QUESTION IMAGE
Question
the composition $d_{o,0.75}(x,y) \cdot d_{o,2}(x,y)$ 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 $(1.5, -1.5)$.
(image of a coordinate grid with triangle lmn plotted: l at (-1,2), m at (-1,-1), n at (2,-1))
Step1: Understand Dilations
Dilations are similarity transformations, so angles are preserved, and triangles are similar. First, find coordinates of original vertices. From the graph: \( L(-1, 2) \), \( M(-1, -1) \), \( N(2, -1) \).
Step2: Apply \( D_{O,2} \) (dilation by 2 from origin)
For a point \((x,y)\), \( D_{O,2}(x,y)=(2x, 2y) \).
- \( L' = D_{O,2}(-1,2)=(-2, 4) \)
- \( M' = D_{O,2}(-1,-1)=(-2, -2) \)
- \( N' = D_{O,2}(2,-1)=(4, -2) \)
Step3: Apply \( D_{O,0.75} \) (dilation by 0.75 from origin) to \( L', M', N' \)
\( D_{O,0.75}(x,y)=(0.75x, 0.75y) \).
- \( L'' = D_{O,0.75}(-2,4)=(-1.5, 3) \)? Wait, no—wait, original \( L \) is \((-1,2)\)? Wait, maybe I misread the graph. Let's recheck: \( L \) is at \((-1, 2)\)? Wait, the grid: \( x=-1 \), \( y=2 \) for \( L \); \( M(-1, -1) \); \( N(2, -1) \). Wait, no, maybe \( L(-1, 2) \), \( M(-1, 0) \)? Wait, the graph shows \( M \) at \( (-1, -1) \)? Wait, no, the shaded triangle: \( M \) is at \( (-1, -1) \)? Wait, maybe I made a mistake. Wait, the composition is \( D_{O,0.75} \circ D_{O,2} \), which is equivalent to \( D_{O, 2 \times 0.75} = D_{O, 1.5} \)? Wait, no: composition of dilations with the same center is multiplication of scales. So \( D_{O,0.75} \cdot D_{O,2} = D_{O, 2 \times 0.75} = D_{O, 1.5} \). Oh! That's a key point: composing dilations with center \( O \) is multiplying the scale factors. So scale factor is \( 2 \times 0.75 = 1.5 \). So we can directly apply \( D_{O,1.5} \) to original vertices.
Step4: Correct Coordinates and Dilation
Original vertices:
- \( L(-1, 2) \)
- \( M(-1, -1) \)
- \( N(2, -1) \)
Apply \( D_{O,1.5} \) (scale factor 1.5):
- \( L'' = 1.5 \times (-1, 2) = (-1.5, 3) \)? Wait, but the option says \( (-3, 1.5) \). Wait, maybe I misread the original coordinates. Wait, looking at the graph: \( L \) is at \( (-1, 2) \)? Wait, no, maybe \( L(-2, 2) \)? Wait, the grid: \( x=-1 \) is between -2 and 0. Wait, the x-axis: -4, -2, 0, 2, 4. So \( L \) is at \( (-1, 2) \)? No, maybe \( L(-2, 2) \)? Wait, the first vertical line left of y-axis is -1? No, the grid lines: each square is 1 unit. So \( L \) is at \( (-1, 2) \), \( M(-1, -1) \), \( N(2, -1) \). Wait, but the option for \( L'' \) is \( (-3, 1.5) \). Wait, maybe original \( L \) is \( (-2, 2) \). Let's re-express:
If \( L(-2, 2) \), \( M(-2, -1) \), \( N(2, -1) \):
Then \( D_{O,2} \) on \( L(-2,2) \) is \( (-4, 4) \), then \( D_{O,0.75} \) is \( (-3, 3) \)? No. Wait, maybe the original \( L \) is \( (-2, 2) \), \( M(-2, 0) \), \( N(2, 0) \)? Wait, the shaded triangle: \( M \) is at \( (-2, 0) \), \( N(2, 0) \), \( L(-2, 2) \). Ah! That makes more sense. Let's correct:
Original vertices (from graph, looking at the grid):
- \( L(-2, 2) \) (x=-2, y=2)
- \( M(-2, 0) \) (x=-2, y=0)
- \( N(2, 0) \) (x=2, y=0)
Now, apply \( D_{O,2} \) (scale 2):
- \( L' = (2 \times -2, 2 \times 2) = (-4, 4) \)
- \( M' = (2 \times -2, 2 \times 0) = (-4, 0) \)
- \( N' = (2 \times 2, 2 \times 0) = (4, 0) \)
Then apply \( D_{O,0.75} \) (scale 0.75):
- \( L'' = (0.75 \times -4, 0.75 \times 4) = (-3, 3) \)? No, the option is \( (-3, 1.5) \). Wait, maybe the original \( L \) is \( (-2, 2) \), but the first dilation is \( D_{O,2} \) then \( D_{O,0.75} \), but maybe I mixed up the order. Wait, composition is \( D_{O,0.75} \cdot D_{O,2} \), which means apply \( D_{O,2} \) first, then \( D_{O,0.75} \). Wait, no: \( (D_{O,0.75} \cdot D_{O,2})(x,y) = D_{O,0.75}(D_{O,2}(x,y)) \). So first \( D_{O,2} \), then \( D_{O,0.75} \).
Wait, let's take the correct original coordinates from the graph:
- \( L(-1, 2) \)? No, the g…
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- \( \angle M \cong \angle M'' \) (True)
- \( \triangle LMN \sim \triangle L''M''N'' \) (True)
- The coordinates of vertex \( L'' \) are (-3, 1.5) (True)
- The coordinates of vertex \( N'' \) are (3, -1.5) (True)
(Note: The option "The coordinates of vertex \( M'' \) are (1.5, -1.5)" is false, as shown.)
So the correct options are:
- \( \angle M \cong \angle M'' \)
- \( \triangle LMN \sim \triangle L''M''N'' \)
- The coordinates of vertex \( L'' \) are (-3, 1.5)
- The coordinates of vertex \( N'' \) are (3, -1.5)