- Read the coordinates of each original vertex carefully from the grid. - A translation changes every point by the same horizontal and vertical amounts. - Keep each image label matched with its original vertex.

1. Read the vertices from the graph: \(A=(-4,3)\), \(B=(-1,3)\), and \(C=(-2,1)\). 2. A translation by \(\langle 3,-2\rangle\) sends \((x,y)\) to \((x+3,y-2)\). 3. Therefore, \(A'=(-1,1)\), \(B'=(2,1)\), and \(C'=(1,-1)\).

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

- Compare the x-coordinates of the point and its image. - Compare the y-coordinates in the same direction, from preimage to image.

1. From the graph, \(P=(-4,1)\) and \(P'=(1,4)\). 2. The horizontal change is \(1-(-4)=5\), and the vertical change is \(4-1=3\). 3. Therefore, the translation vector is \(\langle 5,3\rangle\).

\(\langle 5,3\rangle\)

- What happens to a point already on the mirror line? - Check each point's location relative to the dashed line.

1. A point on a reflection line is fixed by the reflection. 2. Points \(A\) and \(C\) lie on \(m\); \(B\) is to its right. 3. Therefore \(A\) and \(C\) stay in place, but \(B\) moves.

\(A\) and \(C\) stay fixed because they lie on the mirror line.

- A reflection line lies halfway between a point and its image. - Look for a point directly above or below \(P\), not diagonally across.

1. The mirror line is \(y=1\). Point \(P=(2,3)\) is \(2\) units above it. 2. The reflection must keep the x-coordinate \(2\) and land \(2\) units below the line, at \((2,-1)\). 3. That is point \(M\). Point \(N\) has the wrong x-coordinate, and \(T\) lies on the mirror line.

\(M\), because it is directly across the line from \(P\), at the same perpendicular distance.

- Which coordinate measures left or right position? - What happens to the vertical position when a point reflects across the x-axis?

1. From the graph, \(P=(-3,2)\) and \(Q=(2,-1)\). 2. Reflection across the x-axis keeps each x-coordinate and reverses the sign of its y-coordinate. 3. Therefore \(P'=(-3,-2)\) and \(Q'=(2,1)\).

\(P'=(-3,-2)\), \(Q'=(2,1)\)

- Track how the coordinates of a point change under each reflection. - Find the new rule by changing the appropriate coordinate, not by guessing the line's tilt. - Which axis contains the point in part c)?

1. Reflection across the y-axis sends \((x,y)\) to \((-x,y)\). Thus \(h_1(x)=h(-x)=1.5x+6\). 2. Reflection across the x-axis sends \((x,y)\) to \((x,-y)\). Thus \(h_2(x)=-h(x)=1.5x-6\). 3. The point \((0,6)\) lies on the y-axis, so reflection across that axis leaves it at \((0,6)\). 4. Reflection across the x-axis changes the sign of its y-coordinate, giving \((0,-6)\).

a) \(h_1(x)=1.5x+6\) b) \(h_2(x)=1.5x-6\) c) Across the y-axis: \((0,6)\); across the x-axis: \((0,-6)\). The first point stays fixed because it lies on the mirror line.

- A mirror line passes through the midpoint of each segment joining a point to its image. - Look at whether the joining segments are horizontal or vertical. - Check the proposed line with the second point pair, not just the first.

1. The midpoint of \(PP'\) is \((3,2)\). Segment \(PP'\) is horizontal, so its perpendicular bisector is the vertical line \(x=3\). 2. The midpoint of \(QQ'\) is also on \(x=3\), since \((-2+8)/2=3\). Both point pairs lie at equal horizontal distances on opposite sides of that line. 3. Therefore, both mappings are reflections across \(x=3\).

a) \(x=3\) b) Both joining segments are horizontal and have midpoints on \(x=3\). Their endpoints are equally far from opposite sides of the line.

- A translation adds the same horizontal and vertical changes to every point. - Find each direction vector by subtracting the starting point from the ending point. - What cancels when the same displacement is added to both endpoints?

1. Add \(2\) to each x-coordinate and subtract \(3\) from each y-coordinate: \(A'=(3,-1)\), \(B'=(7,1)\). 2. The vector from \(A\) to \(B\) is \(\langle 5-1,4-2\rangle=\langle 4,2\rangle\). 3. The vector from \(A'\) to \(B'\) is \(\langle 7-3,1-(-1)\rangle=\langle 4,2\rangle\). 4. The lines have the same direction. They are distinct parallel lines because their points differ by a translation not along the original line.

a) \(A'=(3,-1)\), \(B'=(7,1)\) b) Both direction vectors are \(\langle 4,2\rangle\), so \(AB\) and \(A'B'\) lie on parallel lines.

- What happens to the x-coordinate of each point in a rightward translation? - Express the old x-coordinate using the new one before rewriting the equation. - Test the translated point in the image equation rather than the original equation.

1. Shifting the graph right by \(6\) replaces \(x\) in the rule with \(x-6\): \(y=\frac{1}{2}(x-6)-4=\frac{1}{2}x-7\). 2. The point \((0,-4)\) moves to \((6,-4)\). 3. Substituting \(x=6\) into the new equation gives \(y=\frac{1}{2}\cdot6-7=-4\), as required.

a) \(y=\frac{1}{2}x-7\) b) \((6,-4)\) lies on the image line because \(\frac{1}{2}\cdot6-7=-4\).

- Compare how a point and its reflection are positioned relative to the \(y\)-axis. - Track what changes and what stays the same in each ordered pair. - After locating the image vertices, follow the labeled vertices in order around each triangle.

1. Read the vertices from the graph: \(P=(-3,1)\), \(Q=(-1,4)\), and \(R=(1,1)\). 2. Reflection across the \(y\)-axis sends \((x,y)\) to \((-x,y)\). 3. Thus, \(P'=(3,1)\), \(Q'=(1,4)\), and \(R'=(-1,1)\). 4. A reflection reverses orientation, so the order \(P\to Q\to R\) and the corresponding order \(P'\to Q'\to R'\) have opposite orientations.

\(P'=(3,1)\), \(Q'=(1,4)\), \(R'=(-1,1)\); the orientation is reversed.

- Where does a mirror line cross the segment joining a point and its image? - What angle must the mirror line make with that segment? - Look for a single line with both properties for both point pairs.

1. The mirror line passes through the midpoint of \(AA'\) and is perpendicular to \(AA'\). 2. Segment \(AA'\) is horizontal, so its perpendicular bisector is vertical. 3. Segment \(BB'\) is also horizontal, and its midpoint lies on the same vertical line as the midpoint of \(AA'\). 4. Therefore, the vertical line through both midpoints is the common mirror line.

The mirror line is vertical, halfway between each point and its image. It is the perpendicular bisector of both \(AA'\) and \(BB'\).

- Trace the labeled vertices in order around each quadrilateral rather than comparing only side lengths. - Which rigid transformations preserve clockwise versus counterclockwise orientation? - Separate properties shared by all rigid motions from the property that distinguishes these two transformations.

1. Following \(A\to B\to C\to D\) around the first quadrilateral gives the opposite orientation from following \(A'\to B'\to C'\to D'\) around the image. 2. A translation preserves orientation, so no translation can produce this labeled image. 3. A reflection reverses orientation, so a reflection is consistent with the diagram. 4. Both translations and reflections preserve distances and angle measures.

Iker's claim is not possible. The labeled vertex order has reversed orientation, which a translation cannot do but a reflection can. Both transformations preserve distances and angle measures.

- Measure each vertex's horizontal distance from the line \(x=2\). - A reflection keeps the y-coordinate unchanged for a vertical mirror line. - Place each image the same horizontal distance on the opposite side of the mirror.

1. From the graph, \(A=(-2,3)\), \(B=(0,1)\), and \(C=(-3,-1)\). 2. Reflection across \(x=2\) keeps each y-coordinate and places the x-coordinates equally far from \(2\) on the other side. Equivalently, \(x'=4-x\). 3. Therefore, \(A'=(6,3)\), \(B'=(4,1)\), and \(C'=(7,-1)\).

\(A'=(6,3)\), \(B'=(4,1)\), \(C'=(7,-1)\)

- Compare each endpoint's vertical distance from \(y=-1\). - A horizontal reflection leaves the x-coordinate unchanged. - The mirror line must bisect the vertical segment joining a point and its image.

1. From the graph, \(M=(-3,3)\) and \(N=(2,1)\). 2. Reflection across \(y=-1\) keeps each x-coordinate and sends \(y\) to \(-2-y\). 3. Thus, \(M'=(-3,-5)\) and \(N'=(2,-3)\). 4. The midpoint of each point-image pair lies on the reflection line, so its y-coordinate is \(-1\).

\(M'=(-3,-5)\), \(N'=(2,-3)\); each point-image midpoint has y-coordinate \(-1\).

- Find the horizontal and vertical changes from \(A\) to \(A'\). - Apply those changes, rather than the final coordinates of \(A'\), to \(B\).

1. From \(A\) to \(A'\), the horizontal change is \(3-(-1)=4\), and the vertical change is \(1-4=-3\). 2. The translation vector is \(\langle4,-3\rangle\). 3. Apply the same changes to \(B\): \(B'=(2+4,-2-3)=(6,-5)\).

Vector \(\langle4,-3\rangle\); \(B'=(6,-5)\). Every point is shifted by the same vector.

- Compare the movement from each original vertex to its image. Is the displacement the same for all three vertices? - For the other image, which coordinate stays the same and which changes sign? - Trace the vertices around each triangle in order. Which motion changes clockwise to counterclockwise?

1. From the graph, \(A=(1,1)\), \(B=(3,1)\), \(C=(1,3)\). Their first images are \(A'=(5,2)\), \(B'=(7,2)\), \(C'=(5,4)\). 2. All first-image vertices move \(4\) units right and \(1\) unit up. Thus \(A'B'C'\) is a translation. 3. The second image has \(A''=(-1,1)\), \(B''=(-3,1)\), \(C''=(-1,3)\), reversing every x-coordinate. It is the y-axis reflection. The order \(A,B,C\) runs counterclockwise, but \(A'',B'',C''\) runs clockwise, so the reflection reverses orientation. 4. Both translations and reflections are rigid motions, so both preserve every side length and angle measure.

a) \(A'B'C'\) is a translation by \(\langle4,1\rangle\). \(A''B''C''\) is a reflection across the y-axis. b) \(A''B''C''\) reverses orientation. c) Yes. Both preserve side lengths and angles.

- Find the slope from the two given points. - Which points stay fixed under a reflection across \(y = 2\)? - Reflect the origin by using its distance from \(y = 2\). - Compare the horizontal and vertical changes before and after the reflection.

1. The slope through \((0, 0)\) and \((4, 2)\) is \(\frac{2}{4} = 0.5\), so \(k(x) = 0.5x\). 2. Point \(B(4, 2)\) lies on the line of reflection, so it stays fixed. The origin is 2 units below \(y = 2\), so it reflects to \((0, 4)\). 3. The image line through \((0, 4)\) and \((4, 2)\) has slope \(\frac{2 - 4}{4 - 0} = -0.5\), so \(k'(x) = -0.5x + 4\). 4. A horizontal reflection keeps horizontal changes the same but reverses vertical changes. Therefore, \(\frac{\Delta y}{\Delta x}\) changes sign.

a) \(k(x) = 0.5x\) b) \(k'(x) = -0.5x + 4\) c) The reflection reverses \(\Delta y\) while keeping \(\Delta x\) unchanged, so the slope changes sign.

- Identify the reflection symmetry of the outer isosceles triangle. - Track a point on one congruent side by its distance from the corresponding base vertex. - What happens to the intersection of two lines when the reflection swaps those two lines?

1. The reflection across the symmetry axis of isosceles \(\triangle ADC\)—the line through \(D\) and the midpoint of \(AC\)—maps \(A\leftrightarrow C\) and side \(AD\) onto side \(CD\). 2. Reflection preserves distance. The image of \(E\) must lie on \(CD\) at the same distance from \(C\) that \(E\) lies from \(A\). Since \(AE=CF\), the image of \(E\) is \(F\). 3. Therefore, line \(CE\) reflects to line \(AF\), and line \(AF\) reflects to line \(CE\). 4. Their intersection \(B\) must map to itself, so \(B\) lies on the mirror line. 5. Since the reflection maps \(A\) to \(C\) and fixes \(B\), it maps \(\overline{BA}\) to \(\overline{BC}\). Reflections preserve distance, so \(BA=BC\). 6. Therefore, \(\triangle ABC\) is isosceles.

Reflect across the symmetry axis through \(D\) and the midpoint of \(AC\). The reflection maps \(A\leftrightarrow C\) and, because \(AE=CF\), maps \(E\leftrightarrow F\). Thus, lines \(AF\) and \(CE\) are images of each other, so their intersection \(B\) is fixed. Hence, \(BA=BC\), and \(\triangle ABC\) is isosceles.

- Reflection preserves distance to points on the mirror line. - Replace the broken path to \(B\) with a path to its reflected point. - The shortest distance between two points is a straight segment. - Use the triangle inequality to compare any other point on the bank.

1. Reflect \(B\) across line \(g\) to point \(B'\). 2. For every point \(X\) on \(g\), reflection preserves distance and fixes \(X\), so \(XB = XB'\). 3. Therefore, \(AX + XB = AX + XB'\). 4. Draw \(\overline{AB'}\). Let its intersection with \(g\) be \(T\). Since \(A\), \(T\), and \(B'\) are collinear, \(AT + TB = AT + TB' = AB'\). 5. For any other point \(Q\) on \(g\), the triangle inequality gives \(AQ + QB' > AB'\). 6. Since \(QB' = QB\), \(AQ + QB > AT + TB\). Thus, \(T\) is the unique minimizing location.

Reflect \(B\) across \(g\) to \(B'\). Place the water station at \(T = \overline{AB'} \cap g\). This location minimizes \(AT + TB\).

- Begin with the two endpoints of the base that is already bisected by \(s\). - Track the image of a ray using the given equal base angles. - Use the equal leg lengths to identify the image of the remaining vertex.

1. Reflection across \(s\) exchanges \(A\) and \(B\) because \(s\) is the perpendicular bisector of \(AB\). 2. Equal base angles mean the reflection maps ray \(AD\) to ray \(BC\). 3. The lengths \(AD\) and \(BC\) are equal, so the point \(D\) on the first ray maps to \(C\) on the second. 4. A reflection line is the perpendicular bisector of every segment joining a point to its image. Thus \(s\) perpendicularly bisects \(CD\).

Reflection across \(s\) sends \(A\) to \(B\). The equal base angles send ray \(AD\) to ray \(BC\), and \(AD=BC\) sends \(D\) to \(C\). Therefore \(s\) is also the perpendicular bisector of \(CD\).

- What can you say about \(XB\) and \(XB'\) when \(X\) lies on the mirror line? - Compare a path from \(A\) to \(B'\) through \(X\) with the direct segment \(AB'\). - When does a route made of two segments have the same length as the straight segment joining its endpoints?

1. For any point \(X\) on \(\ell\), reflection preserves distance and fixes \(X\), so \(XB=XB'\). 2. The length of a route through \(X\) is \(AX+XB=AX+XB'\). 3. By the triangle inequality, \(AX+XB'\ge AB'\). Equality holds when \(A\), \(X\), and \(B'\) lie on the same segment. 4. Point \(P\) is on \(AB'\) and \(\ell\). Hence \(AP+PB=AP+PB'=AB'\), so no other point on \(\ell\) gives a shorter path.

For any \(X\) on \(\ell\), \(AX+XB=AX+XB'\ge AB'\). At \(P\), equality holds because \(A,P,B'\) lie on one segment. Thus \(AP+PB=AB'\) is the shortest possible length.

- A vertical translation changes the constant term of the boundary. - Keep the same inequality direction because the shaded side moves with the line. - Test the point separately in both regions.

1. Translating upward \(5\) units changes the boundary from \(y=-x+2\) to \(y=-x+7\), with the same inclusive side. The new inequality is \(y\le-x+7\). 2. For the original region at \((3, 4)\), \(4\le-3+2=-1\) is false. 3. For the translated region, \(4\le-3+7=4\) is true. 4. The point belongs only to the translated region.

a) \(y\le-x+7\) b) \((3, 4)\) belongs only to the translated region.

- An x-axis reflection changes the sign of the vertical coordinate. - Substitute the transformed coordinate before simplifying. - Multiplying by a negative number reverses the comparison.

1. Under reflection across the x-axis, a reflected point \((x,y)\) comes from \((x,-y)\) in the original region. 2. Substitute \(-y>-x+1\). 3. Multiply by \(-1\) and reverse the comparison: \(y<x-1\). 4. The inequality remains strict, so the boundary is dashed.

The reflected region is \(y<x-1\), with a dashed boundary.

- Apply the same translation to every boundary and every vertex. - For the slanted boundary, express the old coordinates in terms of the new coordinates. - Check one translated vertex in all three new inequalities.

1. Translating \(x\le0\) right by \(4\) gives \(x\le4\). 2. Translating \(y\ge0\) down by \(1\) gives \(y\ge-1\). 3. Substitute \(x-4\) for x and \(y+1\) for y in \(y\le x+3\): \(y+1\le x-4+3\), so \(y\le x-2\). 4. The original vertices \((-3, 0)\), \((0, 0)\), and \((0, 3)\) translate to \((1, -1)\), \((4, -1)\), and \((4, 2)\).

The translated system is \(x\le4\), \(y\ge-1\), and \(y\le x-2\). Its vertices are \((1, -1)\), \((4, -1)\), and \((4, 2)\).