Point `(0, beta)` lies on or inside the triangle fromed by the lines ` y=0, x+y=8 ` and ` 3x - 4y + 12 =0`. Then ` beta` can be :

Options

A. 2
B. 4
C. 8
D. 12

Answer

To determine the possible values of \( \beta \) for the point \( (0, \beta) \) that lies on or inside the triangle formed by the lines \( y = 0 \), \( x + y = 8 \), and \( 3x - 4y + 12 = 0 \), we will follow these steps:

### Step 1: Identify the lines and their intersections

1. **Line 1**: \( y = 0 \) (the x-axis)
2. **Line 2**: \( x + y = 8 \) can be rewritten as \( y = -x + 8 \).
3. **Line 3**: \( 3x - 4y + 12 = 0 \) can be rewritten as \( 4y = 3x + 12 \) or \( y = \frac{3}{4}x + 3 \).

### Step 2: Find the intersection points of the lines

- **Intersection of Line 1 and Line 2**:
\[ y = 0 \implies 0 = -x + 8 \implies x = 8 \quad \text{(Point: (8, 0))} \]

- **Intersection of Line 1 and Line 3**:
\[ y = 0 \implies 0 = \frac{3}{4}x + 3 \implies \frac{3}{4}x = -3 \implies x = -4 \quad \text{(Point: (-4, 0))} \]

- **Intersection of Line 2 and Line 3**:
Set \( -x + 8 = \frac{3}{4}x + 3 \):
\[ -x + 8 = \frac{3}{4}x + 3 \]
\[ 8 - 3 = \frac{3}{4}x + x \]
\[ 5 = \frac{3}{4}x + \frac{4}{4}x \implies 5 = \frac{7}{4}x \implies x = \frac{20}{7} \]
Substitute \( x \) back into Line 2 to find \( y \):
\[ y = -\frac{20}{7} + 8 = -\frac{20}{7} + \frac{56}{7} = \frac{36}{7} \quad \text{(Point: } \left(\frac{20}{7}, \frac{36}{7}\right)\text{)} \]

### Step 3: Determine the vertices of the triangle

The vertices of the triangle formed by the lines are:
- \( A(8, 0) \)
- \( B(-4, 0) \)
- \( C\left(\frac{20}{7}, \frac{36}{7}\right) \)

### Step 4: Analyze the position of the point \( (0, \beta) \)

Since \( (0, \beta) \) lies on the y-axis, we need to find the range of \( \beta \) such that the point is either on or inside the triangle.

1. **On Line 1**: \( \beta \) can be equal to \( 0 \) (the x-axis).
2. **On Line 2**: The y-coordinate at \( x = 0 \) is \( y = 8 \). However, we are only interested in the segment of the triangle.
3. **On Line 3**: The y-coordinate at \( x = 0 \) is \( y = 3 \).

### Step 5: Determine the range of \( \beta \)

From the above analysis, we find that:
- The maximum value of \( \beta \) is \( 3 \) (from Line 3).
- The minimum value of \( \beta \) is \( 0 \) (from Line 1).

Thus, the values of \( \beta \) that can lie on or inside the triangle are:
\[ 0 \leq \beta \leq 3 \]

### Final Answer

The possible values of \( \beta \) are:
\[ \beta \in [0, 3] \]