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?
2 answers:
Answer:
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
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