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?
options:
(-5, -3)
(3, -3)
(-1, -6)
(4, -9)

Explanation:

Step1: Find translation vector

Translation vector = \( S'(x) - S(x), S'(y) - S(y) = (-4 - 3, 1 - (-5)) = (-7, 6) \)

Step2: Calculate pre-image of S

Pre-image \( S(x,y) = S'(x) - \text{trans_x}, S'(y) - \text{trans_y} = (-4 - (-7), 1 - 6) = (3, -5) \)
(Note: The question asks for the pre-image point S which is already given as (3,-5), but checking options, the only matching coordinate is (3,-3)? Wait correction: Wait the problem says "point S has coordinates of (3,-5), which point lies on a side of the pre-image square RSTU?" Wait no, original problem: "Square RSTU is translated to form R'S'T'U', 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?"

Correct Step1: Find translation vector using R and R': Translation vector \( (h,k) = R'(x)-R(x) \)? No, pre-image to image: \( R'(x) = R(x)+h, R'(y)=R(y)+k \). Given image vertices R'(-8,1), S'(-4,1), T'(-4,-3), U'(-8,-3). So pre-image RSTU: \( R(x)=R'(x)-h, R(y)=R'(y)-k \). We know image S'(-4,1) corresponds to pre-image S(3,-5). So \( 3 = -4 - h \Rightarrow h = -7 \); \( -5 = 1 - k \Rightarrow k = 6 \). So translation vector is (-7,6) (pre-image + (-7,6) = image). So pre-image vertices: \( R = R' - (-7,6) = (-8 +7,1 -6)=(-1,-5) \); \( S=(3,-5) \); \( T = T' +7,-6=(-4+7,-3-6)=(3,-9) \); \( U=U'+7,-6=(-8+7,-3-6)=(-1,-9) \). So pre-image square RSTU has vertices (-1,-5), (3,-5), (3,-9), (-1,-9). Check which option is on its sides:

Option A (-5,-3): No. Option B (3,-3): No. Option C (-1,-6): Yes (lies on side RU or ST? RU is from (-1,-5) to (-1,-9), x=-1, y from -5 to -9. (-1,-6) is on RU). Option D (4,-9): No.

Answer:

C. (-1,-6)