Question

PLEASE HELP ME!! A hexagon has vertices (3,1) and (4,1). The hexagon is dilated. The new hexagon has vertices (6,1) and (10,1). {In the same spots as the old hexagon}. What is the center of dilation? What is the dilation factor? I can try to add information.

Answer 1

Answer:

( 2,1) is the center of dilation and 4 is the scale factor

Step-by-step explanation:

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

3,1 becomes 6,1

6,1 = k( 3-a) +a, k( 1-b)+b

6 = 3k -ka+a

1 = k -kb +b

4,1 becomes 10,1

10,1 = k( 4-a) +a, k( 1-b)+b

10 = 4k -ka+a

1 = k -kb +b

Using these two equations

6 = 3k -ka+a

10 = 4k -ka+a

Subtracting the top from the bottom

10 = 4k -ka+a

-6 = -3k +ka-a

------------------------

4 = k

Now solving for a

6 = 3k -ka+a

6 = 3(4) -4a+a

6 =12 -3a

Subtract 12

6-12 = -3a

-6 = -3a

Divide by -3

-6/-3 = -3a/-3

2 =a

Now finding b

1 = k -kb +b

1 = 4 - 4b+b

1 =4 -3b

Subtract 4

-3 = -3b

Divide by -3

1 = b

Answer 2

Answer:

Dilation factor: 4.  

Center of dilation: (2, 1).  

Step-by-step explanation:

The distance between the old vertices was 4 - 3 = 1. The distance between the new vertices is 10 - 6 = 4. 4 / 1 = 4. That means that the dilation factor is 4.  

Now that we have a dilation factor, we can use the formulas x1 = d(x-a) +a and y1 = d( y-b)+b to solve for the center of dilation.  

In this case, d = 4, x1 = 10, x = 4, y1 = 1, and y = 1.  

10 = 4(4 – a) + a

10 = 16 – 4a + a

10 = 16 – 3a

-3a + 16 = 10

-3a = -6

a = 2

1 = 4(1 – b) + b

1 = 4 – 4b + b

1 = 4 – 3b

-3b + 4 = 1

-3b = -3

b = 1

And so, your center of dilation will be (2, 1).  

Hope this helps!