Question

On a coordinate plane, a square and a point are shown. The square has points R prime (negative 8, 1), S prime (negative 4, 1), T prime (negative 4, negative 3), and U prime (negative 8, negative 3). Point S is at (3, negative 5). 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? (–5, –3) (3, –3) (–1, –6) (4, –9)

Answer 1

Answer:

(–1, –6)

Step-by-step explanation:

Given that S(3, –5) was translated to S'(–4, 1), the transformation rule is (x-7, y+6). Then, the coordinates of square RSTU are:

R'(–8, 1) -> (-8+7, 1-6) -> R(-1, -5)

S'(–4, 1) -> (-4+7, 1-6) -> S(3, -5)

T'(–4, –3) -> (-4+7, -3-6) -> T(3, -9)

U'(–8, –3) -> (-8+7, -3-6) -> U(-1, -9)

The point (–1, –6) lies on the segment RU

Answer 2

The point that lies on a side of the pre-image is mathematically given as

x=(-1,-6)

What is a Pre-Image?

A pre-image is simply a the group of chosen elements from a mapped domain to give a given subset of the codomain.

What points can be found on the pre-image, square RSTU?

Question Parameter(s):

which has vertices

R'(–8, 1),

S'(–4, 1),

T'(–4, –3),

U'(–8,

Generally, the equation for the transformation  is mathematically given as

(x-7, y+6).

Therefore

R'(–8, 1) --> (-8+7, 1-6)

R'(–8, 1) --> R(-1, -5)

S'(–4, 1) -=> (-4+7, 1-6)

S'(–4, 1)-> S(3, -5)

T'(–4, –3) --> (-4+7, -3-6)

T'(–4, –3 -> T(3, -9)

U'(–8, –3) --> (-8+7, -3-6)

U'(–8, –3)-> U(-1, -9)

In conclusion, point S with coordinates of (3, –5) lies within RU

x=(-1,-6)

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