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

C: -1, -6

EDGE 2021

Explanation:

Answer 2

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)`