Explore a vast range of topics and get informed answers at IDNLearn.com. Get the information you need from our experts, who provide reliable and detailed answers to all your questions.
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
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Thank you for choosing IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more solutions.