To find the range of values of \( \alpha \) such that the point \( (0, \alpha) \) lies on or inside the triangle formed by the lines:
1. \( y + 3x + 2 = 0 \)
2. \( 3y - 2x - 5 = 0 \)
3. \( 4y + x - 14 = 0 \)
we will first determine the vertices of the triangle formed by these lines.
### Step 1: Find the intersection points (vertices of the triangle)
**Finding the intersection of the first two lines:**
1. The first line: \( y + 3x + 2 = 0 \) can be rewritten as:
\[
y = -3x - 2
\]
2. The second line: \( 3y - 2x - 5 = 0 \) can be rewritten as:
\[
3y = 2x + 5 \quad \Rightarrow \quad y = \frac{2}{3}x + \frac{5}{3}
\]
Now, we set the two equations for \( y \) equal to find \( x \):
\[
-3x - 2 = \frac{2}{3}x + \frac{5}{3}
\]
Multiplying through by 3 to eliminate the fraction:
\[
-9x - 6 = 2x + 5
\]
Combining like terms:
\[
-11x = 11 \quad \Rightarrow \quad x = -1
\]
Substituting \( x = -1 \) back into the first line to find \( y \):
\[
y = -3(-1) - 2 = 3 - 2 = 1
\]
Thus, the first vertex \( A \) is \( (-1, 1) \).
**Finding the intersection of the second and third lines:**
1. The second line: \( 3y - 2x - 5 = 0 \) is already rewritten as:
\[
y = \frac{2}{3}x + \frac{5}{3}
\]
2. The third line: \( 4y + x - 14 = 0 \) can be rewritten as:
\[
4y = -x + 14 \quad \Rightarrow \quad y = -\frac{1}{4}x + \frac{14}{4} = -\frac{1}{4}x + \frac{7}{2}
\]
Setting the two equations for \( y \) equal:
\[
\frac{2}{3}x + \frac{5}{3} = -\frac{1}{4}x + \frac{7}{2}
\]
Multiplying through by 12 to eliminate fractions:
\[
8x + 20 = -3x + 42
\]
Combining like terms:
\[
11x = 22 \quad \Rightarrow \quad x = 2
\]
Substituting \( x = 2 \) back into the second line to find \( y \):
\[
y = \frac{2}{3}(2) + \frac{5}{3} = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3
\]
Thus, the second vertex \( B \) is \( (2, 3) \).
**Finding the intersection of the first and third lines:**
1. The first line: \( y + 3x + 2 = 0 \) is:
\[
y = -3x - 2
\]
2. The third line: \( 4y + x - 14 = 0 \) is:
\[
y = -\frac{1}{4}x + \frac{7}{2}
\]
Setting the two equations for \( y \) equal:
\[
-3x - 2 = -\frac{1}{4}x + \frac{7}{2}
\]
Multiplying through by 4 to eliminate fractions:
\[
-12x - 8 = -x + 14
\]
Combining like terms:
\[
-11x = 22 \quad \Rightarrow \quad x = -2
\]
Substituting \( x = -2 \) back into the first line to find \( y \):
\[
y = -3(-2) - 2 = 6 - 2 = 4
\]
Thus, the third vertex \( C \) is \( (-2, 4) \).
### Step 2: Identify the vertices
The vertices of the triangle are:
- \( A(-1, 1) \)
- \( B(2, 3) \)
- \( C(-2, 4) \)
### Step 3: Determine the range of \( \alpha \)
To find the range of \( \alpha \) such that the point \( (0, \alpha) \) lies within or on the triangle, we need to find the \( y \)-coordinates of the triangle's edges where \( x = 0 \).
**Finding the \( y \)-coordinate on line \( AB \):**
The line \( AB \) can be found using the two points \( A(-1, 1) \) and \( B(2, 3) \):
- Slope \( m = \frac{3 - 1}{2 - (-1)} = \frac{2}{3} \)
- Using point-slope form:
\[
y - 1 = \frac{2}{3}(x + 1) \quad \Rightarrow \quad y = \frac{2}{3}x + \frac{5}{3}
\]
At \( x = 0 \):
\[
y = \frac{5}{3}
\]
**Finding the \( y \)-coordinate on line \( BC \):**
Using points \( B(2, 3) \) and \( C(-2, 4) \):
- Slope \( m = \frac{4 - 3}{-2 - 2} = \frac{1}{-4} = -\frac{1}{4} \)
- Using point-slope form:
\[
y - 3 = -\frac{1}{4}(x - 2) \quad \Rightarrow \quad y = -\frac{1}{4}x + \frac{7}{2}
\]
At \( x = 0 \):
\[
y = \frac{7}{2}
\]
### Step 4: Conclusion
Thus, the range of \( \alpha \) such that the point \( (0, \alpha) \) lies within or on the triangle is:
\[
\frac{5}{3} \leq \alpha \leq \frac{7}{2}
\]
