To find the coordinates of point $X$ on line segment $\overline{AB}$ such that the ratio of $AX$ to $XB$ is 3:5, we can use the section formula.

The section formula states that if a point divides a line segment in a given ratio, the coordinates of that point can be found using:

$X = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right)$

Where:

• $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of points $A$ and $B$,
• $m:n$ is the given ratio of $AX$ to $XB$.

In this case:

• $A(-1, 2)$, $B(3, 6)$,
• $m:n = 3:5$.

Now, applying the section formula:

$X = \left( \frac{5(3) + 3(-1)}{3+5}, \frac{5(6) + 3(2)}{3+5} \right)$

First, calculate the $x$-coordinate:

$x = \frac{5(3) + 3(-1)}{8} = \frac{15 - 3}{8} = \frac{12}{8} = 1.5$

Next, calculate the $y$-coordinate:

$y = \frac{5(6) + 3(2)}{8} = \frac{30 + 6}{8} = \frac{36}{8} = 4.5$

Thus, the coordinates of point $X$ are $(1.5, 4.5)$.

Summary:

The coordinates of point $X$ are $\boxed{(1.5, 4.5)}$.

Would you like any further details or explanations?

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Here are some related questions for practice:

1. Find the coordinates of point $X$ dividing the line segment in the ratio 2:3.
2. How would the section formula change if $AX$ to $XB$ was 5:3?
3. What are the coordinates of the midpoint of $\overline{AB}$?
4. How do you apply the section formula when the ratio involves negative numbers?
5. Can you derive the section formula from basic geometry?

Tip: Always label your points clearly in geometry problems, as it helps to avoid confusion when applying formulas.