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?

The point T'=(-4,-3) lies on a side of the pre-image, square RSTU

Step-by-step explanation:

S=(3,-5)=(xs,ys)→xs=3, ys=-5

S'=(-4,1)=(xs',ys')→xs'=-4, ys'=1

xs'=xs+a; ys'=ys+b

Replacing the known values in the formulas above:

xs'=xs+a→-4=3+a

Solving for a: Subtracting 3 both sides of the equation:

-4-3=3+a-3

Subtracting:

-7=a→a=-7

ys'=ys+b→1=-5+b

Solving for b: Adding 5 both sides of the equation:

1+5=-5+b+5

Adding:

6=b→b=6

R'=(-8,1)=(xr+a,yr+b)

-8=xr+a; 1=yr+b

Replacing the known values in the formulas above:

-8=xr+a→-8=xr+(-7)→-8=xr-7

Solving for xr: Adding 7 both sides of the equation:

-8+7=xr-7+7

Adding:

-1=xr→xr=-1

1=yr+b→1=yr+6

Solving for yr: Subtracting 6 both sides of the equation:

1-6=yr+6-6

Subtracting:

-5=yr→yr=-5

R=(xr,yr)→R=(-1,-5)

T'=(-4,-3)=(xt+a,yt+b)

-4=xt+a; -3=yt+b

Replacing the known values in the formulas above:

-4=xt+a→-4=xt+(-7)→-4=xt-7

Solving for xt: Adding 7 both sides of the equation:

-4+7=xt-7+7

Adding:

3=xt→xt=3

-3=yt+b→-3=yt+6

Solving for yt: Subtracting 6 both sides of the equation:

-3-6=yt+6-6

Subtracting:

-9=yt→yt=-9

T=(xt,yt)→T=(3,-9)

U'=(-8,-3)=(xu+a,yu+b)

-8=xu+a; -3=yu+b

Replacing the known values in the formulas above:

-8=xu+a→-8=xu+(-7)→-8=xu-7

Solving for xu: Adding 7 both sides of the equation:

-8+7=xu-7+7

Adding:

-1=xu→xu=-1

-3=yu+b→-3=yu+6

Solving for yu: Subtracting 6 both sides of the equation:

-3-6=yu+6-6

Subtracting:

-9=yu→yu=-9

U=(xu,yu)→U=(-1,-9)

Graphing the points (See the attached graph)

The point (-4,-3)=T' lies on a side of the pre-image, square RSTU

i got 10.28 but that rounds up to 10.3