Question

the composition ( d_{o,0.75}(x,y) circ d_{o,2}(x,y) ) is applied to ( \triangle lmn ) to create ( \triangle lmn ). which statements must be true regarding the two triangles? check all that apply. ( square angle m cong angle m ) ( square \triangle lmn sim \triangle lmn ) ( square \triangle lmn cong \triangle lmn ) ( square ) the coordinates of vertex ( l ) are ( (-3, 1.5) ). ( square ) the coordinates of vertex ( n ) are ( (3, -1.5) ). ( square ) the coordinates of vertex ( m ) are ( (1.5, -1.5) ).

Explanation:

Step1: Recall Dilation Properties

Dilation is a similarity transformation, so angles are preserved (congruent), and triangles are similar (not necessarily congruent). The composition \( D_{O,0.75}(x,y) \circ D_{O,2}(x,y) \) is a single dilation with scale factor \( 2 \times 0.75 = 1.5 \)? Wait, no: composition of dilations \( D_{O,k_1} \circ D_{O,k_2} \) is \( D_{O,k_1 \times k_2} \)? Wait, actually, \( D_{O,2} \) first scales by 2, then \( D_{O,0.75} \) scales by 0.75, so overall scale factor is \( 2 \times 0.75 = 1.5 \)? Wait, no: dilation composition: \( D_{O, k_2}(D_{O, k_1}(x,y)) = D_{O, k_1 \times k_2}(x,y) \). So first apply \( D_{O,2} \), then \( D_{O,0.75} \), so the combined scale factor is \( 2 \times 0.75 = 1.5 \)? Wait, no, order: \( D_{O,0.75} \circ D_{O,2} \) means apply \( D_{O,2} \) first, then \( D_{O,0.75} \). So for a point \( (x,y) \), \( D_{O,2}(x,y) = (2x, 2y) \), then \( D_{O,0.75}(2x, 2y) = (0.75 \times 2x, 0.75 \times 2y) = (1.5x, 1.5y) \). So the overall scale factor is 1.5.

First, find coordinates of original triangle:
- From graph: \( L(-1, 2) \), \( M(-1, -1) \), \( N(2, -1) \) (wait, check graph: M is at (-1, -1)? Wait, the grid: M is at (-1, -1)? Wait, L is at (-1, 2), M at (-1, -1), N at (2, -1). Let's confirm:

Original coordinates:
- \( L(-1, 2) \)
- \( M(-1, -1) \)
- \( N(2, -1) \)

Now apply \( D_{O,2} \) first: scale each coordinate by 2:
- \( L_1 = (2 \times -1, 2 \times 2) = (-2, 4) \)
- \( M_1 = (2 \times -1, 2 \times -1) = (-2, -2) \)
- \( N_1 = (2 \times 2, 2 \times -1) = (4, -2) \)

Then apply \( D_{O,0.75} \): scale each coordinate by 0.75:
- \( L'' = (0.75 \times -2, 0.75 \times 4) = (-1.5, 3) \)? Wait, wait, maybe I misread the original coordinates. Wait, looking at the graph: L is at (-1, 2)? Wait, the x-axis: -2, -1, 0, 1, 2. Y-axis: -2, -1, 0, 1, 2, 3, 4. Wait, L is at (-1, 2)? M is at (-1, -1)? N is at (2, -1)? Wait, maybe I made a mistake. Let's re-examine:

Looking at the grid: L is at (-1, 2) (x=-1, y=2), M is at (-1, -1) (x=-1, y=-1), N is at (2, -1) (x=2, y=-1). Wait, but when we apply \( D_{O,2} \), then \( D_{O,0.75} \), let's recalculate:

Wait, maybe the original coordinates are: L(-1, 2), M(-1, -1), N(2, -1). Let's check the options:

Option 1: \( \angle M \cong \angle M'' \). Dilation preserves angles, so this is true.

Option 2: \( \triangle LMN \sim \triangle L''M''N'' \). Dilation is a similarity transformation, so triangles are similar. True.

Option 3: \( \triangle LMN \cong \triangle L''M''N'' \). Congruent means same size and shape. Dilation changes size (scale factor 1.5), so not congruent. False.

Option 4: Coordinates of L'': Let's recalculate. Wait, maybe I messed up the order. Wait, \( D_{O,0.75} \circ D_{O,2} \) is apply \( D_{O,2} \) first, then \( D_{O,0.75} \). So original L: (-1, 2). \( D_{O,2}(-1,2) = (-2,4) \). Then \( D_{O,0.75}(-2,4) = (-1.5, 3) \). But the option says (-3, 1.5). That's different. Wait, maybe original L is (-2, 2)? Let's check the graph again. The grid: L is at (-2, 2)? Wait, the x-axis: -4, -2, 0, 2, 4. So L is at (-2, 2), M at (-2, -1), N at (2, -1). Ah, that's probably the mistake. Let's correct:

Original coordinates:
- \( L(-2, 2) \)
- \( M(-2, -1) \)
- \( N(2, -1) \)

Now apply \( D_{O,2} \) (scale by 2):
- \( L_1 = (2 \times -2, 2 \times 2) = (-4, 4) \)
- \( M_1 = (2 \times -2, 2 \times -1) = (-4, -2) \)
- \( N_1 = (2 \times 2, 2 \times -1) = (4, -2) \)

Then apply \( D_{O,0.75} \) (scale by 0.75):
- \( L'' = (0.75 \times -4, 0.75 \times 4) = (-3, 3) \)? No, that's not matching. Wait, maybe the order is reversed? \( D_{O,0.75} \circ D_{O,2} \) is \( D_{O,0.75}(D_{O,2}(x,y)) \), but maybe the dilation is about the origin, so coordinates (x,y) scaled by k: (kx, ky). Wait, maybe the original L is (-2, 2), M(-2, -1), N(2, -1). Let's check the option for L'': (-3, 1.5). Let's see: if we first apply \( D_{O,2} \), then \( D_{O,0.75} \), but maybe I got the scale factor wrong. Wait, 0.75 * 2 = 1.5? No, 2 * 0.75 = 1.5. Wait, maybe the original L is (-2, 2). Then \( D_{O,2} \) gives (-4, 4), then \( D_{O,0.75} \) gives (-3, 3). Not (-3, 1.5). Wait, maybe the original L is (-2, 2), and the first dilation is \( D_{O,0.75} \) then \( D_{O,2} \)? No, the composition is \( D_{O,0.75} \circ D_{O,2} \), so \( D_{O,2} \) first. Wait, maybe the original coordinates are different. Let's look at the graph again: L is at (-1, 2)? No, the grid lines: x=-2 is a vertical line, x=-1, 0, 1, 2. Y=2, 1, 0, -1, -2. So L is at (-1, 2)? M at (-1, -1)? N at (2, -1). Then:

\( D_{O,2}(-1,2) = (-2,4) \), \( D_{O,0.75}(-2,4) = (-1.5, 3) \). Not matching the option. Wait, the option says (-3, 1.5). Let's see: if original L is (-2, 2), then \( D_{O,0.75}(-2,2) = (-1.5, 1.5) \), then \( D_{O,2}(-1.5,1.5) = (-3, 3) \). No. Wait, maybe the composition is \( D_{O,2} \circ D_{O,0.75} \)? But the problem says \( D_{O,0.75} \circ D_{O,2} \). Wait, maybe I misread the dilation notation: \( D_{O, k} \) is dilation with center O and scale factor k. So \( D_{O,0.75} \circ D_{O,2} \) means apply \( D_{O,2} \) first, then \( D_{O,0.75} \). So for a point (x,y), first (2x, 2y), then (0.75*2x, 0.75*2y) = (1.5x, 1.5y). So scale factor 1.5.

Original coordinates: Let's check M: original M is (-1, -1)? Then M'' would be (1.5*(-1), 1.5*(-1)) = (-1.5, -1.5). But the option says (1.5, -1.5). Wait, maybe original M is (1, -1)? No, the graph shows M at (-1, -1), N at (2, -1), L at (-1, 2). Wait, maybe the original coordinates are:

L(-1, 2), M(-1, -1), N(2, -1). Then:

After \( D_{O,2} \): L( -2, 4), M(-2, -2), N(4, -2).

After \( D_{O,0.75} \): L(-1.5, 3), M(-1.5, -1.5), N(3, -1.5).

Now check options:

1. \( \angle M \cong \angle M'' \): Dilation preserves angles, so true.

2. \( \triangle LMN \sim \triangle L''M''N'' \): Dilation is similarity, so true.

3. \( \triangle LMN \cong \triangle L''M''N'' \): Not congruent (scale factor 1.5), false.

4. Coordinates of L'': (-3, 1.5)? No, we have (-1.5, 3). So false.

5. Coordinates of N'': (3, -1.5)? Yes, because N(2, -1) → \( D_{O,2}(2,-1)=(4,-2) \) → \( D_{O,0.75}(4,-2)=(3, -1.5) \). True.

6. Coordinates of M'': (1.5, -1.5)? M(-1, -1) → \( D_{O,2}(-1,-1)=(-2,-2) \) → \( D_{O,0.75}(-2,-2)=(-1.5, -1.5) \). So the option says (1.5, -1.5), which is false. Wait, maybe original M is (1, -1)? No, the graph shows M at (-1, -1). Wait, maybe I flipped the x-coordinate. Wait, if original M is (1, -1), then \( D_{O,2}(1,-1)=(2,-2) \) → \( D_{O,0.75}(2,-2)=(1.5, -1.5) \), which matches the option. But the graph shows M at (-1, -1). Wait, maybe the graph is mirrored? No, the x-axis is left to right, y-axis up to down.

Wait, maybe the original coordinates are:

L(-2, 2), M(-2, -1), N(2, -1). Then:

\( D_{O,2}(-2,2)=(-4,4) \), \( D_{O,0.75}(-4,4)=(-3,3) \) → L''(-3,3) – not matching option.

\( D_{O,2}(2,-1)=(4,-2) \), \( D_{O,0.75}(4,-2)=(3, -1.5) \) → N''(3, -1.5) – matches option 5.

\( D_{O,2}(-2,-1)=(-4,-2) \), \( D_{O,0.75}(-4,-2)=(-3, -1.5) \) → M''(-3, -1.5) – not matching option 6.

Wait, maybe the original M is (1, -1), L(-2, 2), N(2, -1). Then:

\( D_{O,2}(1,-1)=(2,-2) \), \( D_{O,0.75}(2,-2)=(1.5, -1.5) \) → M''(1.5, -1.5) – matches option 6.

\( D_{O,2}(-2,2)=(-4,4) \), \( D_{O,0.75}(-4,4)=(-3,3) \) → L''(-3,3) – not matching option 4.

\( D_{O,2}(2,-1)=(4,-2) \), \( D_{O,0.75}(4,-2)=(3, -1.5) \) → N''(3, -1.5) – matches option 5.

\( \angle M \cong \angle M'' \): true.

\( \triangle LMN \sim \triangle L''M''N'' \): true.

So let's re-express:

Original coordinates (correcting M to (1, -1)? No, the graph shows M at (-1, -1). Wait, maybe the grid is such that each square is 1 unit, so L is at (-1, 2), M at (-1, -1), N at (2, -1). Then:

- \( \angle M \cong \angle M'' \): true (dilation preserves angles).

- \( \triangle LMN \sim \triangle L''M''N'' \): true (dilation is similarity).

- \( \triangle LMN \cong \triangle L''M''N'' \): false (different scale).

- L'': (-1.5, 3) – option 4 says (-3, 1.5) – false.

- N'': (3, -1.5) – true (since N(2, -1) → 2*2=4, 4*0.75=3; -1*2=-2, -2*0.75=-1.5).

- M'': (-1.5, -1.5) – option 6 says (1.5, -1.5) – false. Wait, unless M is (1, -1). Then M(1, -1) → 1*2=2, 2*0.75=1.5; -1*2=-2, -2*0.75=-1.5 → M''(1.5, -1.5) – true. So maybe the original M is (1, -1). Let's check the graph again: M is at (-1, -1) or (1, -1)? The graph shows M to the left of the y-axis, so x=-1. So M(-1, -1). Then M'' is (-1.5, -1.5), not (1.5, -1.5). So option 6 is false.

So correct options:

- \( \angle M \cong \angle M'' \)

- \( \triangle LMN \sim \triangle L''M''N'' \)

- The coordinates of vertex N'' are (3, -1.5)

Wait, let's verify N'' again:

Original N: (2, -1)

First dilation \( D_{O,2} \): (2*2, -1*2) = (4, -2)

Second dilation \( D_{O,0.75} \): (4*0.75, -2*0.75) = (3, -1.5) – correct.

M'':

Original M: (-1, -1)

\( D_{O,2} \): (-2, -2)

\( D_{O,0.75} \): (-1.5, -1.5) – option 6 says (1.5, -1.5) – incorrect.

L'':

Original L: (-1, 2)

\( D_{O,2} \): (-2, 4)

Answer:

Step1: Recall Dilation Properties

Dilation is a similarity transformation, so angles are preserved (congruent), and triangles are similar (not necessarily congruent). The composition \( D_{O,0.75}(x,y) \circ D_{O,2}(x,y) \) is a single dilation with scale factor \( 2 \times 0.75 = 1.5 \)? Wait, no: composition of dilations \( D_{O,k_1} \circ D_{O,k_2} \) is \( D_{O,k_1 \times k_2} \)? Wait, actually, \( D_{O,2} \) first scales by 2, then \( D_{O,0.75} \) scales by 0.75, so overall scale factor is \( 2 \times 0.75 = 1.5 \)? Wait, no: dilation composition: \( D_{O, k_2}(D_{O, k_1}(x,y)) = D_{O, k_1 \times k_2}(x,y) \). So first apply \( D_{O,2} \), then \( D_{O,0.75} \), so the combined scale factor is \( 2 \times 0.75 = 1.5 \)? Wait, no, order: \( D_{O,0.75} \circ D_{O,2} \) means apply \( D_{O,2} \) first, then \( D_{O,0.75} \). So for a point \( (x,y) \), \( D_{O,2}(x,y) = (2x, 2y) \), then \( D_{O,0.75}(2x, 2y) = (0.75 \times 2x, 0.75 \times 2y) = (1.5x, 1.5y) \). So the overall scale factor is 1.5.

First, find coordinates of original triangle:

• From graph: \( L(-1, 2) \), \( M(-1, -1) \), \( N(2, -1) \) (wait, check graph: M is at (-1, -1)? Wait, the grid: M is at (-1, -1)? Wait, L is at (-1, 2), M at (-1, -1), N at (2, -1). Let's confirm:

Original coordinates:

• \( L(-1, 2) \)
• \( M(-1, -1) \)
• \( N(2, -1) \)

Now apply \( D_{O,2} \) first: scale each coordinate by 2:

• \( L_1 = (2 \times -1, 2 \times 2) = (-2, 4) \)
• \( M_1 = (2 \times -1, 2 \times -1) = (-2, -2) \)
• \( N_1 = (2 \times 2, 2 \times -1) = (4, -2) \)

Then apply \( D_{O,0.75} \): scale each coordinate by 0.75:

• \( L'' = (0.75 \times -2, 0.75 \times 4) = (-1.5, 3) \)? Wait, wait, maybe I misread the original coordinates. Wait, looking at the graph: L is at (-1, 2)? Wait, the x-axis: -2, -1, 0, 1, 2. Y-axis: -2, -1, 0, 1, 2, 3, 4. Wait, L is at (-1, 2)? M is at (-1, -1)? N is at (2, -1)? Wait, maybe I made a mistake. Let's re-examine:

Looking at the grid: L is at (-1, 2) (x=-1, y=2), M is at (-1, -1) (x=-1, y=-1), N is at (2, -1) (x=2, y=-1). Wait, but when we apply \( D_{O,2} \), then \( D_{O,0.75} \), let's recalculate:

Wait, maybe the original coordinates are: L(-1, 2), M(-1, -1), N(2, -1). Let's check the options:

Option 1: \( \angle M \cong \angle M'' \). Dilation preserves angles, so this is true.

Option 2: \( \triangle LMN \sim \triangle L''M''N'' \). Dilation is a similarity transformation, so triangles are similar. True.

Option 3: \( \triangle LMN \cong \triangle L''M''N'' \). Congruent means same size and shape. Dilation changes size (scale factor 1.5), so not congruent. False.

Option 4: Coordinates of L'': Let's recalculate. Wait, maybe I messed up the order. Wait, \( D_{O,0.75} \circ D_{O,2} \) is apply \( D_{O,2} \) first, then \( D_{O,0.75} \). So original L: (-1, 2). \( D_{O,2}(-1,2) = (-2,4) \). Then \( D_{O,0.75}(-2,4) = (-1.5, 3) \). But the option says (-3, 1.5). That's different. Wait, maybe original L is (-2, 2)? Let's check the graph again. The grid: L is at (-2, 2)? Wait, the x-axis: -4, -2, 0, 2, 4. So L is at (-2, 2), M at (-2, -1), N at (2, -1). Ah, that's probably the mistake. Let's correct:

Original coordinates:

• \( L(-2, 2) \)
• \( M(-2, -1) \)
• \( N(2, -1) \)

Now apply \( D_{O,2} \) (scale by 2):

• \( L_1 = (2 \times -2, 2 \times 2) = (-4, 4) \)
• \( M_1 = (2 \times -2, 2 \times -1) = (-4, -2) \)
• \( N_1 = (2 \times 2, 2 \times -1) = (4, -2) \)

Then apply \( D_{O,0.75} \) (scale by 0.75):

• \( L'' = (0.75 \times -4, 0.75 \times 4) = (-3, 3) \)? No, that's not matching. Wait, maybe the order is reversed? \( D_{O,0.75} \circ D_{O,2} \) is \( D_{O,0.75}(D_{O,2}(x,y)) \), but maybe the dilation is about the origin, so coordinates (x,y) scaled by k: (kx, ky). Wait, maybe the original L is (-2, 2), M(-2, -1), N(2, -1). Let's check the option for L'': (-3, 1.5). Let's see: if we first apply \( D_{O,2} \), then \( D_{O,0.75} \), but maybe I got the scale factor wrong. Wait, 0.75 2 = 1.5? No, 2 0.75 = 1.5. Wait, maybe the original L is (-2, 2). Then \( D_{O,2} \) gives (-4, 4), then \( D_{O,0.75} \) gives (-3, 3). Not (-3, 1.5). Wait, maybe the original L is (-2, 2), and the first dilation is \( D_{O,0.75} \) then \( D_{O,2} \)? No, the composition is \( D_{O,0.75} \circ D_{O,2} \), so \( D_{O,2} \) first. Wait, maybe the original coordinates are different. Let's look at the graph again: L is at (-1, 2)? No, the grid lines: x=-2 is a vertical line, x=-1, 0, 1, 2. Y=2, 1, 0, -1, -2. So L is at (-1, 2)? M at (-1, -1)? N at (2, -1). Then:

\( D_{O,2}(-1,2) = (-2,4) \), \( D_{O,0.75}(-2,4) = (-1.5, 3) \). Not matching the option. Wait, the option says (-3, 1.5). Let's see: if original L is (-2, 2), then \( D_{O,0.75}(-2,2) = (-1.5, 1.5) \), then \( D_{O,2}(-1.5,1.5) = (-3, 3) \). No. Wait, maybe the composition is \( D_{O,2} \circ D_{O,0.75} \)? But the problem says \( D_{O,0.75} \circ D_{O,2} \). Wait, maybe I misread the dilation notation: \( D_{O, k} \) is dilation with center O and scale factor k. So \( D_{O,0.75} \circ D_{O,2} \) means apply \( D_{O,2} \) first, then \( D_{O,0.75} \). So for a point (x,y), first (2x, 2y), then (0.752x, 0.752y) = (1.5x, 1.5y). So scale factor 1.5.

Original coordinates: Let's check M: original M is (-1, -1)? Then M'' would be (1.5(-1), 1.5(-1)) = (-1.5, -1.5). But the option says (1.5, -1.5). Wait, maybe original M is (1, -1)? No, the graph shows M at (-1, -1), N at (2, -1), L at (-1, 2). Wait, maybe the original coordinates are:

L(-1, 2), M(-1, -1), N(2, -1). Then:

After \( D_{O,2} \): L( -2, 4), M(-2, -2), N(4, -2).

After \( D_{O,0.75} \): L(-1.5, 3), M(-1.5, -1.5), N(3, -1.5).

Now check options:

1. \( \angle M \cong \angle M'' \): Dilation preserves angles, so true.

1. \( \triangle LMN \sim \triangle L''M''N'' \): Dilation is similarity, so true.

1. \( \triangle LMN \cong \triangle L''M''N'' \): Not congruent (scale factor 1.5), false.

1. Coordinates of L'': (-3, 1.5)? No, we have (-1.5, 3). So false.

1. Coordinates of N'': (3, -1.5)? Yes, because N(2, -1) → \( D_{O,2}(2,-1)=(4,-2) \) → \( D_{O,0.75}(4,-2)=(3, -1.5) \). True.

1. Coordinates of M'': (1.5, -1.5)? M(-1, -1) → \( D_{O,2}(-1,-1)=(-2,-2) \) → \( D_{O,0.75}(-2,-2)=(-1.5, -1.5) \). So the option says (1.5, -1.5), which is false. Wait, maybe original M is (1, -1)? No, the graph shows M at (-1, -1). Wait, maybe I flipped the x-coordinate. Wait, if original M is (1, -1), then \( D_{O,2}(1,-1)=(2,-2) \) → \( D_{O,0.75}(2,-2)=(1.5, -1.5) \), which matches the option. But the graph shows M at (-1, -1). Wait, maybe the graph is mirrored? No, the x-axis is left to right, y-axis up to down.

Wait, maybe the original coordinates are:

L(-2, 2), M(-2, -1), N(2, -1). Then:

\( D_{O,2}(-2,2)=(-4,4) \), \( D_{O,0.75}(-4,4)=(-3,3) \) → L''(-3,3) – not matching option.

\( D_{O,2}(2,-1)=(4,-2) \), \( D_{O,0.75}(4,-2)=(3, -1.5) \) → N''(3, -1.5) – matches option 5.

\( D_{O,2}(-2,-1)=(-4,-2) \), \( D_{O,0.75}(-4,-2)=(-3, -1.5) \) → M''(-3, -1.5) – not matching option 6.

Wait, maybe the original M is (1, -1), L(-2, 2), N(2, -1). Then:

\( D_{O,2}(1,-1)=(2,-2) \), \( D_{O,0.75}(2,-2)=(1.5, -1.5) \) → M''(1.5, -1.5) – matches option 6.

\( D_{O,2}(-2,2)=(-4,4) \), \( D_{O,0.75}(-4,4)=(-3,3) \) → L''(-3,3) – not matching option 4.

\( D_{O,2}(2,-1)=(4,-2) \), \( D_{O,0.75}(4,-2)=(3, -1.5) \) → N''(3, -1.5) – matches option 5.

\( \angle M \cong \angle M'' \): true.

\( \triangle LMN \sim \triangle L''M''N'' \): true.

So let's re-express:

Original coordinates (correcting M to (1, -1)? No, the graph shows M at (-1, -1). Wait, maybe the grid is such that each square is 1 unit, so L is at (-1, 2), M at (-1, -1), N at (2, -1). Then:

• \( \angle M \cong \angle M'' \): true (dilation preserves angles).

• \( \triangle LMN \sim \triangle L''M''N'' \): true (dilation is similarity).

• \( \triangle LMN \cong \triangle L''M''N'' \): false (different scale).

• L'': (-1.5, 3) – option 4 says (-3, 1.5) – false.

• N'': (3, -1.5) – true (since N(2, -1) → 22=4, 40.75=3; -12=-2, -20.75=-1.5).

• M'': (-1.5, -1.5) – option 6 says (1.5, -1.5) – false. Wait, unless M is (1, -1). Then M(1, -1) → 12=2, 20.75=1.5; -12=-2, -20.75=-1.5 → M''(1.5, -1.5) – true. So maybe the original M is (1, -1). Let's check the graph again: M is at (-1, -1) or (1, -1)? The graph shows M to the left of the y-axis, so x=-1. So M(-1, -1). Then M'' is (-1.5, -1.5), not (1.5, -1.5). So option 6 is false.

So correct options:

• \( \angle M \cong \angle M'' \)

• \( \triangle LMN \sim \triangle L''M''N'' \)

• The coordinates of vertex N'' are (3, -1.5)

Wait, let's verify N'' again:

Original N: (2, -1)

First dilation \( D_{O,2} \): (22, -12) = (4, -2)

Second dilation \( D_{O,0.75} \): (40.75, -20.75) = (3, -1.5) – correct.

M'':

Original M: (-1, -1)

\( D_{O,2} \): (-2, -2)

\( D_{O,0.75} \): (-1.5, -1.5) – option 6 says (1.5, -1.5) – incorrect.

L'':

Original L: (-1, 2)

\( D_{O,2} \): (-2, 4)