Area of triangle formed by straight lines `3x - 4y = 0, x = 4` and x-axis

Options

A. 6 sq. unit
B. 8 sq. unit
C. 12 sq. unit
D. 16 sq. unit

Answer

To find the area of the triangle formed by the lines \(3x - 4y = 0\), \(x = 4\), and the x-axis, we can follow these steps:

### Step 1: Find the points of intersection
1. **Find the intersection of \(3x - 4y = 0\) and the x-axis.**
- The x-axis is represented by \(y = 0\).
- Substitute \(y = 0\) into the equation \(3x - 4y = 0\):
\[ 3x - 4(0) = 0 \implies 3x = 0 \implies x = 0 \]
- Thus, the point of intersection is \((0, 0)\).

2. **Find the intersection of \(3x - 4y = 0\) and \(x = 4\).**
- Substitute \(x = 4\) into the equation \(3x - 4y = 0\):
\[ 3(4) - 4y = 0 \implies 12 - 4y = 0 \implies 4y = 12 \implies y = 3 \]
- Thus, the point of intersection is \((4, 3)\).

3. **Identify the intersection of \(x = 4\) with the x-axis.**
- The x-axis intersects \(x = 4\) at the point \((4, 0)\).

### Step 2: Identify the vertices of the triangle
The vertices of the triangle formed by these lines are:
- Point A: \((0, 0)\)
- Point B: \((4, 3)\)
- Point C: \((4, 0)\)

### Step 3: Calculate the area of the triangle
The area \(A\) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
- Here, we can take the base as the distance from point C to point A along the x-axis, which is \(4\) units (from \((0, 0)\) to \((4, 0)\)).
- The height is the vertical distance from point B to the x-axis, which is \(3\) units (from \((4, 0)\) to \((4, 3)\)).

Substituting the values into the area formula:
\[ A = \frac{1}{2} \times 4 \times 3 = \frac{12}{2} = 6 \text{ square units} \]

### Final Answer
The area of the triangle formed by the lines is \(6\) square units.

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