To tackle this problem, let's first understand dilation and how it affects points and lines on a plane. Given that the center of dilation [tex]\(W\)[/tex] is at [tex]\((3, 2)\)[/tex] and point [tex]\(X\)[/tex] is at [tex]\((7, 5)\)[/tex], let's dilate point [tex]\(X\)[/tex] with a scale factor of 3.
### Calculating the coordinates of [tex]\(X'\)[/tex]
1. Find the relative coordinates of [tex]\(X\)[/tex] with respect to [tex]\(W\)[/tex]:
[tex]\[
\Delta x = 7 - 3 = 4
\][/tex]
[tex]\[
\Delta y = 5 - 2 = 3
\][/tex]
2. Apply the scale factor to these relative coordinates:
[tex]\[
\Delta x' = 4 \times 3 = 12
\][/tex]
[tex]\[
\Delta y' = 3 \times 3 = 9
\][/tex]
3. Determine the coordinates of [tex]\(X'\)[/tex] by adding these scaled relative coordinates back to [tex]\(W\)[/tex]:
[tex]\[
X' = (3 + 12, 2 + 9) = (15, 11)
\][/tex]
### Calculating the slopes
4. Slope of the line segment [tex]\(\overline{WX}\)[/tex]:
[tex]\[
\text{slope}_{WX} = \frac{5 - 2}{7 - 3} = \frac{3}{4}
\][/tex]
5. Slope of the line segment [tex]\(\overline{W'X'}\)[/tex]:
[tex]\[
\text{slope}_{W'X'} = \frac{11 - 2}{15 - 3} = \frac{9}{12} = \frac{3}{4}
\][/tex]
### Calculating the lengths
6. Length of the original segment [tex]\(\overline{WX}\)[/tex]:
[tex]\[
\text{length}_{WX} = \sqrt{(7 - 3)^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\][/tex]
7. Length of the dilated segment [tex]\(\overline{W'X'}\)[/tex]:
[tex]\[
\text{length}_{W'X'} = 3 \times \text{length}_{WX} = 3 \times 5 = 15
\][/tex]
### Conclusion
With these calculations, we see that:
- The slope of [tex]\(\overline{W'X'}\)[/tex] is [tex]\(\frac{3}{4}\)[/tex].
- The length of [tex]\(\overline{W'X'}\)[/tex] is 15.
Thus, the correct statement is:
C. The slope of [tex]\(\overline{W^{\prime} X^{\prime}}\)[/tex] is [tex]\(\frac{3}{4}\)[/tex], and the length of [tex]\(\overline{W^{\prime} X^{\prime}}\)[/tex] is 15.
