Question

XY is dilated with a scale factor of 2 and a center of cilation at the origin. Find points X' and Y' | to prove that a dilation of 2 has a length that is twice the length of overline XY

Answer 1

To perform a dilation with a scale factor of 2 and a center of dilation at the origin, we take any point \( (x, y) \) and transform it into a new point \( (x', y') \) using the following formulas:

\[
x' = k \cdot x
\]
\[
y' = k \cdot y
\]

where \( k \) is the scale factor. In this case, \( k = 2 \).

Now, let's begin by determining the original coordinates of points \( X \) and \( Y \). Assume the coordinates are as follows:

- \( X(x_1, y_1) \)
- \( Y(x_2, y_2) \)

The new points \( X' \) and \( Y' \) after dilation by a factor of 2 will be:

\[
X' = (2x_1, 2y_1)
\]
\[
Y' = (2x_2, 2y_2)
\]

Next, we need to calculate the length of the original segment \( \overline{XY} \) and the length of the dilated segment \( \overline{X'Y'} \).

The length of \( \overline{XY} \) can be calculated using the distance formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

The length of the dilated segment \( \overline{X'Y'} \) will be:

\[
d' = \sqrt{(2x_2 - 2x_1)^2 + (2y_2 - 2y_1)^2}
\]

We can simplify \( d' \):

\[
d' = \sqrt{2^2 \cdot (x_2 - x_1)^2 + 2^2 \cdot (y_2 - y_1)^2}
\]
\[
d' = \sqrt{4\left((x_2 - x_1)^2 + (y_2 - y_1)^2\right)}
\]
\[
d' = 2 \cdot \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
\[
d' = 2d
\]

This shows that the length of the segment \( \overline{X'Y'} \) after dilation is indeed twice the length of the segment \( \overline{XY} \).

To illustrate with an example, let's use specific points:

Let:
- \( X(1, 2) \)
- \( Y(3, 4) \)

Calculate the length of \( \overline{XY} \):

\[
d = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

Now, find the dilated points \( X' \) and \( Y' \):

\[
X' = (2 \cdot 1, 2 \cdot 2) = (2, 4)
\]
\[
Y' = (2 \cdot 3, 2 \cdot 4) = (6, 8)
\]

Calculate the length of \( \overline{X'Y'} \):

\[
d' = \sqrt{(6 - 2)^2 + (8 - 4)^2} = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

Now we verify the relationship:

\[
d' = 2d \Longrightarrow 4\sqrt{2} = 2(2\sqrt{2})
\]

Hence, it is confirmed that the dilation effectively doubles the length of the segment.