Question

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?

Answer 1

 The options of the question are the points

$A(-5,-3)\\B(3,-3)\\C(-1,-6)\\D(4,-9)$

Step 1

Find the rule of the translation of the pre-image to the image

we know that

the point S has coordinates of $(3,-5)$

the point S' has coordinates of  $(-4,1)$

so

a) the rule of the translation of the pre-image to the image is

$(x.y)----> (x-7,y+6)$

that means

the translation is $7$ units to the left and $6$ units up

Step 2

Find the inverse rule of the translation of the image to the pre-image

a) the inverse rule of the translation of the image to the pre-image is

$(x',y')---> (x'+7,y'-6)$  

that means

the translation is $7$ units to the right and $6$ units down

Step 3

Find the coordinates of the vertices of the pre-image

Applying the inverse rule of the translation of the image to the pre-image

a) Point $R'(-8,1)$

$(x',y')---> (x'+7,y'-6)$  

$R'(-8,1)---> R(-8+7,1-6)$  

$R'(-8,1)---> R(-1,-5)$  

b) Point $T'(-4,-3)$

$(x',y')---> (x'+7,y'-6)$  

$T'(-4,-3)---> T(-4+7,-3-6)$  

$T'(-4,-3)---> T(3,-9)$  

c) Point  $U'(-8,-3)$

$(x',y')---> (x'+7,y'-6)$  

$U'(-8,-3)---> U(-8+7,-3-6)$  

$U'(-8,-3)---> U(-1,-9)$  

Step 4

Using a graphing tool

graph the points of the pre-image and the points A,B,C and D to determine the solution of the problem

we have

$R(-1,-5)\\S(3,-5)\\T(3,-9)\\U(-1,-9)$

$A(-5,-3)\\B(3,-3)\\C(-1,-6)\\D(4,-9)$

see the attached figure        

therefore        

the answer is

$C(-1,-6)$

Answer 2

C: -1, -6

EDGE 2021

Step-by-step explanation: