Free math worksheets

- A rotation fixes its center and preserves distance from that center. - The direction of the angle matters as well as its size. - Once the correct ray is fixed, ask how many points on that ray can be the required distance from \(O\).

1. From ray \(\overrightarrow{OP}\), construct a ray at \(O\) that is \(75^\circ\) counterclockwise from \(\overrightarrow{OP}\). 2. On the new ray, locate \(P'\) so that \(OP'=OP\). 3. These two conditions, \(m\angle POP'=75^\circ\) counterclockwise and \(OP'=OP\), are exactly the defining conditions for the rotation image. 4. A ray contains exactly one point at a specified positive distance from its endpoint, so \(P'\) is unique.

Construct the ray \(75^\circ\) counterclockwise from \(\overrightarrow{OP}\), then place \(P'\) on it so that \(OP'=OP\). These conditions determine one point, so the rotation image is unique.

- Note the point's horizontal and vertical distances from the origin. - A quarter-turn keeps the distance from the origin unchanged. - Check that the image lies in the quadrant reached by turning counterclockwise.

1. From the graph, \(P=(2,1)\). 2. A \(90^\circ\) counterclockwise rotation about the origin sends \((x,y)\) to \((-y,x)\). 3. Therefore, \(P'=(-1,2)\).

\(P'=(-1,2)\)

- The origin is the fixed center of the dilation. - A scale factor between \(0\) and \(1\) moves each point toward the center. - Apply the same factor to both coordinates.

1. From the graph, \(A=(4,2)\), \(B=(2,6)\), and \(C=(-2,4)\). 2. A dilation centered at the origin with scale factor \(\frac{1}{2}\) multiplies both coordinates by \(\frac{1}{2}\). 3. Therefore, \(A'=(2,1)\), \(B'=(1,3)\), and \(C'=(-1,2)\).

\(A'=(2,1)\), \(B'=(1,3)\), \(C'=(-1,2)\)

- 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) \(m(x) = -0.5x - 2\)

- What does being the midpoint of a segment tell you about a half-turn centered there? - Track both endpoints of \(\overline{AB}\). - Recall which properties of a segment and its supporting line a rotation preserves.

1. Since \(K\) is the midpoint of \(\overline{AC}\), a \(180^\circ\) rotation about \(K\) maps \(A\) to \(C\). 2. Since \(K\) is also the midpoint of \(\overline{BD}\), the same rotation maps \(B\) to \(D\). 3. Therefore, segment \(\overline{AB}\) maps to segment \(\overline{CD}\). 4. A rotation preserves length, so \(AB=CD\). 5. A \(180^\circ\) rotation maps a line not passing through its center to a parallel line, so \(AB\parallel CD\).

The half-turn about \(K\) maps \(A\to C\) and \(B\to D\), so \(\overline{AB}\) maps to \(\overline{CD}\). Therefore, \(AB=CD\) and \(AB\parallel CD\).

- Read the coordinates of \(A\) and \(B\) from the grid before applying the dilation. - A dilation centered at the origin multiplies both coordinates by the scale factor. - Compare the two slopes after finding the image coordinates.

1. From the graph, \(O=(0, 0)\), \(A=(2, 1)\), and \(B=(1, 3)\). 2. Dilating from the origin by scale factor \(2\) gives \(A'=(4, 2)\) and \(B'=(2, 6)\). 3. The slope of \(AB\) is \(\frac{3-1}{1-2}=-2\). 4. The slope of \(A'B'\) is \(\frac{6-2}{2-4}=-2\). 5. The lines have equal slopes and are distinct, so \(A'B'\parallel\ell\).

\(A'=(4, 2)\) and \(B'=(2, 6)\). Both \(AB\) and \(A'B'\) have slope \(-2\), so \(A'B'\parallel\ell\).

- Count how many equal vertex positions occur in one full turn. - The smallest rotational symmetry moves each vertex to the next matching position. - Follow the vertex labels counterclockwise around the center.

1. A regular hexagon has \(6\) equally spaced vertices around its center. 2. The smallest such rotational symmetry is one sixth of a full turn: \(360^\circ \div 6 = 60^\circ\). 3. Using the counterclockwise direction shown by the vertex order, \(A\) maps to \(B\), and \(C\) maps to \(D\).

A \(60^\circ\) counterclockwise rotation; \(A\to B\) and \(C\to D\).

- A rotation is determined by more than the magnitude of its angle; the direction from a point to its image matters. - Follow ray \(OA\) to ray \(OA'\) around the center. - Apply exactly that same directed turn and preserve the distance from \(O\) when locating the image of \(C\).

1. From ray \(OA\) to ray \(OA'\), the indicated quarter-turn is counterclockwise, so the mapping \(A\to A'\) fixes the rotation as \(90^\circ\) counterclockwise. 2. The same rotation must be applied to \(C\), preserving its distance from \(O\) and turning ray \(OC\) through the same directed angle. 3. The resulting point is \(X\). Therefore, \(C'=X\).

The rotation is \(90^\circ\) counterclockwise, and \(C'=X\).

- Use \(S\), not the origin, as the reference point. - Find each displacement vector from \(S\) to an endpoint. - A scale factor of \(\frac{1}{2}\) halves those displacement vectors.

1. From the graph, \(S=(2,-1)\), \(A=(6,3)\), and \(B=(-2,5)\). 2. From \(S\) to \(A\) the displacement is \(\langle4,4\rangle\); halving it gives \(\langle2,2\rangle\), so \(A'=(4,1)\). 3. From \(S\) to \(B\) the displacement is \(\langle-4,6\rangle\); halving it gives \(\langle-2,3\rangle\), so \(B'=(0,2)\).

\(A'=(4,1)\), \(B'=(0,2)\)

- Which point defines where all dilation distances are measured from? - Think about what multiplying a distance from the center by a positive number can change and what it cannot reverse. - Apply the general rule separately to \(P\mapsto P'\) and \(Q\mapsto Q'\).

1. The center of a dilation does not move, so \(C\) is the fixed point. 2. A positive scale factor multiplies a point's distance from the center without reversing its direction from the center. Therefore, the point and its image remain on the same ray starting at the center. 3. Thus, \(P\) and \(P'\) lie on ray \(\overrightarrow{CP}\), and \(Q\) and \(Q'\) lie on ray \(\overrightarrow{CQ}\).

\(C\) is fixed. A positive dilation changes distance from \(C\) without reversing direction, so \(P\) and \(P'\) lie on the same ray from \(C\), and \(Q\) and \(Q'\) lie on the same ray from \(C\).

- Separate what a dilation does to lengths from what it does to angles. - Use the scale factor only on the length. - Compare the angle at \(B\) with its corresponding angle at \(B'\).

1. A dilation multiplies every length by the scale factor, so \(A'B'=1.5\cdot4\,\text{cm}=6\,\text{cm}\). 2. A dilation preserves angle measure, so \(\angle A'B'C'=52^\circ\).

\(A'B'=6\,\text{cm}\) and \(\angle A'B'C'=52^\circ\).

- Recall the defining distance property of a rigid motion. - Compare the two given corresponding side lengths. - Which transformation preserves angles but can multiply all lengths by the same factor?

1. A rigid motion preserves all distances, but the corresponding side changes from \(3\,\text{cm}\) to \(6\,\text{cm}\). Therefore, no rigid motion can map the first triangle to the second. 2. The corresponding side length is multiplied by \(\frac{6}{3}=2\), while corresponding angles are preserved. 3. Those properties are consistent with a dilation of scale factor \(2\).

No rigid motion can map the triangles because the corresponding side lengths differ. A dilation with scale factor \(2\) is consistent with the change.

- Compare the two sides that meet at \(A\) in each square. - A square fixes both a right-angle turn and an equal distance from its vertex. - Track the two endpoints of \(\overline{EC}\) under the same rotation.

1. In square \(ABDE\), ray \(\overrightarrow{AE}\) turns \(90^\circ\) counterclockwise to ray \(\overrightarrow{AB}\), and \(AE=AB\). Therefore, a \(90^\circ\) counterclockwise rotation about \(A\) maps \(E\) to \(B\). 2. In square \(ACFG\), the same \(90^\circ\) counterclockwise rotation about \(A\) maps \(C\) to \(G\). 3. Therefore, the rotation maps segment \(\overline{EC}\) to segment \(\overline{BG}\). 4. Rotations preserve distance, so \(EC=BG\).

A \(90^\circ\) counterclockwise rotation about \(A\) maps \(E\to B\) and \(C\to G\). Thus, \(\overline{EC}\) maps to \(\overline{BG}\), and distance preservation gives \(EC=BG\).

- 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.

- Apply the coordinate rule to both coordinates of every vertex. - Separate the change in direction from the change in distance from the origin. - Which positive dilation factor changes each distance from the origin by the required amount? - Area scales by the square of the positive dilation factor.

1. Applying the rule gives \(A'=(4,2)\), \(B'=(-2,2)\), \(C'=(-2,-2)\), and \(D'=(4,-2)\). 2. First, a \(180^\circ\) rotation about the origin sends \((x,y)\) to \((-x,-y)\). 3. Then a dilation centered at the origin with scale factor \(2\) sends \((-x,-y)\) to \((-2x,-2y)\). 4. The dilation doubles every side length. Area is multiplied by \(2^2=4\).

a) \(A'=(4,2)\), \(B'=(-2,2)\), \(C'=(-2,-2)\), \(D'=(4,-2)\) b) Rotate \(180^\circ\) about the origin, then dilate from the origin with scale factor \(2\). Side lengths double and area is multiplied by \(4\).

- Start with what the diagonals of the original parallelogram tell you about \(S\). - Compare the remaining distances from \(P\) and \(Q\) to \(S\). - A half-turn pairs points on opposite rays when they are the same distance from the center. - Connect those point mappings to the diagonals of the new quadrilateral.

1. The diagonals of parallelogram \(ABCD\) bisect each other, so \(S\) is the midpoint of both \(\overline{AC}\) and \(\overline{BD}\). Therefore, a \(180^\circ\) rotation about \(S\) maps \(A\leftrightarrow C\) and \(B\leftrightarrow D\). 2. Because \(BS=DS\) and \(BP=DQ\), subtracting equal lengths gives \(SP=SQ\). 3. Points \(P\) and \(Q\) lie on opposite rays of line \(BD\) from \(S\), so the half-turn maps \(P\leftrightarrow Q\). 4. Thus, the rotation maps \(\overline{AP}\) to \(\overline{CQ}\) and \(\overline{AQ}\) to \(\overline{CP}\). 5. The half-turn therefore maps quadrilateral \(APCQ\) onto itself with opposite vertices paired. Its diagonals \(AC\) and \(PQ\) have the same midpoint \(S\), so they bisect each other. 6. A quadrilateral whose diagonals bisect each other is a parallelogram. Therefore, \(APCQ\) is a parallelogram.

The \(180^\circ\) rotation about \(S\) maps \(A\leftrightarrow C\). From \(BS=DS\) and \(BP=DQ\), it follows that \(SP=SQ\), so the same half-turn maps \(P\leftrightarrow Q\). Hence, \(AC\) and \(PQ\) bisect each other at \(S\), and \(APCQ\) is a parallelogram.

- Begin with the half-turn symmetry of the given parallelogram. - Compare the distances from \(B\) and \(D\) to the rotation center by adding equal collinear pieces. - A half-turn pairs points on opposite rays when their distances from the center agree.

1. The diagonals of parallelogram \(AKCF\) bisect each other at \(O\). Therefore, a \(180^\circ\) rotation about \(O\) maps \(A\leftrightarrow C\) and \(K\leftrightarrow F\). 2. Because \(KO=FO\) and \(BK=FD\), segment addition gives \(BO=DO\). 3. Points \(B\) and \(D\) lie on opposite rays of line \(KF\) from \(O\), so the half-turn maps \(B\leftrightarrow D\). 4. Thus, the rotation maps \(\overline{AB}\) to \(\overline{CD}\) and \(\overline{BC}\) to \(\overline{DA}\). 5. In particular, \(O\) is the midpoint of both \(\overline{AC}\) and \(\overline{BD}\). Therefore, the diagonals of \(ABCD\) bisect each other. 6. Hence, \(ABCD\) is a parallelogram.

The \(180^\circ\) rotation about \(O\) maps \(A\leftrightarrow C\), \(K\leftrightarrow F\), and, using \(BK=FD\), \(B\leftrightarrow D\). Thus, \(O\) is the midpoint of both diagonals \(AC\) and \(BD\), so \(ABCD\) is a parallelogram.

- 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.

- The parallel segments create two triangles with a shared vertex \(S\). - Compare their corresponding angles before comparing their side lengths. - Use the side ratio and the ray directions to interpret the similarity as a dilation.

1. Lines \(AC\) and \(BD\) meet at \(S\). 2. Since \(AB\parallel CD\), triangles \(SAB\) and \(SCD\) are similar by AA. 3. Therefore, \(\frac{SC}{SA}=\frac{SD}{SB}=\frac{CD}{AB}=\frac{9}{6}=\frac{3}{2}\). 4. Because each image point lies on the same ray from \(S\) as its preimage, the scale factor is positive. 5. A dilation centered at \(S\) with scale factor \(\frac{3}{2}\) sends \(A\) to \(C\) and \(B\) to \(D\). 6. Thus, \(S=AC\cap BD\) is the required center of dilation.

The dilation center is \(S=AC\cap BD\), and the scale factor is \(\frac{3}{2}\).

- Treat the center \(S\), not the origin, as the fixed reference point. - Compare the horizontal and vertical displacement from \(S\) to each vertex. - A scale factor of \(2\) doubles each displacement from the center.

1. From the graph, \(S=(1,-1)\), \(A=(2,0)\), \(B=(4,0)\), and \(C=(2,2)\). 2. For each vertex, start at \(S\) and double the displacement from \(S\) to that vertex. 3. This gives \(A'=(3,1)\), \(B'=(7,1)\), and \(C'=(3,5)\).

\(A'=(3,1)\), \(B'=(7,1)\), \(C'=(3,5)\)

- For points on \(m\), look at the rays that start at \(O\). - Ask whether moving a point farther along such a ray can take it off a line that already contains the ray. - Contrast that with a line that never passes through the center. - Use the scale factor on the perpendicular distance from \(O\) to \(n\).

1. Every point of \(m\) lies on a ray from \(O\) that is contained in \(m\). Dilating those points keeps them on \(m\), so line \(m\) maps to itself as a set. 2. From the graph, \(n\) is the line \(x=2\). Its distance from the center \(O\) is \(2\) units. 3. A dilation with scale factor \(2\) doubles that perpendicular distance while preserving the line's direction, so the image is the parallel line \(x=4\).

Line \(m\) maps to itself. The image of \(n\) is \(x=4\), a distinct line parallel to \(n\).

- 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 \(M_4\).

- Start with the rotational symmetry of an equilateral triangle. - Track a point on one side by its distance from a vertex. - Compare the directions of the two intersecting segments after the rotation.

1. A \(120^\circ\) counterclockwise rotation about the center of equilateral triangle \(ABC\) maps \(A\to B\), \(B\to C\), and \(C\to A\). 2. Point \(M\) lies on \(AB\) at distance \(AM\) from \(A\). Its image lies on \(BC\) at the same distance from \(B\). Since \(AM=BN\), the image of \(M\) is \(N\). 3. Therefore, the directed line from \(M\) toward \(C\) rotates onto the directed line from \(N\) toward \(A\). 4. Ray \(\overrightarrow{PC}\) has the same direction as \(\overrightarrow{MC}\), and ray \(\overrightarrow{PA}\) has the same direction as \(\overrightarrow{NA}\). The rotation between those directions is \(120^\circ\). 5. Hence, \(m\angle APC=120^\circ\).

The \(120^\circ\) counterclockwise rotation maps \(A\to B\to C\to A\) and, because \(AM=BN\), maps \(M\to N\). It maps the direction of \(MC\) to the direction of \(NA\), so \(m\angle APC=120^\circ\).

- 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.

- Avoid choosing the parallel radii along the line of centers; that choice makes the endpoint line coincide with the center line. - Use the two parallel radii to identify a pair of similar triangles. - Show that the intersection is the unique external division point of \(\overline{O_1O_2}\) in the ratio \(R:r\).

1. The two radii satisfy \(O_1A\parallel O_2B\), point in the same direction, and have lengths \(R\) and \(r\). 2. Because \(R\ne r\) and the radius direction is not parallel to \(O_1O_2\), line \(AB\) is not parallel to line \(O_1O_2\), so the lines meet at \(S\). 3. Triangles \(SO_1A\) and \(SO_2B\) are similar because \(O_1A\parallel O_2B\) and the other corresponding sides lie on the same two intersecting lines. 4. Thus \(\frac{SO_1}{SO_2}=\frac{O_1A}{O_2B}=\frac{R}{r}\). 5. Because the parallel radii point in the same direction and \(R\ne r\), point \(S\) lies outside segment \(\overline{O_1O_2}\). It is the unique external point on the line of centers whose distances to \(O_1\) and \(O_2\) have ratio \(R:r\). 6. That external division point depends only on \(O_1\), \(O_2\), \(R\), and \(r\), not on the chosen valid radius direction. Therefore, every valid choice produces the same point \(S\).

Choose same-direction parallel radii that are not parallel to the line of centers, and intersect the line through their endpoints with line \(O_1O_2\). The intersection \(S\) satisfies \(SO_1:SO_2=R:r\) and lies outside \(\overline{O_1O_2}\). Since there is exactly one external division point with that ratio, every valid radius direction gives the same external center of similarity.

- If \(P\) is a midpoint, the endpoints are images under a half-turn about \(P\). - Determine where the image of line \(\ell\) lies after that half-turn. - Use the fact that nonparallel lines intersect exactly once.

1. The app creates the image \(\ell'\) of line \(\ell\) under a \(180^\circ\) rotation about \(P\). The image line satisfies \(\ell'\parallel\ell\). 2. Since \(\ell\) is not parallel to \(m\), line \(\ell'\) meets \(m\) at exactly one point \(B\). 3. The app reflects \(B\) across \(P\) to point \(A\), so \(P\) is the midpoint of \(\overline{AB}\). 4. Because \(B\) lies on the rotated image \(\ell'\), its half-turn image \(A\) lies on \(\ell\). 5. The unique intersection \(B=\ell'\cap m\) makes the segment unique.

The app rotates \(\ell\) by \(180^\circ\) about \(P\) to obtain \(\ell'\parallel\ell\). Let \(B=\ell'\cap m\), then reflect \(B\) across \(P\) to obtain \(A\). The result is unique because \(\ell'\) and \(m\) have exactly one intersection.