Let the base of a triangle lie along the line x = a and be of length a. 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, we need to find the equation of the line on which the vertex of the triangle lies, given that the base lies along the line \( x = a \) and has a length of \( a \). The area of the triangle is given as \( a^2 \).

### Step-by-Step Solution:

1. **Understanding the Triangle's Base**:
- The base of the triangle lies on the line \( x = a \) and has a length of \( a \).
- Therefore, the coordinates of the endpoints of the base can be taken as \( (a, 0) \) and \( (a, a) \).

2. **Vertex of the Triangle**:
- Let the coordinates of the vertex be \( (x, y) \).

3. **Area of the Triangle**:
- The area \( A \) of a triangle formed by the vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- Substituting the coordinates of our triangle:
- \( (x_1, y_1) = (a, 0) \)
- \( (x_2, y_2) = (a, a) \)
- \( (x_3, y_3) = (x, y) \)

4. **Substituting into the Area Formula**:
- The area is given as \( a^2 \):
\[ a^2 = \frac{1}{2} \left| a(a - y) + a(y - 0) + x(0 - a) \right| \]
- Simplifying the expression:
\[ a^2 = \frac{1}{2} \left| a^2 - ay + ay - ax \right| \]
- This simplifies to:
\[ a^2 = \frac{1}{2} \left| a^2 - ax \right| \]

5. **Removing the Absolute Value**:
- We can consider two cases for the absolute value.

6. **Case 1: Positive Case**:
- Assume \( a^2 - ax \geq 0 \):
\[ a^2 = \frac{1}{2}(a^2 - ax) \]
- Multiplying both sides by 2:
\[ 2a^2 = a^2 - ax \]
- Rearranging gives:
\[ 2a^2 + ax = a^2 \]
\[ ax = 3a^2 \implies x = 3a \]

7. **Case 2: Negative Case**:
- Assume \( a^2 - ax < 0 \):
\[ a^2 = \frac{1}{2}(-a^2 + ax) \]
- Multiplying both sides by 2:
\[ 2a^2 = -a^2 + ax \]
- Rearranging gives:
\[ 3a^2 = ax \implies x = -a \]

8. **Final Equations**:
- Therefore, the vertex of the triangle can lie on the lines:
\[ x = 3a \quad \text{or} \quad x = -a \]

### Summary of the Solution:
The vertex of the triangle can lie on the lines \( x = 3a \) or \( x = -a \).