The point `P(alpha,alpha +1)` will lie inside the triangle whose vertices are `A(0,3), B(-2,0)` and `C(6,1)` if

Options

A. `alpha =- 1`
B. `alpha =- (1)/(2)`
C. `alpha = (1)/(2)`
D. `-(6)/(7) lt alpha lt (3)/(2)`

Explanation

To determine the conditions under which the point \( P(\alpha, \alpha + 1) \) lies inside the triangle formed by the vertices \( A(0, 3) \), \( B(-2, 0) \), and \( C(6, 1) \), we can follow these steps:

### Step 1: Find the equations of the sides of the triangle

1. **Equation of line AC**:
- Points: \( A(0, 3) \) and \( C(6, 1) \)
- Slope \( m_{AC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{6 - 0} = -\frac{1}{3} \)
- Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 3 = -\frac{1}{3}(x - 0) \implies y = -\frac{1}{3}x + 3 \]
- Rearranging gives:
\[ x + 3y - 9 = 0 \quad \text{(Equation 1)} \]

2. **Equation of line BC**:
- Points: \( B(-2, 0) \) and \( C(6, 1) \)
- Slope \( m_{BC} = \frac{1 - 0}{6 - (-2)} = \frac{1}{8} \)
- Using point-slope form:
\[ y - 0 = \frac{1}{8}(x + 2) \implies y = \frac{1}{8}x + \frac{1}{4} \]
- Rearranging gives:
\[ x - 8y + 2 = 0 \quad \text{(Equation 2)} \]

3. **Equation of line AB**:
- Points: \( A(0, 3) \) and \( B(-2, 0) \)
- Slope \( m_{AB} = \frac{0 - 3}{-2 - 0} = \frac{3}{2} \)
- Using point-slope form:
\[ y - 3 = \frac{3}{2}(x - 0) \implies y = \frac{3}{2}x + 3 \]
- Rearranging gives:
\[ 3x - 2y + 6 = 0 \quad \text{(Equation 3)} \]

### Step 2: Determine the conditions for point \( P(\alpha, \alpha + 1) \)

To find if the point \( P(\alpha, \alpha + 1) \) lies inside the triangle, we need to check the following inequalities:

1. **For line AC**:
\[ \alpha + 1 < -\frac{1}{3}\alpha + 3 \implies \frac{4}{3}\alpha < 2 \implies \alpha < \frac{3}{2} \quad \text{(Inequality 1)} \]

2. **For line BC**:
\[ \alpha + 1 > \frac{1}{8}\alpha + \frac{1}{4} \implies \frac{7}{8}\alpha > -\frac{3}{4} \implies \alpha > -\frac{6}{7} \quad \text{(Inequality 2)} \]

3. **For line AB**:
\[ \alpha + 1 < \frac{3}{2}\alpha + 3 \implies -\frac{1}{2}\alpha < 2 \implies \alpha > -4 \quad \text{(Inequality 3)} \]

### Step 3: Combine the inequalities

From the inequalities derived:

- From Inequality 1: \( \alpha < \frac{3}{2} \)
- From Inequality 2: \( \alpha > -\frac{6}{7} \)
- From Inequality 3: \( \alpha > -4 \)

The most restrictive conditions are:
\[ -\frac{6}{7} < \alpha < \frac{3}{2} \]

### Final Answer

The point \( P(\alpha, \alpha + 1) \) will lie inside the triangle if:
\[ -\frac{6}{7} < \alpha < \frac{3}{2} \]