Area of triangle formed by straight lines `3x + 4y = 24, x = 8 and y =1 `

Options

A. 12 sq. unit
B. 24 sq. unit
C. 48 sq. unit
D. None of these

Answer

To find the area of the triangle formed by the lines \(3x + 4y = 24\), \(x = 8\), and \(y = 1\), we will follow these steps:

### Step 1: Find the intersection points of the lines
1. **Intersection of \(3x + 4y = 24\) and \(x = 8\)**:
- Substitute \(x = 8\) into the equation \(3x + 4y = 24\):
\[ 3(8) + 4y = 24 \implies 24 + 4y = 24 \implies 4y = 0 \implies y = 0 \]
- So, the intersection point is \((8, 0)\).

2. **Intersection of \(3x + 4y = 24\) and \(y = 1\)**:
- Substitute \(y = 1\) into the equation \(3x + 4y = 24\):
\[ 3x + 4(1) = 24 \implies 3x + 4 = 24 \implies 3x = 20 \implies x = \frac{20}{3} \]
- So, the intersection point is \(\left(\frac{20}{3}, 1\right)\).

3. **Intersection of \(x = 8\) and \(y = 1\)**:
- This gives us the point \((8, 1)\).

### Step 2: Identify the vertices of the triangle
The vertices of the triangle formed by the lines are:
- \(A(8, 0)\)
- \(B\left(\frac{20}{3}, 1\right)\)
- \(C(8, 1)\)

### Step 3: Calculate the base and height of the triangle
- **Base**: The base can be considered as the line segment between points \(A(8, 0)\) and \(C(8, 1)\). The length of the base is:
\[ \text{Base} = |y_C - y_A| = |1 - 0| = 1 \]

- **Height**: The height can be considered as the horizontal distance from point \(B\left(\frac{20}{3}, 1\right)\) to the line \(x = 8\). The height is:
\[ \text{Height} = |x_C - x_B| = |8 - \frac{20}{3}| = \left| \frac{24}{3} - \frac{20}{3} \right| = \frac{4}{3} \]

### Step 4: Calculate the area of the triangle
The area \(A\) of the triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Substituting the values we found:
\[ A = \frac{1}{2} \times 1 \times \frac{4}{3} = \frac{2}{3} \]

### Conclusion
The area of the triangle formed by the lines \(3x + 4y = 24\), \(x = 8\), and \(y = 1\) is \(\frac{2}{3}\) square units.