5. Find the point on the directed segment from $$ (-3,-2) $$ to $$ (4,8) $$ that divides it into a ratio of $$ 3: 2 $$.

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Let $$ A=(-3,-2) $$ and $$ B=(4,8) $$. We want to find the point $$ P $$ on $$ \overline{AB} $$ such that
$$
AP : PB = 3 : 2.
$$

Using the section formula for internal division, if a point $$ P $$ divides $$ \overline{AB} $$ in the ratio $$ m : n $$ (with $$ m $$ for $$ PB $$ and $$ n $$ for $$ AP $$), then
$$
P = \left(\frac{n x_1 + m x_2}{m+n}, \frac{n y_1 + m y_2}{m+n}\right),
$$
where $$ A=(x_1,y_1) $$ and $$ B=(x_2,y_2) $$.

In our problem, since $$ AP : PB = 3 : 2 $$, we have $$ n=2 $$ (weight for $$ A $$) and $$ m=3 $$ (weight for $$ B $$).

Substitute the coordinates:
$$
x_1 = -3,\quad y_1 = -2,\quad x_2 = 4,\quad y_2 = 8,
$$
we get
$$
P = \left(\frac{2(-3) + 3(4)}{3+2}, \frac{2(-2) + 3(8)}{3+2}\right).
$$

Calculate the coordinates step by step:

1. For the $$ x $$-coordinate:
   $$
   x = \frac{2(-3) + 3(4)}{5} = \frac{-6 + 12}{5} = \frac{6}{5}.
   $$

2. For the $$ y $$-coordinate:
   $$
   y = \frac{2(-2) + 3(8)}{5} = \frac{-4 + 24}{5} = \frac{20}{5} = 4.
   $$

Thus, the point $$ P $$ is
$$
P = \left(\frac{6}{5}, 4\right).
$$
The point that divides the segment from $$ (-3,-2) $$ to $$ (4,8) $$ in a ratio of $$ 3:2 $$ is $$ \left(\frac{6}{5}, 4\right) $$.

5. Find the point on the directed segment from \( (-3,-2) \) to \( (4,8) \) that divides it into a ratio of \( 3: 2 \).

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