To find the coordinates of point [tex]\( Q \)[/tex] that lies along the directed line segment from [tex]\( R(-2, 4) \)[/tex] to [tex]\( S(18, -6) \)[/tex] and partitions the segment in the ratio of [tex]\( 3:7 \)[/tex], we can use the section formula. The section formula for a point dividing a line segment in the ratio [tex]\( \frac{m}{n} \)[/tex] is given by:
[tex]\[
Q \left( x_Q, y_Q \right) = \left( \frac{m \cdot x_2 + n \cdot x_1}{m + n}, \frac{m \cdot y_2 + n \cdot y_1}{m + n} \right)
\][/tex]
For our given problem:
- [tex]\( x_1 = -2 \)[/tex]
- [tex]\( y_1 = 4 \)[/tex]
- [tex]\( x_2 = 18 \)[/tex]
- [tex]\( y_2 = -6 \)[/tex]
- [tex]\( m = 3 \)[/tex]
- [tex]\( n = 7 \)[/tex]
Plugging these values into the section formula, we get:
[tex]\[
x_Q = \frac{3 \cdot 18 + 7 \cdot (-2)}{3 + 7}
\][/tex]
[tex]\[
y_Q = \frac{3 \cdot (-6) + 7 \cdot 4}{3 + 7}
\][/tex]
Now we calculate [tex]\( x_Q \)[/tex]:
[tex]\[
x_Q = \frac{54 + (-14)}{10} = \frac{54 - 14}{10} = \frac{40}{10} = 4
\][/tex]
And calculate [tex]\( y_Q \)[/tex]:
[tex]\[
y_Q = \frac{-18 + 28}{10} = \frac{28 - 18}{10} = \frac{10}{10} = 1
\][/tex]
Therefore, the coordinates of point [tex]\( Q \)[/tex] are [tex]\( (4, 1) \)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{(4, 1)}
\][/tex]
