If two vertical pale AB and CD of height 15 m and 10 m and A and C are on ground. P is the point of intersection of BC and AD. What is height of P from the ground in m.

Options

A. `20/3`
B. `10/3`
C. 5
D. `-6`

Answer

To solve the problem, we need to find the height of point P, which is the intersection of lines BC and AD formed by two vertical poles AB and CD.

### Step-by-Step Solution:

1. **Define the Coordinates**:
- Let point A be at the origin (0, 0).
- Since pole AB has a height of 15 m, point B will be at (0, 15).
- Let point C be at a distance L from A on the ground, so C is at (L, 0).
- Since pole CD has a height of 10 m, point D will be at (L, 10).

2. **Find the Equations of Lines BC and AD**:
- **Line BC**: The slope of line BC can be calculated as follows:
\[ \text{slope of BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{15 - 0}{0 - L} = -\frac{15}{L} \]
Using point-slope form, the equation of line BC is:
\[ y - 0 = -\frac{15}{L}(x - L) \]
Simplifying this gives:
\[ y = -\frac{15}{L}x + 15 \]

- **Line AD**: The slope of line AD is:
\[ \text{slope of AD} = \frac{10 - 0}{L - 0} = \frac{10}{L} \]
Using point-slope form, the equation of line AD is:
\[ y - 0 = \frac{10}{L}(x - 0) \]
Simplifying this gives:
\[ y = \frac{10}{L}x \]

3. **Set the Equations Equal to Find Intersection Point P**:
- To find the height of point P, we set the two equations equal to each other:
\[ -\frac{15}{L}x + 15 = \frac{10}{L}x \]
- Rearranging gives:
\[ 15 = \frac{15}{L}x \]
- Multiplying both sides by L:
\[ 15L = 15x \]
- Dividing both sides by 15:
\[ x = L \]

4. **Substituting x back to find y**:
- Substitute \(x = L\) back into either equation to find y. Using the equation of line AD:
\[ y = \frac{10}{L}L = 10 \]

5. **Conclusion**:
- The height of point P from the ground is 10 m.

### Final Answer:
The height of point P from the ground is **10 m**.