Question

square rstu is translated to form rstu which has vertices r(-8, 1), s(-4, 1), t(-4, -3), and u(-8, -3). if point s has coordinates of (3, -5), which point lies on a side of the pre - image, square rstu?
(-5, -3)
(3, -3)
(-1, -6)
(4, -9)

Explanation:

Step1: Find the translation rule

We know that \(S(3,-5)\) is translated to \(S'(-4,1)\).
The translation rule for the \(x\) - coordinate: \(x'=x + a\), so \(-4=3 + a\), then \(a=-7\).
The translation rule for the \(y\) - coordinate: \(y'=y + b\), so \(1=-5 + b\), then \(b = 6\). The translation rule is \((x,y)\to(x - 7,y + 6)\).

Step2: Reverse the translation for each option

For option A: If \((x,y)=(-5,-3)\), then applying the reverse translation (i.e., \((x,y)\to(x + 7,y - 6)\)), we get \((-5 + 7,-3-6)=(2,-9)\)
For option B: If \((x,y)=(3,-3)\), then applying the reverse translation \((x,y)\to(x + 7,y - 6)\), we get \((3 + 7,-3-6)=(10,-9)\).
For option C: If \((x,y)=(-1,-6)\), then applying the reverse translation \((x,y)\to(x + 7,y - 6)\), we get \((-1+7,-6 - 6)=(6,-12)\)
For option D: If \((x,y)=(4,-9)\), then applying the reverse translation \((x,y)\to(x + 7,y - 6)\), we get \((4 + 7,-9-6)=(11,-15)\)

We can also use another approach.
The side of the square \(R'S'T'U'\) has length \(|-4-(-8)| = 4\) (using the \(x\) - coordinates of \(R'\) and \(S'\)) and \(|1-(-3)|=4\) (using the \(y\) - coordinates of \(S'\) and \(T'\)).
Since \(S(3,-5)\), and the side - length of the square is \(4\).
If we consider the vertical and horizontal distances.
The square \(RSTU\) (pre - image) has side - length \(4\).
Let's check the distance from the given points to the known point \(S(3,-5)\)
The distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For the point \((3,-3)\):
The distance from \((3,-5)\) to \((3,-3)\) is \(d=\sqrt{(3 - 3)^2+(-3+5)^2}=\sqrt{0 + 4}=2\).
Since the side - length of the square is \(4\), and if we consider the vertical sides of the square (parallel to the \(y\) - axis), a point at a \(y\) - coordinate \(2\) units above \((3,-5)\) (with the same \(x\) - coordinate) can lie on the side of the square.
The side of a square is a straight line segment. If we assume the square has vertical and horizontal sides (after translation, \(R'S'T'U'\) has vertical and horizontal sides as \(R'(-8,1)\), \(S'(-4,1)\) (horizontal side) and \(S'(-4,1)\), \(T'(-4,-3)\) (vertical side)), and using the translation rule \((x,y)\to(x - 7,y + 6)\) in reverse.
The original square \(RSTU\) has side - length \(4\). If we consider a vertical side passing through \(S(3,-5)\), the \(y\) - coordinates of the endpoints of the vertical side are \(y=-5\pm2\) (because the side - length is \(4\)). When \(y=-3\) (with \(x = 3\)), the point \((3,-3)\) lies on the side of the square \(RSTU\).

Answer:

B. \((3, - 3)\)