B\). 4. Reflections preserve length, so the median from \(B\) has the same length as the labeled median from \(A\): \(8\,\text{cm}\).

Answer

a) Reflect across the perpendicular bisector of \(\overline{AB}\) through \(C\). The midpoint of \(\overline{BC}\) maps to the midpoint of \(\overline{AC}\). b) The median from \(B\) has length \(8\,\text{cm}\).
5546098
Triangle \(ABC\) and center \(O\) are shown on the coordinate plane. Rotate \(ABC\) \(90^\circ\) counterclockwise about \(O\). Give the coordinates of \(A'\), \(B'\), and \(C'\), and explain how you used the center \(O\).
Figure for problem 554609

Hints

- Read the rotation center and all three vertices from the graph. - Describe each vertex by its horizontal and vertical displacement from \(O\). - Rotate those displacements first, then return to the coordinate plane using the same center.

Solution

1. From the graph, \(O(2, 1)\), \(A(4, 1)\), \(B(4, 3)\), and \(C(3, 4)\). 2. Relative to \(O\), the vectors to \(A\), \(B\), and \(C\) are \((2, 0)\), \((2, 2)\), and \((1, 3)\). 3. A \(90^\circ\) counterclockwise turn sends these vectors to \((0, 2)\), \((-2, 2)\), and \((-3, 1)\). 4. Add the rotated vectors back to \(O(2, 1)\): \(A'(2, 3)\), \(B'(0, 3)\), and \(C'(-1, 2)\).

Answer

\(A'(2, 3)\), \(B'(0, 3)\), and \(C'(-1, 2)\). The rotation is applied to each vertex's displacement from \(O\), not to its coordinates as if the origin were the center.
5546108
The graph shows congruent triangles \(ABC\) and \(A'B'C'\). Give one sequence of rigid motions that maps \(ABC\) onto \(A'B'C'\). State the rotation and the translation precisely.
Figure for problem 554610

Hints

- Compare the orientation of corresponding sides before deciding which kind of rigid motion is needed first. - After matching the orientation, compare one pair of corresponding vertices to determine the remaining shift. - Check the same sequence on all three vertices, not just one.

Solution

1. From the graph, \(A(0, 1)\), \(B(2, 1)\), \(C(0, 4)\), \(A'(4, 5)\), \(B'(4, 3)\), and \(C'(7, 5)\). 2. Rotate \(ABC\) \(90^\circ\) clockwise about the origin. The vertices become \(A_1(1, 0)\), \(B_1(1, -2)\), and \(C_1(4, 0)\). 3. Translate by vector \(\langle3, 5\rangle\). This sends \(A_1\) to \(A'(4, 5)\), \(B_1\) to \(B'(4, 3)\), and \(C_1\) to \(C'(7, 5)\). 4. Therefore, one valid sequence is a \(90^\circ\) clockwise rotation about the origin followed by translation by \(\langle3, 5\rangle\).

Answer

One valid sequence is: rotate \(90^\circ\) clockwise about the origin, then translate by \(\langle3, 5\rangle\).
5124088
In parallelogram \(ABCD\), the diagonals intersect at \(M\). A line \(g\) passes through \(M\) and divides the parallelogram into two regions. a) If \(g\) does not pass through a vertex, what shapes are the two regions? b) Use the parallelogram's rotational symmetry to explain why the regions are always congruent. c) Under what conditions are the two regions congruent rectangles?

Hints

- What symmetry does every parallelogram have about the intersection of its diagonals? - Where does a line through the center go under a \(180^\circ\) rotation? - What special parallelogram has four right angles?

Solution

1. If \(g\) avoids the vertices, it intersects a pair of opposite sides and forms two quadrilaterals. Usually they are trapezoids; if \(g\) is parallel to a pair of sides, they are parallelograms. 2. A \(180^\circ\) rotation about \(M\) maps the parallelogram to itself and maps the line \(g\) to itself. 3. The rotation therefore maps one region exactly onto the other, so the regions are congruent. 4. For both regions to be rectangles, the original parallelogram must be a rectangle and \(g\) must be parallel to one pair of opposite sides.

Answer

a) Two quadrilaterals: generally trapezoids, or parallelograms when the cut is parallel to a pair of sides. b) A \(180^\circ\) rotation about \(M\) maps one region onto the other, so they are congruent. c) The original figure must be a rectangle, and \(g\) must be parallel to either pair of opposite sides.
5191098
Kite \(ABCD\) is symmetric across diagonal \(\overline{AC}\). Its diagonals meet at \(M\), so reflection across \(\overline{AC}\) maps \(B\) to \(D\). a) Suppose \(AM=CM\). Explain why a \(180^\circ\) rotation about \(M\) maps \(A\) to \(C\) and \(B\) to \(D\). b) What quadrilateral does the kite become under this additional half-turn symmetry? Justify your answer using the two rigid motions. c) If \(AM\ne CM\), explain why the half-turn fails to map \(A\) to \(C\).

Hints

- A half-turn maps a point to the point directly opposite it through the rotation center. - Determine what the two symmetries say about the endpoints of both diagonals. - For part b, compare side lengths by tracking where each side maps.

Solution

1. If \(AM=CM\), then \(M\) is the midpoint of \(\overline{AC}\). Reflection symmetry already makes \(M\) the midpoint of \(\overline{BD}\). 2. A \(180^\circ\) rotation about a midpoint swaps the endpoints of the segment, so the half-turn maps \(A\leftrightarrow C\) and \(B\leftrightarrow D\). 3. The reflection gives \(AB=AD\) and \(CB=CD\). The half-turn gives \(AB=CD\) and \(AD=CB\). Together, all four sides are congruent, so the kite is a rhombus. 4. If \(AM\ne CM\), then \(M\) is not the midpoint of \(\overline{AC}\), so a half-turn about \(M\) cannot send \(A\) to \(C\).

Answer

a) Since \(M\) is the midpoint of both diagonals, a \(180^\circ\) rotation about \(M\) maps \(A\leftrightarrow C\) and \(B\leftrightarrow D\). b) The kite is a rhombus; the reflection and half-turn together force all four sides to be congruent. c) If \(AM\ne CM\), then \(M\) is not the midpoint of \(\overline{AC}\), so the half-turn does not map \(A\) to \(C\).
5546118
The graph shows triangle \(PQR\) and a proposed final image \(P'Q'R'\). The stated transformation sequence is: translate \(5\) units right and \(2\) units down, then rotate \(180^\circ\) about the origin. a) Is the proposed final image correct? b) If not, identify the incorrect vertex and give its correct coordinate. c) Use one preserved side length to check your conclusion.
Figure for problem 554611

Hints

- Apply the two motions in the stated order to each original vertex. - A rigid-motion sequence must preserve all side lengths. - If one plotted vertex is suspicious, compare a side touching that vertex with its corresponding original side.

Solution

1. From the graph, \(P(-4, 1)\), \(Q(-1, 1)\), and \(R(-4, 3)\). 2. After translating, the vertices are \(P_1(1, -1)\), \(Q_1(4, -1)\), and \(R_1(1, 1)\). 3. A \(180^\circ\) rotation sends these to \(P'(-1, 1)\), \(Q'(-4, 1)\), and \(R'(-1, -1)\). 4. The proposed graph places \(R'\) at \((-1, -2)\), so the proposed final image is not correct. 5. In the original triangle, \(PR=2\). With the correct point \(R'(-1, -1)\), \(P'R'=2\), as required by rigid motions. The proposed point \((-1, -2)\) would give length \(3\), confirming the error.

Answer

a) No. b) \(R'\) is incorrect; it should be \((-1, -1)\). c) \(PR=2\) and the correct \(P'R'=2\); the proposed \(P'R'\) would be \(3\).

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