To solve the problem, follow these steps:
1. Identify the coordinates of points W and X:
[tex]\[
W = (3, 2) \quad \text{and} \quad X = (7, 5)
\][/tex]
2. Calculate the slope of line segment [tex]\( \overline{WX} \)[/tex]:
The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
\text{slope}_{WX} = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Substituting the values:
[tex]\[
\text{slope}_{WX} = \frac{5 - 2}{7 - 3} = \frac{3}{4}
\][/tex]
3. Dilate point X with respect to W with a scale factor of 3:
The formula to find the coordinates of the dilated point [tex]\(X'\)[/tex] is given by:
[tex]\[
X' = \left(W_x + \text{scale factor} \times (X_x - W_x), W_y + \text{scale factor} \times (X_y - W_y)\right)
\][/tex]
Plugging in the values:
[tex]\[
X' = \left(3 + 3 \times (7 - 3), 2 + 3 \times (5 - 2)\right) = (3 + 12, 2 + 9) = (15, 11)
\][/tex]
4. Calculate the slope of line segment [tex]\( \overline{W'X'} \)[/tex]:
Since W' coincides with W (because they are the center of dilation) the slope of [tex]\( \overline{W'X'} \)[/tex] is the same as the slope of [tex]\( \overline{WX} \)[/tex]:
[tex]\[
\text{slope}_{W'X'} = \frac{11 - 2}{15 - 3} = \frac{9}{12} = \frac{3}{4}
\][/tex]
So, the slope of [tex]\( \overline{W'X'} \)[/tex] remains:
[tex]\[
\text{slope}_{W'X'} = \frac{3}{4}
\][/tex]
5. Calculate the length of [tex]\( \overline{W'X'} \)[/tex]:
Using the distance formula to calculate the length of [tex]\( \overline{W'X'} \)[/tex]:
[tex]\[
\text{length}_{W'X'} = \sqrt{(15 - 3)^2 + (11 - 2)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15
\][/tex]
6. Identify the correct statement:
We need to match the slope of [tex]\( \overline{W'X'} \)[/tex] and the length of [tex]\( \overline{W'X'} \)[/tex] with the given options.
- We found 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:
[tex]\[
\boxed{\text{C. The slope of \(\overline{W'X'}\) is \(\frac{3}{4}\), and the length of \(\overline{W'X'}\) is 15.}}
\][/tex]
