Answer:
Please, see the answer in the attached files.
Step-by-step explanation:
1.) Suppose:
A=(-3,3)=(xa,ya)→xa=-3, ya=3
B=(0,6)=(xb,yb)→xb=0, yb=6
C=(2,4)=(xc,yc)→xc=2, yc=4
Centered at P=(2,2)=(xp,yp)→xp=2, yp=2
Scale factor: f=2
xa'=xp + f (xa-xp)=2+2(-3-2)=2+2(-5)=2-10→xa'=-8
ya'=yp + f (ya-yp)=2+2(3-2)=2+2(1)=2+2→ya'=4
A'=(xa',ya')→A'=(-8,4)
xb'=xp + f (xb-xp)=2+2(0-2)=2+2(-2)=2-4→xb'=-2
yb'=yp + f (yb-yp)=2+2(6-2)=2+2(4)=2+8→yb'=10
B'=(xb',ya')→B'=(-2,10)
xc'=xp + f (xc-xp)=2+2(2-2)=2+2(0)=2+0→xc'=2
yc'=yp + f (yc-yp)=2+2(4-2)=2+2(2)=2+4→yc'=6
C'=(xc',yc')→C'=(2,6)
Please, see the graph in the attached files.
2.) Suppose:
A=(-5,4)=(xa,ya)→xa=-5, ya=4
B=(3,8)=(xb,yb)→xb=3, yb=8
C=(5,4)=(xc,yc)→xc=5, yc=4
Centered at P=(-3,0)=(xp,yp)→xp=-3, yp=0
Scale factor: f=1/2
xa'=xp + f (xa-xp)=-3+(1/2)(-5-(-3))=-3+(1/2)(-5+3)=-3+(1/2)(-2)=-3-1→xa'=-4
ya'=yp + f (ya-yp)=0+(1/2)(4-0)=0+(1/2)(4)=0+2→ya'=2
A'=(xa',ya')→A'=(-4,2)
xb'=xp + f (xb-xp)=-3+(1/2)(3-(-3))=-3+(1/2)(3+3)=-3+(1/2)(6)=-3+3→xb'=0
yb'=yp + f (yb-yp)=0+(1/2)(8-0)=0+(1/2)(8)=0+4→yb'=4
B'=(xb',yb')→B'=(0,4)
xc'=xp + f (xc-xp)=-3+(1/2)(5-(-3))=-3+(1/2)(5+3)=-3+(1/2)(8)=-3+4→xc'=1
yc'=yp + f (yc-yp)=0+(1/2)(4-0)=0+(1/2)(4)=0+2→yc'=2
C'=(xc',yc')→C'=(1,2)
Please, see the graph in the attached files.
