Given the points A(- 5, 3) and B(10, 0), find the coordinates of the point P that divides AB externally in the ratio 6:3.

To find the coordinates of the point \( P \) that divides the line segment \( AB \) externally in the ratio \( 6:3 \), we can use the section formula for external division. The formula is given by:

\[
P(x, y) = \left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2} \right)
\]

where:

- \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of points \( A \) and \( B \), respectively,
- \( m_1:m_2 \) is the given ratio (in this case, 6:3).

Here, \( A(-5, 3) \), \( B(10, 0) \), and the ratio is \( 6:3 \). Let's plug the values into the formula.

### Step 1: Apply the coordinates and the ratio.

- \( x_1 = -5 \), \( y_1 = 3 \)
- \( x_2 = 10 \), \( y_2 = 0 \)
- \( m_1 = 6 \), \( m_2 = 3 \)

### Step 2: Use the section formula for external division.

\[
P(x) = \frac{6 \cdot 10 - 3 \cdot (-5)}{6 - 3} = \frac{60 + 15}{3} = \frac{75}{3} = 25
\]

\[
P(y) = \frac{6 \cdot 0 - 3 \cdot 3}{6 - 3} = \frac{0 - 9}{3} = \frac{-9}{3} = -3
\]

Thus, the coordinates of the point \( P \) are \( (25, -3) \).

### Final Answer:
The point \( P \) that divides the line \( AB \) externally in the ratio \( 6:3 \) is \( P(25, -3) \).

Would you like further details on how the section formula works or have any other questions?

Here are 5 related questions you might consider:
1. How is the section formula different for internal division?
2. Can the section formula be extended to 3D coordinates?
3. What is the geometric interpretation of dividing a line externally?
4. How would you find the midpoint of a line segment?
5. How does the ratio affect the location of the dividing point?

**Tip**: Always pay attention to whether a problem asks for internal or external division, as the formulas are slightly different!

Learn how to find the coordinates of point P that divides the line segment AB externally in the ratio 6:3. This solution uses the section formula for external division, applied to points A(-5, 3) and B(10, 0). Suitable for students in Grades 9-11.

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