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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?
A.(–5, –3)
B.(3, –3)
C.(–1, –6)
D.(4, –9)???????????????

Square RSTU is translated to form RSTU which has vertices R8 1 S4 1 T4 3 and U8 3 If point S has coordinates of 3 5 which point lies on a side of the preimage s class=

Respuesta :

calculista

Step 1

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

we know that

the point S has coordinates of ([tex](3,-5)[/tex]

the point S' has coordinates of ([tex](-4,1)[/tex]

so

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

[tex](x,y)------> (x-7,y+6)[/tex]

Step 2

Find the 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

[tex](x',y')------> (x'+7,y'-6)[/tex]

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 [tex]R'(-8,1)[/tex]

[tex]R(-8+7,1-6)=R(-1,-5)[/tex]

b) Point [tex]T'(-4,-3)[/tex]

[tex]T(-4+7,-3-6)=T(3,-9)[/tex]

c) Point [tex]U'(-8,-3)[/tex]

[tex]U(-8+7,-3-6)=U(-1,-9)[/tex]

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

[tex]R(-1,-5) \\S(3,-5)\\T(3,-9)\\U(-1,-9)\\ \\A(-5,-3)\\B(3,-3)\\C(-1,-6)\\D(4,-9)[/tex]

see the attached figure        

therefore        

the answer is the option

[tex]C(-1,-6)[/tex]


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