Answer :
To determine which point lies on a side of the pre-image square RSTU, we need to follow these steps:
1. Identify the translation vector:
We know that point [tex]\( S \)[/tex] in the pre-image has coordinates [tex]\( (3, -5) \)[/tex]. The corresponding point in the translated image, [tex]\( S' \)[/tex], has coordinates [tex]\( (-4, 1) \)[/tex]. To find the translation vector, subtract the coordinates of [tex]\( S \)[/tex] from [tex]\( S' \)[/tex]:
[tex]\[ \text{Translation vector} = (S' - S) = (-4 - 3, 1 + 5) = (-7, 6) \][/tex]
2. Find the position of [tex]\( R \)[/tex] in the original square:
Given that [tex]\( R' \)[/tex] (translated position of [tex]\( R \)[/tex]) is [tex]\( (-8, 1) \)[/tex], we can find [tex]\( R \)[/tex] by applying the inverse of the translation vector:
[tex]\[ R = R' - \text{Translation vector} = (-8, 1) - (-7, 6) = (-8 + 7, 1 - 6) = (-1, -5) \][/tex]
3. Vertices of pre-image square RSTU:
Since [tex]\( S \)[/tex] in the pre-image is [tex]\( (3, -5) \)[/tex] and [tex]\( R \)[/tex] is [tex]\( (-1, -5) \)[/tex], we have two points. To find the other vertices [tex]\( T \)[/tex] and [tex]\( U \)[/tex]:
- [tex]\( T \)[/tex] must be directly down from [tex]\( S \)[/tex] by the same distance as [tex]\( S' \)[/tex] to [tex]\( T' \)[/tex]. Since [tex]\( S' \)[/tex] to [tex]\( T' \)[/tex] is 4 units down ([tex]\( 1 \)[/tex] to [tex]\( -3 \)[/tex]):
[tex]\[ T = (3, -5 - 4) = (3, -9) \][/tex]
- [tex]\( U \)[/tex] must be directly down from [tex]\( R \)[/tex] by the same distance:
[tex]\[ U = (-1, -5 - 4) = (-1, -9) \][/tex]
4. Check which point lies on the side of RSTU:
We check if any of the given points lies on the sides of the pre-image square RSTU.
- Point [tex]\( (-5, -3) \)[/tex]:
Neither [tex]\( x = -5 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -3 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
- Point [tex]\( (3, -3) \)[/tex]:
Neither [tex]\( x = 3 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -3 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
- Point [tex]\( (-1, -6) \)[/tex]:
Neither [tex]\( x = -1 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -6 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
- Point [tex]\( (4, -9) \)[/tex]:
Neither [tex]\( x = 4 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -9 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
Thus, after checking all given options, we find that none of the given points [tex]\( (-5,-3), (3,-3), (-1,-6), (4,-9) \)[/tex] lies on a side of the pre-image square.
Hence, the final result is:
[tex]\[ \text{None of the given points lies on a side of the pre-image square RSTU.} \][/tex]
1. Identify the translation vector:
We know that point [tex]\( S \)[/tex] in the pre-image has coordinates [tex]\( (3, -5) \)[/tex]. The corresponding point in the translated image, [tex]\( S' \)[/tex], has coordinates [tex]\( (-4, 1) \)[/tex]. To find the translation vector, subtract the coordinates of [tex]\( S \)[/tex] from [tex]\( S' \)[/tex]:
[tex]\[ \text{Translation vector} = (S' - S) = (-4 - 3, 1 + 5) = (-7, 6) \][/tex]
2. Find the position of [tex]\( R \)[/tex] in the original square:
Given that [tex]\( R' \)[/tex] (translated position of [tex]\( R \)[/tex]) is [tex]\( (-8, 1) \)[/tex], we can find [tex]\( R \)[/tex] by applying the inverse of the translation vector:
[tex]\[ R = R' - \text{Translation vector} = (-8, 1) - (-7, 6) = (-8 + 7, 1 - 6) = (-1, -5) \][/tex]
3. Vertices of pre-image square RSTU:
Since [tex]\( S \)[/tex] in the pre-image is [tex]\( (3, -5) \)[/tex] and [tex]\( R \)[/tex] is [tex]\( (-1, -5) \)[/tex], we have two points. To find the other vertices [tex]\( T \)[/tex] and [tex]\( U \)[/tex]:
- [tex]\( T \)[/tex] must be directly down from [tex]\( S \)[/tex] by the same distance as [tex]\( S' \)[/tex] to [tex]\( T' \)[/tex]. Since [tex]\( S' \)[/tex] to [tex]\( T' \)[/tex] is 4 units down ([tex]\( 1 \)[/tex] to [tex]\( -3 \)[/tex]):
[tex]\[ T = (3, -5 - 4) = (3, -9) \][/tex]
- [tex]\( U \)[/tex] must be directly down from [tex]\( R \)[/tex] by the same distance:
[tex]\[ U = (-1, -5 - 4) = (-1, -9) \][/tex]
4. Check which point lies on the side of RSTU:
We check if any of the given points lies on the sides of the pre-image square RSTU.
- Point [tex]\( (-5, -3) \)[/tex]:
Neither [tex]\( x = -5 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -3 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
- Point [tex]\( (3, -3) \)[/tex]:
Neither [tex]\( x = 3 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -3 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
- Point [tex]\( (-1, -6) \)[/tex]:
Neither [tex]\( x = -1 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -6 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
- Point [tex]\( (4, -9) \)[/tex]:
Neither [tex]\( x = 4 \)[/tex] lies between the [tex]\( x \)[/tex]-coordinates of [tex]\( R \)[/tex] and [tex]\( S \)[/tex] nor does [tex]\( y = -9 \)[/tex] match any [tex]\( y \)[/tex]-coordinate of the pre-image vertices.
Thus, after checking all given options, we find that none of the given points [tex]\( (-5,-3), (3,-3), (-1,-6), (4,-9) \)[/tex] lies on a side of the pre-image square.
Hence, the final result is:
[tex]\[ \text{None of the given points lies on a side of the pre-image square RSTU.} \][/tex]
