Determine the scale factor used in the dilation of line segment AB, (4,-12) (8,-16) centered of origin.

Answer 1

To find the scale factor used in the dilation of line segment AB, we can compare the length of AB before and after the dilation.

The length of AB before the dilation can be found using the distance formula:

AB = √((x2 - x1)^2 + (y2 - y1)^2)

In this case, the coordinates of A are (4, -12) and the coordinates of B are (8, -16).

So, AB = √((8 - 4)^2 + (-16 - (-12))^2)
= √(4^2 + (-4)^2)
= √(16 + 16)
= √32

Now, let's consider the coordinates of the dilated line segment. Since the center of dilation is the origin, the scale factor can be found by comparing the coordinates of A and B to their corresponding dilated coordinates.

The dilated coordinates can be found by multiplying the original coordinates by the scale factor. Let's call the scale factor "k".

For A: (x, y) -> (kx, ky)
So, (4, -12) -> (k * 4, k * -12) = (4k, -12k)

For B: (x, y) -> (kx, ky)
So, (8, -16) -> (k * 8, k * -16) = (8k, -16k)

Now, let's find the length of the dilated line segment AB:

AB' = √((8k - 4k)^2 + (-16k - (-12k))^2)
= √((4k)^2 + (-4k)^2)
= √(16k^2 + 16k^2)
= √(32k^2)
= √(32) * k
= 4√2 * k

Since AB' is the dilated length and AB is the original length, we can equate them:

AB' = AB

4√2 * k = √32

Dividing both sides of the equation by √32, we get:

4√2 * k / √32 = 1

Simplifying this expression, we get:

4 * (√2 / √32) * k = 1

Simplifying further, we get:

4 * (√(2 / 32)) * k = 1
4 * (√(1 / 16)) * k = 1
4 * (1 / 4) * k = 1
k = 1/4

Therefore, the scale factor used in the dilation of line segment AB is 1/4.