To determine which of the given points lies on a side of the pre-image square [tex]\( RSTU \)[/tex], we will map the vertices of the given translated image [tex]\( R'S'T'U' \)[/tex] back to the pre-image using the given coordinates.
First, let’s find the translation vector by comparing the coordinates of point [tex]\( S \)[/tex] in the pre-image and its corresponding point [tex]\( S' \)[/tex] in the image:
- Coordinates of [tex]\( S' \)[/tex] are [tex]\( (-4,1) \)[/tex].
- Coordinates of [tex]\( S \)[/tex] are [tex]\( (3,-5) \)[/tex].
The translation vector [tex]\( \mathbf{t} \)[/tex] can be computed as follows:
[tex]\[ \mathbf{t} = (S'[0] - S[0], S'[1] - S[1]) \][/tex]
[tex]\[ \mathbf{t} = (-4 - 3, 1 - (-5)) \][/tex]
[tex]\[ \mathbf{t} = (-7, 6) \][/tex]
Now we will use this translation vector [tex]\( \mathbf{t} = (-7, 6) \)[/tex] to find the coordinates of the remaining points [tex]\( R, T, \)[/tex] and [tex]\( U \)[/tex] in the pre-image [tex]\( RSTU \)[/tex].
Finding the coordinates of [tex]\( R \)[/tex]:
- Coordinates of [tex]\( R' \)[/tex] are [tex]\( (-8,1) \)[/tex].
[tex]\[ R = (R'[0] - \mathbf{t}[0], R'[1] - \mathbf{t}[1]) \][/tex]
[tex]\[ R = (-8 - (-7), 1 - 6) \][/tex]
[tex]\[ R = (-8 + 7, 1 - 6) \][/tex]
[tex]\[ R = (-1, -5) \][/tex]
Finding the coordinates of [tex]\( T \)[/tex]:
- Coordinates of [tex]\( T' \)[/tex] are [tex]\( (-4,-3) \)[/tex].
[tex]\[ T = (T'[0] - \mathbf{t}[0], T'[1] - \mathbf{t}[1]) \][/tex]
[tex]\[ T = (-4 - (-7), -3 - 6) \][/tex]
[tex]\[ T = (-4 + 7, -3 - 6) \][/tex]
[tex]\[ T = (3, -9) \][/tex]
Finding the coordinates of [tex]\( U \)[/tex]:
- Coordinates of [tex]\( U' \)[/tex] are [tex]\( (-8,-3) \)[/tex].
[tex]\[ U = (U'[0] - \mathbf{t}[0], U'[1] - \mathbf{t}[1]) \][/tex]
[tex]\[ U = (-8 - (-7), -3 - 6) \][/tex]
[tex]\[ U = (-8 + 7, -3 - 6) \][/tex]
[tex]\[ U = (-1, -9) \][/tex]
Therefore, the coordinates of the vertices of the pre-image [tex]\( RSTU \)[/tex] are:
- [tex]\( R (-1, -5) \)[/tex]
- [tex]\( S (3, -5) \)[/tex]
- [tex]\( T (3, -9) \)[/tex]
- [tex]\( U (-1, -9) \)[/tex]
Next, we need to determine which of the given points [tex]\((-5,-3), (3,-3), (-1,-6), (4,-9)\)[/tex] lies on one of the sides of the square [tex]\( RSTU \)[/tex].
- Side [tex]\( RS \)[/tex] has endpoints [tex]\( R (-1, -5) \)[/tex] and [tex]\( S (3, -5) \)[/tex]:
No points given share the same y-coordinate [tex]\(-5\)[/tex].
- Side [tex]\( ST \)[/tex] has endpoints [tex]\( S (3, -5) \)[/tex] and [tex]\( T (3, -9) \)[/tex]:
No points given share the same x-coordinate [tex]\(3\)[/tex].
- Side [tex]\( TU \)[/tex] has endpoints [tex]\( T (3, -9) \)[/tex] and [tex]\( U (-1, -9) \)[/tex]:
No points given share the same y-coordinate [tex]\(-9\)[/tex].
- Side [tex]\( UR \)[/tex] has endpoints [tex]\( U (-1, -9) \)[/tex] and [tex]\( R (-1, -5) \)[/tex]:
One point ([tex]\(-1, -6\)[/tex]) shares the same x-coordinate [tex]\(-1\)[/tex] and lies between [tex]\( -9 \)[/tex] and [tex]\( -5 \)[/tex].
Therefore, the point [tex]\( (-1, -6) \)[/tex] lies on a side of the pre-image, square [tex]\( RSTU \)[/tex].
