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

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2   is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of   numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes.     our only condition is that it passes through the point (-2,5)