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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?

User James Ehly
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6.3k points

2 Answers

2 votes

C: -1, -6

EDGE 2021

Explanation:

User James Allardice
by
7.0k points
2 votes

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)

Square RSTU is translated to form R'S'T'U', which has vertices R'(–8, 1), S'(–4, 1), T-example-1
User Bye
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7.5k points