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!