Let the base of a triangle lie along the line x = a and be of length 2a. The area of this triangles is `a^(2)`, if the vertex lies on the line

Options

A. `x = -a`
B. x = 0
C. `x=(a)/(2)`
D. x = 2a

Answer

To solve the problem step by step, we need to analyze the given information about the triangle:

1. **Understanding the Triangle's Base**:
- The base of the triangle lies along the line \( x = a \).
- The length of the base is \( 2a \).
- Since the base is vertical along the line \( x = a \), the endpoints of the base can be represented as \( (a, y_1) \) and \( (a, y_2) \) where \( y_1 \) and \( y_2 \) are the y-coordinates of the endpoints.

2. **Finding the Coordinates of the Base**:
- Since the length of the base is \( 2a \), we can set \( y_1 = y \) and \( y_2 = y + 2a \) (assuming \( y_1 < y_2 \)).
- Thus, the coordinates of the base vertices are \( (a, y) \) and \( (a, y + 2a) \).

3. **Calculating the Height of the Triangle**:
- Let the vertex of the triangle be at point \( (x_v, y_v) \).
- Since the vertex lies on the line \( x = a \), we can set \( x_v = a \).
- The height of the triangle will be the vertical distance from the vertex to the base, which can be calculated as \( h = |y_v - y| \) or \( h = |y_v - (y + 2a)| \).

4. **Using the Area Formula**:
- The area \( A \) of a triangle is given by the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
- Here, the base is \( 2a \) and the height is \( h \). Therefore, we can write:
\[ A = \frac{1}{2} \times 2a \times h = a \times h \]

5. **Setting Up the Equation**:
- We know from the problem that the area of the triangle is \( a^2 \). Thus, we can set up the equation:
\[ a \times h = a^2 \]
- Simplifying this gives:
\[ h = a \]

6. **Finding the Vertex's Y-Coordinate**:
- Since we have established that the height \( h \) is equal to \( a \), we can express this in terms of the y-coordinates:
- If \( y_v \) is above the base, we have \( y_v - y = a \) which gives \( y_v = y + a \).
- If \( y_v \) is below the base, we have \( (y + 2a) - y_v = a \) which gives \( y_v = y + 2a - a = y + a \).

7. **Conclusion**:
- In both cases, the vertex's y-coordinate is \( y + a \). Thus, the vertex of the triangle can be represented as \( (a, y + a) \).

### Summary of the Solution:
- The base of the triangle lies along the line \( x = a \) with endpoints at \( (a, y) \) and \( (a, y + 2a) \).
- The height of the triangle is \( a \), leading to the vertex being at \( (a, y + a) \).