Free Math Worksheets

- 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.

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 \(A(5.2, 8)\).

The original point is \(A(5.2, 8)\). A reflection across the \(y\)-axis changes \((x, y)\) to \((-x, y)\).

- What does it mean for \(Z\) to be the center of a \(180^\circ\) rotation that maps one vertex to the opposite vertex? - Compare the horizontal and vertical changes from \(A\) to \(Z\), then continue those same changes from \(Z\). - Recall that the diagonals of a parallelogram bisect each other.

1. Because \(Z\) is the midpoint of \(\overline{AC}\), use \(x_C=2x_Z-x_A\) and \(y_C=2y_Z-y_A\). 2. Then \(x_C=2\cdot3-1=5\) and \(y_C=2\cdot3-1=5\), so \(C(5, 5)\). 3. Because \(Z\) is the midpoint of \(\overline{BD}\), use \(x_D=2x_Z-x_B\) and \(y_D=2y_Z-y_B\). 4. Then \(x_D=2\cdot3-4=2\) and \(y_D=2\cdot3-2=4\), so \(D(2, 4)\).

The missing vertices are \(C(5, 5)\) and \(D(2, 4)\).

- 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.

1. A \(180^\circ\) rotation about \(S\) maps each vertex to the opposite vertex. 2. Therefore, \(S\) must be the midpoint of each segment joining opposite vertices. 3. The diagonals bisect each other: \(\overline{AS}=\overline{CS}\) and \(\overline{BS}=\overline{DS}\). 4. A quadrilateral whose diagonals bisect each other is a parallelogram. 5. Examples include a parallelogram, rectangle, rhombus, and square.

a) The diagonals bisect each other. b) Examples include a parallelogram, rectangle, rhombus, and square.

- 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.

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.

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

- Imagine folding each quadrilateral so that its edges match exactly. - Picture the quadrilaterals and test possible lines of reflection mentally. - Distinguish lines through vertices from lines through side midpoints. - Which quadrilateral is the most regular?

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 line 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.

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.

- 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.

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.

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.

- Which quadrilaterals can be reflected across a diagonal and map onto themselves? - Imagine folding a rectangle so that its vertices match. Where are the fold lines?

1. If a line of symmetry passes through two opposite vertices, the other two vertices are reflections of each other. This creates two pairs of adjacent congruent sides, so the quadrilateral is a kite. 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.

a) The kite family. b) Two lines of symmetry, each through the midpoints of a pair of opposite sides.

- Which quadrilaterals have four congruent sides? - Compare the angle requirements for a rhombus and a rectangle. - Recall the rotational symmetry of every parallelogram. - What angle condition would turn the figure into a square?

1. A quadrilateral with four congruent sides is a rhombus. For a non-square rhombus, the two diagonals are its lines of symmetry. 2. A rectangle has four right angles, so all four interior angles are congruent. The given quadrilateral does not have four congruent angles, so it is not a rectangle. 3. Every rhombus is a parallelogram, and every parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals.

a) A rhombus. b) A rectangle must have four congruent right angles, but this quadrilateral does not. c) Yes. As a parallelogram, it has \(180^\circ\) rotational symmetry about the intersection of its diagonals.

- 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.

1. True. A rectangle is a parallelogram, and every parallelogram has \(180^\circ\) rotational symmetry about the intersection of its diagonals. 2. False. A non-square rhombus has two lines of symmetry, but a square is also a rhombus and has four. 3. False. A general parallelogram has no line of symmetry. Only special parallelograms, such as rectangles and rhombi, have line symmetry. 4. True. A square has two diagonal lines of symmetry and two lines through the midpoints of opposite sides.

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.

- 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.

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

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

- 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.

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.

a) x-axis b) y-axis c) x-axis

- 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.

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 \(Q = (22, 14)\).

\(Q = (22, 14)\)

- Think about what reflection across \(\overline{AC}\) does to vertices \(B\) and \(D\). - Recall the relationship between a segment and the line of reflection that swaps its endpoints. - Describe both the angle at which the diagonals meet and how they divide each other.

1. Reflection across \(\overline{AC}\) maps point \(B\) to point \(D\). 2. Therefore, the line of reflection \(\overline{AC}\) is the perpendicular bisector of \(\overline{BD}\). 3. Thus, the diagonals are perpendicular, and \(\overline{AC}\) bisects \(\overline{BD}\).

The diagonals are perpendicular, and \(\overline{AC}\) bisects \(\overline{BD}\).

- Which point is halfway between \(A\) and \(A'\)? - Compare \(A(1, 2)\) with \(A'(-1, -2)\). - Apply the same coordinate change to \(B\), \(C\), and \(D\). - Generalize the pattern for \(P(x, y)\).

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

a) \(B'(-4, -1)\), \(C'(-3, -4)\), and \(D'(-1, -3)\). b) Change the sign of both coordinates: \((x, y)\to(-x, -y)\).

- 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?

1. The original vertices are \(P(1, 1)\), \(Q(5, 2)\), and \(R(2, 4)\). 2. Apply \((x, y)\to(-x, y)\): \(P'(-1, 1)\), \(Q'(-5, 2)\), and \(R'(-2, 4)\). 3. Only the horizontal position changes, and each point remains the same distance from the \(y\)-axis. The rule represents a reflection across the \(y\)-axis.

a) \(P'(-1, 1)\), \(Q'(-5, 2)\), and \(R'(-2, 4)\). b) A reflection across the \(y\)-axis.

- 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?

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.

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

- 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?

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

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

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

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.

Yes. Both translations have vector \((3, 2)\).

- Recall how a kite’s line of symmetry divides the other diagonal. - Compare \(QM\) and \(MS\). - Add the two parts of \(\overline{QS}\).

1. In a kite, the symmetry diagonal bisects the other diagonal. 2. Therefore, \(QM = MS = 3.5\,\text{cm}\). 3. Add the two parts: \(QS = 3.5\,\text{cm} + 3.5\,\text{cm} = 7\,\text{cm}\).

\(QS = 7\,\text{cm}\)

- What line of symmetry does an isosceles triangle have? - Under that reflection, where do vertices \(A\) and \(B\) map? - What happens to the two medians and their lengths?

1. An isosceles triangle is symmetric across the perpendicular bisector of its base \(\overline{AB}\). 2. Reflection across this line maps \(A\) to \(B\), maps side \(BC\) to side \(AC\), and maps the midpoint of \(BC\) to the midpoint of \(AC\). 3. Therefore, the median from \(A\) maps to the median from \(B\). 4. Reflections preserve length, so the median from \(B\) also has length \(8\,\text{cm}\).

The median from \(B\) has length \(8\,\text{cm}\).

- 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?

1. Reflect \(P(-12, 7)\) across the \(x\)-axis by changing the sign of the \(y\)-coordinate: \(P'(-12, -7)\). 2. Reflect \(P'(-12, -7)\) across the \(y\)-axis by changing the sign of the \(x\)-coordinate: \(P''(12, -7)\). 3. From \(P\) to \(P''\), both coordinate signs change. 4. Changing \((x, y)\) to \((-x, -y)\) is a \(180^\circ\) rotation about the origin.

a) \(P'(-12, -7)\) and \(P''(12, -7)\). b) A \(180^\circ\) rotation about the origin maps \(P\) directly to \(P''\).

- 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?

1. Reflecting a point through the origin changes \((x, y)\) to \((-x, -y)\), so \(S(-6, 9)\) maps to \(S'(6, -9)\). 2. Any point on the \(y\)-axis has \(x\)-coordinate \(0\). A reflection across the \(x\)-axis leaves the \(x\)-coordinate unchanged, so the image still has \(x=0\) and remains on the \(y\)-axis. 3. A reflection across the \(x\)-axis maps \((x, y)\) to \((x, -y)\). A point is fixed only when \(y=-y\), which gives \(y=0\). 4. Therefore, every point on the \(x\)-axis is fixed by a reflection across the \(x\)-axis.

a) \(S'(6, -9)\). b) A point on the \(y\)-axis has \(x=0\). Reflection across the \(x\)-axis does not change the \(x\)-coordinate, so the image also has \(x=0\). c) All points on the \(x\)-axis, or all points of the form \((x, 0)\), remain fixed.

- 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.

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 \(y=0\).

The claim is incorrect. The line of reflection is \(y=1\), the perpendicular bisector of \(\overline{PQ}\).

- 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.

1. For a reflection across the vertical line \(x=5\), each image point is the same horizontal distance from the line as its original point. 2. Point \(A(1, 1)\) is \(4\) units left of \(x=5\), so \(A'(9, 1)\). 3. Point \(B(4, 1)\) is \(1\) unit left of \(x=5\), so \(B'(6, 1)\). Similarly, \(C'(6, 3)\) and \(D'(9, 3)\). 4. The original rectangle has width \(4-1=3\) and height \(3-1=2\), so its area is \(3\cdot2=6\) square units. 5. The reflected rectangle also has width \(9-6=3\) and height \(3-1=2\), so its area is \(6\) square units. 6. A reflection is a rigid motion, so it preserves lengths and area.

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.

- 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?

1. Reflect across \(x = 7\). For a point \((x, y)\), the reflected x-coordinate is \(14 - x\). This gives \(A_1(12, 1)\), \(B_1(8, 1)\), and \(C_1(10, 4)\). 2. Translate each point \(3\) units up by adding \(3\) to its y-coordinate. Thus, \(A_2(12, 4)\), \(B_2(8, 4)\), and \(C_2(10, 7)\). 3. Reflections and translations are rigid motions. They preserve all distances and angle measures, so the final triangle is congruent to the original triangle.

a) \(A_2(12, 4)\), \(B_2(8, 4)\), and \(C_2(10, 7)\) b) The triangles are congruent because a reflection followed by a translation is a sequence of rigid motions, which preserves lengths and angle measures.

- Where is the intersection of the diagonals relative to opposite vertices of a square? - How do equal, perpendicular diagonals help you locate the other two vertices? - Rotate the vector from \(M\) to \(R\) by \(90^\circ\) in both directions.

1. Point \(M\) 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 either direction gives \((-2, 2)\) and \((2, -2)\). 4. Adding these vectors to \(M(4, 4)\) gives the missing vertices \((2, 6)\) and \((6, 2)\). 5. The labels may be \(Q(6, 2)\), \(S(2, 6)\), or \(Q(2, 6)\), \(S(6, 2)\).

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.

- 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?

1. A kite has a line of symmetry through two opposite vertices. This line is one of the diagonals; call it \(e\). 2. The endpoints of the other diagonal \(f\) are reflections of each other across \(e\). Therefore, \(e\) is the perpendicular bisector of \(f\). 3. Because the kite is not a rhombus, the intersection of the diagonals is not the midpoint of \(e\). 4. Therefore, the diagonal on the line of symmetry bisects the other diagonal but is not itself bisected.

a) The diagonal on the line of symmetry is the perpendicular bisector of the other diagonal. The other diagonal does not bisect the diagonal on the line of symmetry. b) The diagonal on the line of symmetry is the one that bisects the other diagonal.

- 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?

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

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.

- 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.

1. In a parallelogram, consecutive angles are supplementary. In an isosceles trapezoid, each pair of base angles is congruent. Congruent supplementary angles must each measure \(90^\circ\), so the quadrilateral is a rectangle. 2. A general parallelogram that is neither a rectangle nor a rhombus has no lines of symmetry. It has only \(180^\circ\) rotational symmetry. 3. A kite that is not a rhombus or an isosceles trapezoid that is not a rectangle has line symmetry but not \(180^\circ\) rotational symmetry.

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.

- Consider which symmetry maps one diagonal to the other. - Ask whether a half-turn maps each vertex to the opposite vertex. - A square combines the diagonal properties of a rectangle and a rhombus.

1. In both a rectangle and an isosceles trapezoid, a line of symmetry maps one diagonal to the other. Rigid motions preserve length, so the diagonals are congruent. 2. A rectangle has \(180^\circ\) rotational symmetry about the intersection of its diagonals, so the diagonals bisect each other. In an isosceles trapezoid that is not a parallelogram, the diagonals do not bisect each other; if they did, the quadrilateral would be a parallelogram. 3. For the quadrilateral to be a square, its diagonals must be congruent, bisect each other, and be perpendicular.

a) The diagonals are congruent. b) A rectangle's diagonals bisect each other; the isosceles trapezoid's diagonals do not. c) The diagonals are congruent, bisect each other, and are perpendicular.

- 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.

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.

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 \(PB=PD\).

- 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 \(\overline{QR}\).

1. Points \(P\) and \(R\) lie on the vertical line \(x = 5\), which can serve as the kite’s line of symmetry. 2. Point \(Q(9, 6)\) is \(4\) units to the right of \(x = 5\). Reflecting \(Q\) across this line gives \(S(1, 6)\). 3. For \(\overline{PQ}\), the horizontal and vertical changes are both \(4\), so \(PQ^2 = 4^2 + 4^2 = 32\). 4. For \(\overline{QR}\), the horizontal change is \(4\) and the vertical change is \(6\), so \(QR^2 = 4^2 + 6^2 = 52\). 5. Because \(32 \ne 52\), \(PQ \ne QR\). The kite does not have four congruent sides, so it is not a rhombus.

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.

- 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.

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 \(x = 1.5 + 1.7 = 3.2\) and \(y = 7 - 2.5 = 4.5\). Thus, \(B' = (3.2, 4.5)\), which is the location of \(A\).

a) Point \(B\) b) Points \(A\) and \(C\) c) \(B' = (3.2, 4.5)\), which is the same location as \(A\).

- 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.

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

a) The quadrilateral is a square. b) \(A' = (2, 5)\) c) \(A' = (2, 5)\), \(B' = (6, 5)\), \(C' = (6, 9)\), and \(D' = (2, 9)\)

- 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?

1. If \(g\) avoids the vertices, it intersects a pair of opposite sides and forms two quadrilaterals. Usually each is a trapezoid; if \(g\) is parallel to a pair of sides, each region is a parallelogram. 2. A parallelogram has \(180^\circ\) rotational symmetry about the intersection point \(M\) of its diagonals. Because \(g\) passes through \(M\), this rotation maps the cut line to itself and maps one region exactly onto the other. 3. A rotation is a rigid motion, so the two 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.

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.

- Compare how the diagonals divide each other in a rhombus and in a general kite. - Think about what changes when all four side lengths become congruent. - Decide whether the intersection must be the midpoint of \(\overline{AC}\).

1. In a kite, the symmetry diagonal \(\overline{AC}\) is the perpendicular bisector of \(\overline{BD}\). 2. In a rhombus, the diagonals bisect each other. 3. Therefore, to make a kite that is not a rhombus, \(\overline{BD}\) must not bisect \(\overline{AC}\). The intersection of the diagonals must not be the midpoint of \(\overline{AC}\).

The diagonal \(\overline{BD}\) must not bisect \(\overline{AC}\). In other words, the diagonals’ intersection must not be the midpoint of \(\overline{AC}\).