Rigid motions and congruence

Learning goals

Identifies corresponding parts of congruent figures.
Maps figures using translations, reflections, rotations, or sequences of these motions.
Justifies congruence with a logical sequence of statements.
Uses congruence to infer corresponding lengths or angles.

Difficulty: 1 – 5

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Rigid motions and congruence 1

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.

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.

Rigid motions and congruence 2

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.

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 \(O\).

Rigid motions and congruence 3

Triangle \(ABC\) is moved to triangle \(DEF\) on the coordinate grid. What single translation maps \(A\) to \(D\), \(B\) to \(E\), and \(C\) to \(F\)?

Hints

- Choose two vertices that correspond to each other. - Count the horizontal and vertical changes between them. - Check that the same movement works for the other vertices.

Solution

1. Compare any pair of matching vertices: \(A(0,0)\) moves to \(D(3,1)\). 2. The change is \(3\) units right and \(1\) unit up. 3. The same change maps \(B\) to \(E\) and \(C\) to \(F\).

Answer

Translate \(3\) units right and \(1\) unit up.

Rigid motions and congruence 4

Reflect point \(P\) across the \(y\)-axis. Give the coordinates of \(P'\), and explain which coordinate changes.

Hints

- Find the point's horizontal distance from the vertical axis. - The image must be the same distance on the opposite side. - Check whether the height should change.

Solution

1. The grid places \(P\) at \((-3,2)\). 2. A reflection across the \(y\)-axis keeps the height and reverses the horizontal position. 3. Hence \(P'=(3,2)\); only the \(x\)-coordinate changes sign.

Answer

\(P'(3,2)\). The \(x\)-coordinate changes sign; the \(y\)-coordinate stays the same.

Rigid motions and congruence 5

A rigid motion maps triangle \(ABC\) to triangle \(MNP\), with \(A\) matching \(M\), \(B\) matching \(N\), and \(C\) matching \(P\). If \(BC=7\,\text{cm}\), what is \(NP\)? Explain.

Hints

- Match the endpoints of the known side to their images. - What does a rigid motion do to the distance between two points?

Solution

1. Side \(BC\) corresponds to side \(NP\). 2. A rigid motion preserves lengths, so \(NP=BC=7\,\text{cm}\).

Answer

\(NP=7\,\text{cm}\), because corresponding sides have the same length after a rigid motion.

Rigid motions and congruence 6

Point \(P\) makes a half-turn around \(Z\). Where does \(P'\) land? Give its coordinates.

Hints

- A half-turn leaves its center in the same place. - Look along the horizontal line through \(P\) and \(Z\). - The new point must be as far from \(Z\) as \(P\), on the other side.

Solution

1. On the grid, \(Z=(2,1)\) and \(P=(4,1)\). Point \(P\) is \(2\) units right of \(Z\). 2. A half-turn puts \(P'\) \(2\) units left of \(Z\), at \((0,1)\).

Answer

\(P'(0,1)\).

Rigid motions and congruence 7

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 \((-x, y)\).

Rigid motions and congruence 8

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 \(D(2, 4)\).

Rigid motions and congruence 9

Quadrilateral \(ABCD\) has \(180^\circ\) rotational symmetry about the intersection \(S\) of its diagonals. a) What does the symmetry tell you about the two parts of each diagonal? b) Name at least three kinds of quadrilaterals with this property.

Hints

- Identify which two vertices trade places in a half-turn. - What position must the rotation center have between each point and its image? - Which family of quadrilaterals has diagonals that bisect each other?

Solution

1. A half-turn around \(S\) swaps \(A\) with \(C\) and \(B\) with \(D\). 2. Therefore, \(AS=SC\) and \(BS=SD\): each diagonal is bisected at \(S\). 3. Quadrilaterals whose diagonals bisect one another are parallelograms, including rectangles, rhombuses, and squares.

Answer

a) The diagonals bisect each other: \(AS=SC\) and \(BS=SD\). b) Parallelograms, rectangles, rhombuses, and squares are examples.

Rigid motions and congruence 10

Match each quadrilateral to its symmetry description. Here, the parallelogram is neither a rectangle nor a rhombus, the kite is not a rhombus, and the isosceles trapezoid is not a parallelogram. **Quadrilaterals** 1. Square 2. Parallelogram 3. Kite 4. Isosceles trapezoid **Descriptions** A: A half-turn maps it onto itself, but it has no line of symmetry. B: It has one line of symmetry through two opposite vertices. C: It has four lines of symmetry and half-turn symmetry. D: It has one line of symmetry through the midpoints of its parallel bases.

Hints

- Imagine reflecting each figure over a possible symmetry line. - A half-turn and a reflection are different motions. - Ask whether changing a shape's side lengths or angles could give it extra symmetries.

Solution

1. A square has four reflection axes and half-turn symmetry, so it matches C. 2. The specified parallelogram has half-turn symmetry but no reflection axis, so it matches A. 3. The specified kite has one reflection axis through opposite vertices, so it matches B. 4. The specified isosceles trapezoid has one reflection axis through the midpoints of its bases, so it matches D.

Answer

1. C 2. A 3. B 4. D

Rigid motions and congruence 11

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.

Rigid motions and congruence 12

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.

Rigid motions and congruence 13

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 \(A'' = (-35, -42)\)

Rigid motions and congruence 14

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

Rigid motions and congruence 15

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

\(Q = (22, 14)\)

Rigid motions and congruence 16

Kite \(ABCD\) is symmetric across the line through \(A\) and \(C\), which swaps \(B\) and \(D\). Without using a memorized kite rule, explain why this line is the perpendicular bisector of \(\overline{BD}\). State two consequences for the diagonals.

Hints

- What kind of line passes midway between a point and its reflected image? - Apply that property to \(B\) and \(D\). - Separate the word perpendicular from the word bisector.

Solution

1. The segment joining a point to its reflection is perpendicular to the reflection line and bisected by it. 2. Since \(B\) reflects to \(D\), the line through \(A\) and \(C\) is the perpendicular bisector of \(\overline{BD}\). 3. Therefore, \(\overline{AC}\perp\overline{BD}\), and \(\overline{AC}\) passes through the midpoint of \(\overline{BD}\).

Answer

The reflection line through \(A\) and \(C\) bisects \(\overline{BD}\) at a right angle. Thus the diagonals are perpendicular, and \(AC\) bisects \(BD\).

Rigid motions and congruence 17

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 \(A\).

Rigid motions and congruence 18

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 \(D'=(2, 9)\)

Rigid motions and congruence 19

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.

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: \((x, y)\to(-x, -y)\).

Rigid motions and congruence 20

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?

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.

Rigid motions and congruence 21

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 \(F\).

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 \(F'(-5, -4)\).

Rigid motions and congruence 22

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.

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

\(DF=4.5\,\text{cm}\)

Rigid motions and congruence 23

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 \((3, 2)\).

Rigid motions and congruence 24

Quadrilateral \(PQRS\) is a kite, and the line through \(P\) and \(R\) is a reflection line. The diagonals intersect at \(M\). a) Where does \(Q\) go under this reflection? b) Find \(QS\) using the marked length.

Hints

- Look for the vertex opposite \(Q\) across the reflection line. - A reflection line cuts the segment between a point and its image in half. - Use the given length on only one half of the diagonal.

Solution

1. The reflection across line \(PR\) swaps \(Q\) and \(S\). 2. The reflection line bisects \(\overline{QS}\) at \(M\), so \(QM=MS\). 3. The diagram marks \(QM=3.5\,\text{cm}\), hence \(QS=2\cdot3.5=7\,\text{cm}\).

Answer

a) \(Q\) maps to \(S\). b) \(QS=7\,\text{cm}\).

Rigid motions and congruence 25

Triangle \(ABC\) is translated to triangle \(A_1B_1C_1\). a) If \(AB=5\,\text{cm}\), find \(A_1B_1\). b) If \(\angle C=48^\circ\), find \(\angle C_1\). c) Besides equal length, what relationship must \(\overline{AB}\) and \(\overline{A_1B_1}\) have?

Hints

- A translation moves each point by the same amount. - Which size and angle properties stay unchanged? - Compare the directions of corresponding sides.

Solution

1. A translation preserves length, so \(A_1B_1=5\,\text{cm}\). 2. It preserves angles, so \(\angle C_1=48^\circ\). 3. It also preserves direction, so the corresponding segments are parallel.

Answer

a) \(5\,\text{cm}\). b) \(48^\circ\). c) \(\overline{AB}\parallel\overline{A_1B_1}\).

Rigid motions and congruence 26

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?

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 \(y=x\).

Rigid motions and congruence 27

Point \(P\) is turned \(180^\circ\) about \(Z\). How far is \(P\) from \(Z\) horizontally and vertically? Use those distances to find \(P'\).

Hints

- The center is \(Z\), not the origin. - Count how far the first point lies from \(Z\) in each direction. - A half-turn places the image opposite \(P\) across \(Z\).

Solution

1. Read \(Z(1,1)\) and \(P(3,2)\) from the grid. 2. From \(Z\) to \(P\), move \(2\) right and \(1\) up. 3. After a half-turn, move \(2\) left and \(1\) down from \(Z\), giving \(P'(-1,0)\).

Answer

\(P\) is \(2\) units right and \(1\) unit up from \(Z\); \(P'(-1,0)\).

Rigid motions and congruence 28

One transformation sends \((x, y)\) to \((x+4, y-2)\). Another sends \((x, y)\) to \((2x, y)\). Which transformation is rigid? Give a pair of points whose distance changes under the other transformation.

Hints

- A rigid motion keeps every distance unchanged. - Test two points on a horizontal line under the second rule. - Compare the original and new distances.

Solution

1. The first rule is a translation, so all distances stay the same. 2. The second rule doubles horizontal distances: \((0, 0)\) and \((1, 0)\) are \(1\) unit apart, but their images \((0, 0)\) and \((2, 0)\) are \(2\) units apart. 3. Only the first transformation is rigid.

Answer

The translation \((x, y)\to(x+4, y-2)\) is rigid. Under \((x, y)\to(2x, y)\), the distance between \((0, 0)\) and \((1, 0)\) changes from \(1\) to \(2\).

Rigid motions and congruence 29

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 \(P''\).

Rigid motions and congruence 30

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.

Rigid motions and congruence 31

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 \(\overline{PQ}\).

Rigid motions and congruence 32

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.

Rigid motions and congruence 33

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.

Rigid motions and congruence 34

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.

Rigid motions and congruence 35

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.

Rigid motions and congruence 36

In quadrilateral \(ABCD\), the lines through its diagonals are both lines of reflection. Reflection over line \(AC\) fixes \(A\) and \(C\) and swaps \(B\) with \(D\). Reflection over line \(BD\) fixes \(B\) and \(D\) and swaps \(A\) with \(C\). a) Use the first reflection to name two pairs of equal side lengths. b) Do the same for the second reflection. c) Classify \(ABCD\) and give its half-turn symmetry.

Hints

- Follow the endpoints of each side under the first reflection. - Reflections preserve lengths. - Combine the side equalities after using both reflections.

Solution

1. Reflection over line \(AC\) gives \(AB=AD\) and \(CB=CD\). 2. Reflection over line \(BD\) gives \(AB=CB\) and \(AD=CD\). 3. All four sides are equal, so the quadrilateral is a rhombus. 4. A \(180^\circ\) rotation about the intersection of the diagonals swaps \(A\) with \(C\) and \(B\) with \(D\).

Answer

a) \(AB=AD\) and \(CB=CD\). b) \(AB=CB\) and \(AD=CD\). c) A rhombus; a \(180^\circ\) rotation about the intersection of the diagonals maps it onto itself.

Rigid motions and congruence 37

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.

Rigid motions and congruence 38

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.

Rigid motions and congruence 39

Points \(A\) and \(C\) are opposite vertices of a kite. The line through \(A\) and \(C\) is its line of symmetry. Describe where the other two vertices, \(B\) and \(D\), must lie relative to this line and to each other.

Hints

- How must two points sit relative to a line to be reflections of each other? - Where can the segment joining those points meet the reflection line? - Which distances from that line must be equal?

Solution

1. A reflection across line \(AC\) swaps \(B\) and \(D\), so the line must be the perpendicular bisector of \(BD\). 2. The segment \(BD\) meets the line through \(A\) and \(C\) at its midpoint \(P\), so \(PB=PD\). 3. Points \(B\) and \(D\) lie on opposite sides of line \(AC\), the same distance from it.

Answer

\(B\) and \(D\) lie on opposite sides of line \(AC\), the same distance from it. Line \(AC\) cuts segment \(BD\) in half at a right angle.

Rigid motions and congruence 40

Points \(P(5,2)\), \(Q(9,6)\), and \(R(5,12)\) are three vertices of kite \(PQRS\). a) Find \(S\) by reflecting \(Q\) across the line through \(P\) and \(R\). b) Is the kite also a rhombus? Compare the midpoints of its diagonals to explain.

Hints

- Which vertical line passes through both \(P\) and \(R\)? - Put \(S\) as far to the other side of that line as \(Q\). - A rhombus must have diagonals that bisect each other.

Solution

1. The reflection line is \(x=5\). Since \(Q\) is \(4\) units right of the line, \(S\) is \(4\) units left, at \((1,6)\). 2. The midpoint of \(PR\) is \((5,7)\), while the midpoint of \(QS\) is \((5,6)\). 3. The diagonal midpoints differ. Rhombuses are parallelograms with diagonals that bisect each other, so this kite is not a rhombus.

Answer

a) \(S(1,6)\). b) No. The diagonal midpoints are \((5,7)\) and \((5,6)\), so the diagonals do not bisect each other.

Rigid motions and congruence 41

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 \(B\).

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 \(8\,\text{cm}\).

Rigid motions and congruence 42

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.

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: \((x, y)\to(y, -x)\)

Rigid motions and congruence 43

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 \(O\).

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.

Rigid motions and congruence 44

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.

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 \(\langle3, 5\rangle\).

Rigid motions and congruence 45

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.

Rigid motions and congruence 46

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 \(C\).

Rigid motions and congruence 47

Triangle \(PQR\) in a) is translated \(5\) units right and \(2\) units down, then rotated \(180^\circ\) about the origin. The proposed result is in b). a) Is the proposed final image correct? b) If not, which vertex is incorrect, and where should it be? c) Check your answer by comparing a corresponding side length.

Hints

- Apply the translation before the half-turn. - A half-turn changes both coordinate signs about the origin. - Compare a side meeting the suspected incorrect vertex with its original.

Solution

1. The original points are \(P(-4,1)\), \(Q(-1,1)\), and \(R(-4,3)\). 2. The translation gives \((1,-1)\), \((4,-1)\), and \((1,1)\). 3. The half-turn gives \(P'(-1,1)\), \(Q'(-4,1)\), and \(R'(-1,-1)\). 4. The proposed \(R'(-1,-2)\) is incorrect. Original \(PR=2\), correct \(P'R'=2\), but proposed \(P'R'=3\).

Answer

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

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Rigid motions and congruence

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