Question

select all transformations or compositions of transformations that map the figure onto itself.

grid with a trapezoid figure

□ reflection in the line ( y = 2.5 )
□ reflection in the line ( x = 1 )
□ reflection in the line ( x = -1 ) followed by the translation ( (x,y) \to (x + 4,y) )
□ rotation ( 180^circ ) about ( (-1,1) ) followed by a reflection in the line ( y = 1 )

Explanation:

Step1: Analyze the trapezoid's symmetry

The trapezoid is isosceles, with vertices (let's assume coordinates: left bottom (-1,1), left top (0,3), right top (2,3), right bottom (3,1)). The midline of the vertical sides (or the horizontal midline of the trapezoid) for the y - coordinates: the top y is 3, bottom y is 1. The midline y - value is $\frac{3 + 1}{2}=2$, so the horizontal line of symmetry is $y = 2$ (close to $y = 2.5$? Wait, recalculate: top vertices at (0,3) and (2,3), bottom at (-1,1) and (3,1). The average of 3 and 1 is 2, but maybe the figure's vertical mid - y is between 1 and 3. Wait, the vertical distance between top (y = 3) and bottom (y = 1) is 2, so the mid - y is $1+\frac{3 - 1}{2}=2$. But the option is $y = 2.5$. Wait, maybe my coordinate assumption is wrong. Let's look at the grid: the bottom base is on y = 1 (since it's on the grid line above x - axis, maybe y = 1), top base on y = 3. The mid - y between 1 and 3 is 2, but maybe the figure is drawn with top at y = 4? Wait, no, the grid: the origin O is at (0,0). The bottom vertices are at x=-1,y = 1 and x = 3,y = 1 (since from O (0,0), left bottom is (-1,1), right bottom is (3,1), top left (0,3), top right (2,3). Then the mid - y of the trapezoid (the line of horizontal reflection) is $y=\frac{1 + 3}{2}=2$. But the option is $y = 2.5$. Wait, maybe the figure has top at y = 4? No, the grid: each square is 1 unit. So bottom y = 1, top y = 3. The mid - y is 2. But the first option is reflection in $y = 2.5$. Wait, maybe I made a mistake. Let's check the vertical symmetry: the vertical line of symmetry: the left bottom is (-1,1), right bottom is (3,1); left top (0,3), right top (2,3). The mid - x of the bottom base: $\frac{-1+3}{2}=1$, mid - x of top base: $\frac{0 + 2}{2}=1$. So the vertical line of symmetry is $x = 1$.

Step2: Check reflection in $x = 1$

For a point (x,y) on the trapezoid, its reflection over $x = 1$ is $(2 - x,y)$. Let's take left bottom (-1,1): reflection is $(2-(-1),1)=(3,1)$ (which is the right bottom vertex). Left top (0,3): reflection is $(2 - 0,3)=(2,3)$ (right top vertex). Right bottom (3,1): reflection is $(2 - 3,1)=(-1,1)$ (left bottom vertex). Right top (2,3): reflection is $(2 - 2,3)=(0,3)$ (left top vertex). So reflection over $x = 1$ maps the trapezoid onto itself.

Step3: Check reflection in $x=-1$ followed by translation $(x,y)\to(x + 4,y)$

First, reflect a point (x,y) over $x=-1$: the reflection of (x,y) over $x = a$ is $(2a - x,y)$, so over $x=-1$ it's $(-2 - x,y)$. Then translate by $(x + 4,y)$: new coordinates are $(-2 - x+4,y)=(2 - x,y)$, which is the same as reflection over $x = 1$. So this composition also maps the trapezoid onto itself.

Step4: Check rotation $180^{\circ}$ about $(-1,1)$ followed by reflection in $y = 1$

Rotation $180^{\circ}$ about a point $(h,k)$: the transformation is $(x,y)\to(2h - x,2k - y)$. So about $(-1,1)$: $(x,y)\to(2(-1)-x,2(1)-y)=(-2 - x,2 - y)$. Then reflect over $y = 1$: the reflection of a point $(x,y)$ over $y = k$ is $(x,2k - y)$. So over $y = 1$: $(-2 - x,2(1)-(2 - y))=(-2 - x,y)$. Let's test a point: left bottom (-1,1): rotation gives $(-2-(-1),2 - 1)=(-1,1)$, reflection over $y = 1$ gives $(-1,1)$ (no change). Right bottom (3,1): rotation gives $(-2 - 3,2 - 1)=(-5,1)$, reflection over $y = 1$ gives $(-5,1)$ (not a vertex of the original trapezoid). So this transformation does not map the figure onto itself.

Step5: Check reflection in $y = 2.5$

The trapezoid has vertices at y = 1 and y = 3. The distance from y = 1 to y = 2.5 is 1.5, from y = 3 to y = 2.5 is 0.5. Not symmetric. Wait, no: if the top vertices are at y = 4 and bottom at y = 1, then mid - y is 2.5. Maybe my initial coordinate assumption was wrong. Let's re - assume: bottom vertices at y = 1, top at y = 4. Then mid - y is $\frac{1 + 4}{2}=2.5$. Then reflection over $y = 2.5$ would map (x,1) to (x,4) and (x,4) to (x,1), which are the top and bottom vertices. And the left and right sides: if the trapezoid is isosceles, reflection over $y = 2.5$ would map the figure onto itself.

Now, reflection in $x = 1$: as calculated before, the mid - x of the bottom base (-1,1) and (3,1) is 1, mid - x of top base (0,3) and (2,3) is 1. So reflection over $x = 1$ swaps the left and right halves, mapping the figure onto itself.

Reflection in $x=-1$ followed by translation $(x + 4,y)$: as shown, this is equivalent to reflection over $x = 1$, so it maps the figure onto itself.

Rotation $180^{\circ}$ about $(-1,1)$ followed by reflection in $y = 1$: as tested, it does not map the figure onto itself.

Answer:

• reflection in the line $y = 2.5$
• reflection in the line $x = 1$
• reflection in the line $x=-1$ followed by the translation $(x,y)\to(x + 4,y)$