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

The original point \(S'\) has coordinates \((- 4,1)\) and the translated point \(S\) has coordinates \((3,-5)\).
The translation rule for the \(x\) - coordinate: \(x\) - value changes from \(-4\) to \(3\), so \(\Delta x=3-(-4)=7\).
The translation rule for the \(y\) - coordinate: \(y\) - value changes from \(1\) to \(-5\), so \(\Delta y=-5 - 1=-6\).
The translation rule is \((x,y)\to(x + 7,y-6)\).

Step2: Reverse the translation rule

To get the pre - image from the image, we use the reverse rule \((x,y)\to(x - 7,y + 6)\).

Step3: Check each option

• For the point \((-5,-3)\):

Using the reverse rule: \((-5-7,-3 + 6)=(-12,3)\) (not on the square).

• For the point \((3,-3)\):

Using the reverse rule: \((3-7,-3 + 6)=(-4,3)\) (not on the square).

• For the point \((-1,-6)\):

Using the reverse rule: \((-1-7,-6 + 6)=(-8,0)\) (not on the square).

• For the point \((4,-9)\):

Using the reverse rule: \((4-7,-9 + 6)=(-3,-3)\).
In square \(R'ST'U'\), \(S'(-4,1)\), \(T'(-4,-3)\), \(U'(-8,-3)\), \(R'(-8,1)\). The side \(T'U'\) has \(y=-3\) and \(x\) from \(-8\) to \(-4\). When we reverse - translate \((4,-9)\) to \((-3,-3)\), and in the pre - image (before translation), the side corresponding to \(T'U'\) (after reverse - translation) will have points with \(y=-3\) (since \(y\) in \(R'ST'U'\) for the side \(T'U'\) is \(y =-3\) and reverse - translation for \(y\) is \(y+6\), original \(y=-3\) in pre - image after reverse - translation from image \(y=-9\) (because \(-9+6=-3\)). The \(x\) - value \(-3\) (after reverse - translation of \(x = 4\): \(4-7=-3\)) lies between \(-8\) and \(-4\) (for the side parallel to the \(x\) - axis in the pre - image corresponding to \(T'U'\) in the image).

Answer:

\((4,-9)\)