The area of a triangle , whose vertices are (3 , 4) , (5 , 2) and the point of intersection of the lines x = a and y = 5 is 3 square units . What is the value of a ?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the value of \( a \) such that the area of the triangle formed by the vertices \( (3, 4) \), \( (5, 2) \), and \( (a, 5) \) is 3 square units, we can use the formula for the area of a triangle given by the coordinates of its vertices.

### Step-by-Step Solution:

1. **Identify the vertices of the triangle**:
The vertices are \( (3, 4) \), \( (5, 2) \), and \( (a, 5) \).

2. **Use the area formula for a triangle**:
The area \( A \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

3. **Substitute the coordinates into the formula**:
Here, \( (x_1, y_1) = (3, 4) \), \( (x_2, y_2) = (5, 2) \), and \( (x_3, y_3) = (a, 5) \).
\[ A = \frac{1}{2} \left| 3(2 - 5) + 5(5 - 4) + a(4 - 2) \right| \]

4. **Simplify the expression**:
\[ A = \frac{1}{2} \left| 3(-3) + 5(1) + a(2) \right| \]
\[ = \frac{1}{2} \left| -9 + 5 + 2a \right| \]
\[ = \frac{1}{2} \left| 2a - 4 \right| \]

5. **Set the area equal to 3 square units**:
Since we know the area is 3 square units, we set up the equation:
\[ \frac{1}{2} \left| 2a - 4 \right| = 3 \]

6. **Multiply both sides by 2**:
\[ \left| 2a - 4 \right| = 6 \]

7. **Solve the absolute value equation**:
This gives us two cases:
- Case 1: \( 2a - 4 = 6 \)
- Case 2: \( 2a - 4 = -6 \)

**For Case 1**:
\[ 2a - 4 = 6 \implies 2a = 10 \implies a = 5 \]

**For Case 2**:
\[ 2a - 4 = -6 \implies 2a = -2 \implies a = -1 \]

8. **Conclusion**:
The possible values for \( a \) are \( 5 \) and \( -1 \). Since the problem asks for the value of \( a \), we can conclude:
\[ a = 5 \quad \text{(as it is a valid intersection point)} \]