The point which divides the line joining the points (2, 4, 5) and (3, 5, -4) in the ratio -`2:3`, lies on

Options

A. XOY plane
B. YOZ plane
C. ZOX plane
D. XYZ plane

Answer

To find the point that divides the line segment joining the points \( A(2, 4, 5) \) and \( B(3, 5, -4) \) in the ratio \(-2:3\), we will use the section formula for external division.

### Step-by-Step Solution:

1. **Identify the Points and the Ratio:**
- Let \( A(2, 4, 5) \) and \( B(3, 5, -4) \).
- The ratio in which the point divides the line segment is \(-2:3\).

2. **Apply the Section Formula:**
The section formula for external division is given by:
\[ P\left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}, \frac{m_1 z_2 + m_2 z_1}{m_1 + m_2} \right) \]
where \( m_1 \) and \( m_2 \) are the ratios and \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are the coordinates of points \( A \) and \( B \).

3. **Substituting the Values:**
- Here, \( m_1 = -2 \), \( m_2 = 3 \)
- \( (x_1, y_1, z_1) = (2, 4, 5) \)
- \( (x_2, y_2, z_2) = (3, 5, -4) \)

Now, substituting these values into the formula:

- For \( x \):
\[ x = \frac{(-2) \cdot 3 + 3 \cdot 2}{-2 + 3} = \frac{-6 + 6}{1} = \frac{0}{1} = 0 \]

- For \( y \):
\[ y = \frac{(-2) \cdot 5 + 3 \cdot 4}{-2 + 3} = \frac{-10 + 12}{1} = \frac{2}{1} = 2 \]

- For \( z \):
\[ z = \frac{(-2) \cdot (-4) + 3 \cdot 5}{-2 + 3} = \frac{8 + 15}{1} = \frac{23}{1} = 23 \]

4. **Final Coordinates:**
The point that divides the line segment in the ratio \(-2:3\) is:
\[ P(0, 2, 23) \]

5. **Determine the Plane:**
To find which plane this point lies on, we can check if it satisfies any standard plane equations. A common plane is the coordinate planes:
- \( x = 0 \) (yz-plane)
- \( y = 0 \) (xz-plane)
- \( z = 0 \) (xy-plane)

Since \( x = 0 \), the point \( P(0, 2, 23) \) lies on the yz-plane.

### Summary:
The point \( P(0, 2, 23) \) divides the line joining \( (2, 4, 5) \) and \( (3, 5, -4) \) in the ratio \(-2:3\) and lies on the yz-plane.