Free Math Worksheets

- What happens to a line's slope after a \(180^\circ\) rotation? - Rotate one convenient point on the original line about \(Z\). - Use the rotated point and the unchanged slope to write the new equation.

1. A \(180^\circ\) rotation maps a line to a parallel line, so the image line still has slope \(1.5\). 2. The point \((0, 2)\) lies on the original line. Rotating it \(180^\circ\) about \((2, 1)\) gives \((4, 0)\). 3. Write the image line as \(f'(x) = 1.5x + b\). Since \((4, 0)\) lies on it, \(0 = 1.5 \cdot 4 + b\), so \(b = -6\). 4. Therefore, \(f'(x) = 1.5x - 6\).

\(f'(x) = 1.5x - 6\)

- What geometric relationship connects the center of a \(180^\circ\) rotation with a point and its image? - Use the known x-coordinate of the center to find the x-coordinate of an image point. - Then use the image line to find that point's y-coordinate.

1. Choose \(P(0, 5)\) on \(g\). Under a \(180^\circ\) rotation, the center \(Z\) is the midpoint of \(P\) and its image \(P'\). 2. Since \(x_Z = 3\), \(3 = \frac{0 + x_{P'}}{2}\), so \(x_{P'} = 6\). 3. The image point lies on \(h\), so \(y_{P'} = -2 \cdot 6 - 1 = -13\). 4. Therefore, \(y_Z = \frac{5 + (-13)}{2} = -4\).

\(y_Z = -4\)

- How do coordinates change under a \(180^\circ\) rotation about the origin? - What happens to the slope of a line under this rotation? - When do two lines with the same slope intersect?

1. A \(180^\circ\) rotation about the origin maps \((x, y)\) to \((-x, -y)\). The inverse transformation is the same rotation. 2. For example, \((0, -4)\) on \(k'\) maps back to \((0, 4)\). The slope remains \(\frac{1}{3}\), so \(k(x) = \frac{1}{3}x + 4\). 3. The two lines have the same slope but different y-intercepts, so they are distinct parallel lines. Therefore, they do not intersect.

a) \(k(x) = \frac{1}{3}x + 4\) b) No. The lines are distinct and parallel.

- What happens to a line's direction under a \(180^\circ\) rotation? - How do coordinates change under this rotation about the origin? - Where does the original y-intercept move?

1. A \(180^\circ\) rotation maps a line to a parallel line, so the slope remains \(-0.5\). Therefore, the image line also has negative slope. 2. Under a \(180^\circ\) rotation about the origin, \((x, y)\) maps to \((-x, -y)\). The y-intercept \((0, 2)\) maps to \((0, -2)\). With slope \(-0.5\), the image line is \(m(x) = -0.5x - 2\).

a) Negative slope b) \(m(x) = -0.5x - 2\)

- Apply the factor to both coordinates of every vertex. - Use the absolute value of the scale factor to compare lengths. - Area changes by the square of the scale factor.

1. Multiply each coordinate by \(-2\): \(A' = (4, 2)\), \(B' = (-2, 2)\), \(C' = (-2, -2)\), and \(D' = (4, -2)\). 2. The absolute value of the scale factor is \(2\), so every side length doubles. 3. Area is multiplied by the square of the scale factor, so the area is multiplied by \((-2)^2 = 4\). 4. The negative scale factor also places each image point on the opposite ray from the origin, which is equivalent to a \(180^\circ\) rotation about the origin followed by a dilation by factor \(2\).

a) \(A' = (4, 2)\), \(B' = (-2, 2)\), \(C' = (-2, -2)\), and \(D' = (4, -2)\) b) The side lengths double, the area is multiplied by \(4\), and the image is turned \(180^\circ\) about the origin while being enlarged.

- What happens to a line under a \(180^\circ\) rotation? - Which sides remain on the outer boundary after the quadrilaterals are joined? - Match each boundary side with its image under the rotation.

1. A \(180^\circ\) rotation maps every line to a parallel line. 2. The rotation about \(Z\) maps one congruent quadrilateral onto the other. 3. The shared side lies inside the combined figure, so the boundary of the hexagon contains three sides from each quadrilateral. 4. Each boundary side from one quadrilateral maps to the opposite boundary side from the other quadrilateral. 5. Since a side and its image lie on parallel lines, the hexagon has three pairs of opposite parallel sides.

The hexagon has three pairs of opposite parallel sides. Each boundary side from one quadrilateral is mapped by the \(180^\circ\) rotation about \(Z\) to its opposite, parallel boundary side.

- Recall the diagonal properties of a square. - Track how the four smaller triangles move under a \(90^\circ\) rotation. - Identify the lines of symmetry of a square. - A reflection maps the incenter of a triangle to the incenter of its image triangle.

1. The diagonals of a square are perpendicular and bisect each other. Therefore, each smaller triangle has a right angle at \(S\), and its two sides from \(S\) to adjacent vertices are congruent. The triangles are isosceles right triangles. 2. A \(90^\circ\) rotation about \(S\) maps each smaller triangle to the next one and therefore maps each incenter to the next incenter. 3. The four incenters are equally distant from \(S\), and consecutive radius segments differ by a \(90^\circ\) rotation. Thus, \(M_1M_2M_3M_4\) is a square. 4. Reflection across \(\overline{AC}\) maps triangle \(ABS\) to triangle \(DAS\). Therefore, it maps the incenter \(M_1\) to \(M_4\).

a) They are isosceles right triangles: each has a \(90^\circ\) angle at \(S\) and two congruent legs. b) \(M_1M_2M_3M_4\) is a square because the \(90^\circ\) rotational symmetry maps consecutive incenters to one another. c) \(M_1\) maps to \(M_4\).

- Consider a \(180^\circ\) rotation about \(Z\). - Where does \(P\) map under this rotation? - If \(PQ\parallel RS\), what line would be the image of \(PQ\)?

1. Since \(P\), \(Z\), and \(R\) are collinear and \(PZ=ZR\), a \(180^\circ\) rotation about \(Z\) maps \(P\) to \(R\). 2. Suppose \(PQ\parallel RS\). A \(180^\circ\) rotation maps line \(PQ\) to a parallel line through \(R\). 3. There is only one line through \(R\) parallel to \(PQ\), so the image of line \(PQ\) would have to be line \(RS\). 4. Line \(QS\) passes through the center of rotation and maps onto itself. Therefore, the image of \(Q\) would lie on both \(QS\) and \(RS\), so it would be \(S\). 5. A \(180^\circ\) rotation mapping \(Q\) to \(S\) would require \(QZ=ZS\), contradicting the given information. Therefore, \(PQ\not\parallel RS\).

No. If \(PQ\parallel RS\), a \(180^\circ\) rotation about \(Z\) would map \(Q\) to \(S\), which would require \(QZ=ZS\). This contradicts \(QZ\ne ZS\).

- Use the \(180^\circ\) rotational symmetry of a parallelogram about the intersection of its diagonals. - How do the equal extension lengths determine the images of \(A\) and \(B\)?

1. Let \(O\) be the intersection of the diagonals of parallelogram \(MNPK\). A \(180^\circ\) rotation about \(O\) maps \(M\) to \(P\) and \(K\) to \(N\). 2. The extension of \(\overline{MN}\) beyond \(M\) maps to the extension of \(\overline{KP}\) beyond \(P\). Since \(MA=PC\), point \(A\) maps to point \(C\). 3. Similarly, the extension of \(\overline{MK}\) beyond \(K\) maps to the extension of \(\overline{PN}\) beyond \(N\). Since \(KB=ND\), point \(B\) maps to point \(D\). 4. Therefore, \(O\) is the midpoint of both \(\overline{AC}\) and \(\overline{BD}\). The diagonals of \(ABCD\) bisect each other. 5. A quadrilateral whose diagonals bisect each other is a parallelogram, so \(ABCD\) is a parallelogram.

A \(180^\circ\) rotation about the center \(O\) of \(MNPK\) maps \(A\) to \(C\) and \(B\) to \(D\). Thus, \(O\) is the midpoint of both diagonals of \(ABCD\), so the diagonals bisect each other and \(ABCD\) is a parallelogram.