1. **Problem:** Find the coordinates of point P along the directed line segment ST so that the ratio of SP to PT is 1 to 3, where S = (6, 4) and T = (-4, -8). 2. **Formula:** To find point P dividing segment ST in ratio $m:n$, use the section formula: $$P = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right)$$ where $S = (x_1, y_1)$ and $T = (x_2, y_2)$. 3. **Apply values:** Here, $m=1$, $n=3$, $S=(6,4)$, $T=(-4,-8)$. Calculate $x$-coordinate: $$x_P = \frac{1 \times (-4) + 3 \times 6}{1+3} = \frac{-4 + 18}{4} = \frac{14}{4} = 3.5$$ Calculate $y$-coordinate: $$y_P = \frac{1 \times (-8) + 3 \times 4}{1+3} = \frac{-8 + 12}{4} = \frac{4}{4} = 1$$ 4. **Answer:** The coordinates of point P are $\boxed{(3.5, 1)}$. This means point P divides the segment ST so that the length SP is one quarter of ST and PT is three quarters of ST, consistent with the ratio 1:3.