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Sagot :
Let's start by understanding the problem and breaking it down step by step:
We have two points, [tex]\( A = (-4, 3) \)[/tex] and [tex]\( B = (4, 4) \)[/tex]. The given dilation is [tex]\( D_{O, 0.5}(x, y) = \left(\frac{1}{2} x, \frac{1}{2} y\right) \)[/tex].
We are to determine the properties of the image after applying this dilation. Specifically, we need to verify the following statements:
1. [tex]$\overline{AB}$[/tex] is parallel to [tex]$\overline{A'B'}$[/tex].
2. The distance from [tex]$A'$[/tex] to the origin is half the distance from [tex]$A$[/tex] to the origin.
3. The vertices of the image are farther from the origin than those of the pre-image.
4. [tex]$\overline{A'B'}$[/tex] is greater than [tex]$\overline{AB}$[/tex].
### Step 1: Apply the dilation to points [tex]\( A \)[/tex] and [tex]\( B \)[/tex]
For point [tex]\( A \)[/tex]:
[tex]\[ A' = \left( \frac{1}{2} \cdot -4, \frac{1}{2} \cdot 3 \right) = (-2, 1.5) \][/tex]
For point [tex]\( B \)[/tex]:
[tex]\[ B' = \left( \frac{1}{2} \cdot 4, \frac{1}{2} \cdot 4 \right) = (2, 2) \][/tex]
### Step 2: Verify the conditions
Condition 1: [tex]$\overline{AB}$[/tex] is parallel to [tex]$\overline{A'B'}$[/tex]
Dilation preserves the parallelism of lines because it scales all coordinates by the same factor. Thus,
[tex]\[ \overline{AB} \parallel \overline{A'B'} \][/tex]
This statement is true.
Condition 2: The distance from [tex]\( A' \)[/tex] to the origin is half the distance from [tex]\( A \)[/tex] to the origin
Calculate the distance from [tex]\( A \)[/tex] to the origin:
[tex]\[ \text{Distance from } A \text{ to the origin} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
Calculate the distance from [tex]\( A' \)[/tex] to the origin:
[tex]\[ \text{Distance from } A' \text{ to the origin} = \sqrt{(-2)^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \][/tex]
Indeed,
[tex]\[ 2.5 = \frac{1}{2} \times 5 \][/tex]
This statement is true.
Condition 3: The vertices of the image are farther from the origin than those of the pre-image
From the distances calculated:
- Distance from [tex]\( A \)[/tex] to the origin: 5
- Distance from [tex]\( A' \)[/tex] to the origin: 2.5 (which is less than 5)
This statement is false.
Condition 4: [tex]$\overline{A'B'}$[/tex] is greater than [tex]$\overline{AB}$[/tex]
Calculate the length of [tex]\( \overline{AB} \)[/tex]:
[tex]\[ AB = \sqrt{(4 - (-4))^2 + (4 - 3)^2} = \sqrt{8^2 + 1^2} = \sqrt{64 + 1} = \sqrt{65} \][/tex]
Calculate the length of [tex]\( \overline{A'B'} \)[/tex]:
[tex]\[ A'B' = \sqrt{(2 - (-2))^2 + (2 - 1.5)^2} = \sqrt{4^2 + 0.5^2} = \sqrt{16 + 0.25} = \sqrt{16.25} \][/tex]
Notice that [tex]\( \sqrt{16.25} \)[/tex] is smaller than [tex]\( \sqrt{65} \)[/tex].
Thus,
[tex]\[ A'B' < AB \][/tex]
Hence, this statement is false.
### Conclusion:
- [tex]$\overline{AB}$[/tex] is parallel to [tex]$\overline{A'B'}$[/tex]: True
- The distance from [tex]$A'$[/tex] to the origin is half the distance from [tex]$A$[/tex] to the origin: True
- The vertices of the image are farther from the origin than those of the pre-image: False
- [tex]$\overline{A'B'}$[/tex] is greater than [tex]$\overline{AB}$[/tex]: False
We have two points, [tex]\( A = (-4, 3) \)[/tex] and [tex]\( B = (4, 4) \)[/tex]. The given dilation is [tex]\( D_{O, 0.5}(x, y) = \left(\frac{1}{2} x, \frac{1}{2} y\right) \)[/tex].
We are to determine the properties of the image after applying this dilation. Specifically, we need to verify the following statements:
1. [tex]$\overline{AB}$[/tex] is parallel to [tex]$\overline{A'B'}$[/tex].
2. The distance from [tex]$A'$[/tex] to the origin is half the distance from [tex]$A$[/tex] to the origin.
3. The vertices of the image are farther from the origin than those of the pre-image.
4. [tex]$\overline{A'B'}$[/tex] is greater than [tex]$\overline{AB}$[/tex].
### Step 1: Apply the dilation to points [tex]\( A \)[/tex] and [tex]\( B \)[/tex]
For point [tex]\( A \)[/tex]:
[tex]\[ A' = \left( \frac{1}{2} \cdot -4, \frac{1}{2} \cdot 3 \right) = (-2, 1.5) \][/tex]
For point [tex]\( B \)[/tex]:
[tex]\[ B' = \left( \frac{1}{2} \cdot 4, \frac{1}{2} \cdot 4 \right) = (2, 2) \][/tex]
### Step 2: Verify the conditions
Condition 1: [tex]$\overline{AB}$[/tex] is parallel to [tex]$\overline{A'B'}$[/tex]
Dilation preserves the parallelism of lines because it scales all coordinates by the same factor. Thus,
[tex]\[ \overline{AB} \parallel \overline{A'B'} \][/tex]
This statement is true.
Condition 2: The distance from [tex]\( A' \)[/tex] to the origin is half the distance from [tex]\( A \)[/tex] to the origin
Calculate the distance from [tex]\( A \)[/tex] to the origin:
[tex]\[ \text{Distance from } A \text{ to the origin} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
Calculate the distance from [tex]\( A' \)[/tex] to the origin:
[tex]\[ \text{Distance from } A' \text{ to the origin} = \sqrt{(-2)^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \][/tex]
Indeed,
[tex]\[ 2.5 = \frac{1}{2} \times 5 \][/tex]
This statement is true.
Condition 3: The vertices of the image are farther from the origin than those of the pre-image
From the distances calculated:
- Distance from [tex]\( A \)[/tex] to the origin: 5
- Distance from [tex]\( A' \)[/tex] to the origin: 2.5 (which is less than 5)
This statement is false.
Condition 4: [tex]$\overline{A'B'}$[/tex] is greater than [tex]$\overline{AB}$[/tex]
Calculate the length of [tex]\( \overline{AB} \)[/tex]:
[tex]\[ AB = \sqrt{(4 - (-4))^2 + (4 - 3)^2} = \sqrt{8^2 + 1^2} = \sqrt{64 + 1} = \sqrt{65} \][/tex]
Calculate the length of [tex]\( \overline{A'B'} \)[/tex]:
[tex]\[ A'B' = \sqrt{(2 - (-2))^2 + (2 - 1.5)^2} = \sqrt{4^2 + 0.5^2} = \sqrt{16 + 0.25} = \sqrt{16.25} \][/tex]
Notice that [tex]\( \sqrt{16.25} \)[/tex] is smaller than [tex]\( \sqrt{65} \)[/tex].
Thus,
[tex]\[ A'B' < AB \][/tex]
Hence, this statement is false.
### Conclusion:
- [tex]$\overline{AB}$[/tex] is parallel to [tex]$\overline{A'B'}$[/tex]: True
- The distance from [tex]$A'$[/tex] to the origin is half the distance from [tex]$A$[/tex] to the origin: True
- The vertices of the image are farther from the origin than those of the pre-image: False
- [tex]$\overline{A'B'}$[/tex] is greater than [tex]$\overline{AB}$[/tex]: False
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