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To solve this problem, we can use the section formula. The section formula helps find the coordinates of a point that divides a line segment in a given ratio. 

The coordinates of the endpoints of \( \overline{TU} \) are:
- \( T(-2, 7) \)
- \( U(4, -8) \)

The ratio \( TV : UV \) is given as \( 1 : 2 \), meaning point \( V \) divides the line segment \( TU \) in the ratio \( m:n = 1:2 \).

The section formula is:

\[
(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)
\]

Here:
- \( x_1 = -2 \), \( y_1 = 7 \) (coordinates of \( T \))
- \( x_2 = 4 \), \( y_2 = -8 \) (coordinates of \( U \))
- \( m = 1 \), \( n = 2 \)

Substituting these values into the formula:

\[
x = \frac{1 \cdot 4 + 2 \cdot (-2)}{1+2} = \frac{4 - 4}{3} = 0
\]
\[
y = \frac{1 \cdot (-8) + 2 \cdot 7}{1+2} = \frac{-8 + 14}{3} = \frac{6}{3} = 2
\]

Thus, the coordinates of \( V \) are \( (0, 2) \).

Would you like further details on the steps, or do you have any questions?

Here are 5 related questions to expand your understanding:
1. How would the coordinates of \( V \) change if the ratio were \( 2:1 \)?
2. What if the point \( V \) divides \( TU \) externally instead of internally?
3. Can you apply the section formula in three-dimensional space?
4. How would you find the length of the segment \( TV \)?
5. How does the midpoint formula relate to the section formula?

**Tip**: The section formula is highly versatile and works for both internal and external divisions, as long as you adjust the sign for external division.

The coordinates of the endpoints of \( \overline{TU} \) are T(-2, 7) and U(4, -8). Point V is on \( \overline{TU} \) and divides it such that \( TV:UV \) is 1:2. What are the coordinates of V? Write your answers as integers or decimals.

Learn how to find the coordinates of a point dividing a line segment in a given ratio with a detailed step-by-step solution. This problem involves using the section formula to determine the coordinates of point V that divides segment TU in the ratio 1:2. Suitable for students in Grades 9-12.

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