A line passing through `A(2,-1,5)andB(4,3,-10)` meets the xy-plane at `(x,y,z)`, then `(x)/(y)` = __________

Answer

To solve the problem, we need to find the ratio \( \frac{x}{y} \) where the line passing through the points \( A(2, -1, 5) \) and \( B(4, 3, -10) \) meets the xy-plane.

### Step-by-Step Solution:

1. **Identify the Points**:
The points given are \( A(2, -1, 5) \) and \( B(4, 3, -10) \).

2. **Find the Direction Ratios**:
The direction ratios of the line can be calculated as:
\[ \text{Direction Ratios} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) = (4 - 2, 3 - (-1), -10 - 5) = (2, 4, -15) \]

3. **Parametric Equations of the Line**:
The parametric equations of the line can be written as:
\[ x = 2 + 2t, \quad y = -1 + 4t, \quad z = 5 - 15t \]
where \( t \) is a parameter.

4. **Condition for the xy-plane**:
The line meets the xy-plane when \( z = 0 \). So, we set the equation for \( z \) to zero:
\[ 5 - 15t = 0 \]
Solving for \( t \):
\[ 15t = 5 \implies t = \frac{1}{3} \]

5. **Substitute \( t \) back to find \( x \) and \( y \)**:
- For \( x \):
\[ x = 2 + 2\left(\frac{1}{3}\right) = 2 + \frac{2}{3} = \frac{6}{3} + \frac{2}{3} = \frac{8}{3} \]
- For \( y \):
\[ y = -1 + 4\left(\frac{1}{3}\right) = -1 + \frac{4}{3} = -\frac{3}{3} + \frac{4}{3} = \frac{1}{3} \]

6. **Calculate \( \frac{x}{y} \)**:
Now we can find \( \frac{x}{y} \):
\[ \frac{x}{y} = \frac{\frac{8}{3}}{\frac{1}{3}} = \frac{8}{3} \times \frac{3}{1} = 8 \]

### Final Answer:
\[ \frac{x}{y} = 8 \]