The area of a triangle, whose verticles are `(3,4) , (5,2)` and the point of the 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

Explanation

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 its vertices:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 1: Identify the vertices
Let:
- \( (x_1, y_1) = (3, 4) \)
- \( (x_2, y_2) = (5, 2) \)
- \( (x_3, y_3) = (a, 5) \)

### Step 2: Set up the area equation
We know the area is 3 square units, so we set up the equation:

\[ 3 = \frac{1}{2} \left| 3(2 - 5) + 5(5 - 4) + a(4 - 2) \right| \]

### Step 3: Simplify the equation
Multiply both sides by 2 to eliminate the fraction:

\[ 6 = \left| 3(-3) + 5(1) + a(2) \right| \]

Calculating the terms inside the absolute value:

\[ 6 = \left| -9 + 5 + 2a \right| \]

This simplifies to:

\[ 6 = \left| -4 + 2a \right| \]

### Step 4: Solve the absolute value equation
We have two cases to consider due to the absolute value:

**Case 1:**
\[ -4 + 2a = 6 \]
Solving for \( a \):
\[ 2a = 10 \implies a = 5 \]

**Case 2:**
\[ -4 + 2a = -6 \]
Solving for \( a \):
\[ 2a = -2 \implies a = -1 \]

### Step 5: Conclusion
The possible values of \( a \) are \( 5 \) and \( -1 \). However, we need to check which of these values is valid based on the context of the problem.

### Final Answer
The value of \( a \) that satisfies the area condition is \( a = 5 \).

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