PLEASE HELP ME A quadrilateral with vertices (0,0), (4,0), (3,2), (1,1) is mapped to a quadrilateral with vertices (6,3).(-2,3),(0,-1),(4,1). What is the center of the dilation and what is the scale factor.

( 2,1) is the center of dilation and -2 is the scale factor

Step-by-step explanation:

We can use the formula

A' = k( x-a) +a, k( y-b)+b  where ( a,b) is the center of dilation and k is the scale factor

(0,0) becomes (6,3)

( 6,3) =  k( 0-a) +a, k( 0-b)+b

6 = -ka+a

3 = -kb+b

We also have

(4,0) becomes (-2,3)

( -2,3) =  k( 4-a) +a, k( 0-b)+b

-2 =4k -ka+a

3 = -kb+b

Using these two equations

6 = -ka+a

-2 =4k -ka+a

Subtracting the top from the bottom

-2 =4k -ka+a

-6 = ka -a

-8 = 4k

Divide by 4

-8/4 = 4k/4

-2 = k

Now solving for a

6 = -ka +a

6 = - (-2)a +a

6 = 2a+a

6 = 3a

Divide by 3

6/3 =3a/3

2=a

Now finding b

3 = -kb+b

3 = -(-2)b+b

3 = 2b+b

3 = 3b

b=1

hope this helps uh.....

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length