To find the value of `lambda` such that the point `(lambda, lambda + 1)` is an interior point of triangle ABC with vertices A(0, 3), B(-2, 0), and C(6, 1), we will follow these steps:
### Step 1: Find the equations of the sides of triangle ABC.
#### 1.1: Equation of line AB
- The coordinates of A are (0, 3) and B are (-2, 0).
- The slope \( m \) of line AB is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 3}{-2 - 0} = \frac{-3}{-2} = \frac{3}{2}
\]
- Using the point-slope form \( y - y_1 = m(x - x_1) \), we can write the equation of line AB:
\[
y - 3 = \frac{3}{2}(x - 0) \implies y = \frac{3}{2}x + 3
\]
- Rearranging gives:
\[
3x - 2y + 6 = 0 \quad \text{(Equation of line AB)}
\]
#### 1.2: Equation of line BC
- The coordinates of B are (-2, 0) and C are (6, 1).
- The slope \( m \) of line BC is:
\[
m = \frac{1 - 0}{6 - (-2)} = \frac{1}{8}
\]
- Using the point-slope form:
\[
y - 0 = \frac{1}{8}(x + 2) \implies y = \frac{1}{8}x + \frac{1}{4}
\]
- Rearranging gives:
\[
x - 8y + 2 = 0 \quad \text{(Equation of line BC)}
\]
#### 1.3: Equation of line AC
- The coordinates of A are (0, 3) and C are (6, 1).
- The slope \( m \) of line AC is:
\[
m = \frac{1 - 3}{6 - 0} = \frac{-2}{6} = \frac{-1}{3}
\]
- Using the point-slope form:
\[
y - 3 = -\frac{1}{3}(x - 0) \implies y = -\frac{1}{3}x + 3
\]
- Rearranging gives:
\[
x + 3y - 9 = 0 \quad \text{(Equation of line AC)}
\]
### Step 2: Determine conditions for the point `(lambda, lambda + 1)` to be an interior point.
#### 2.1: Condition for point P to be on the same side of line BC as A
- Substitute \( P(\lambda, \lambda + 1) \) into the equation of line BC:
\[
\lambda - 8(\lambda + 1) + 2 > 0
\]
Simplifying gives:
\[
\lambda - 8\lambda - 8 + 2 > 0 \implies -7\lambda - 6 > 0 \implies \lambda < -\frac{6}{7} \quad \text{(Condition 1)}
\]
#### 2.2: Condition for point P to be on the same side of line AB as C
- Substitute \( P(\lambda, \lambda + 1) \) into the equation of line AB:
\[
3\lambda - 2(\lambda + 1) + 6 > 0
\]
Simplifying gives:
\[
3\lambda - 2\lambda - 2 + 6 > 0 \implies \lambda + 4 > 0 \implies \lambda > -4 \quad \text{(Condition 2)}
\]
#### 2.3: Condition for point P to be on the same side of line AC as B
- Substitute \( P(\lambda, \lambda + 1) \) into the equation of line AC:
\[
\lambda + 3(\lambda + 1) - 9 > 0
\]
Simplifying gives:
\[
\lambda + 3\lambda + 3 - 9 > 0 \implies 4\lambda - 6 > 0 \implies \lambda > \frac{3}{2} \quad \text{(Condition 3)}
\]
### Step 3: Combine the conditions
From the conditions derived:
1. \( \lambda < -\frac{6}{7} \)
2. \( \lambda > -4 \)
3. \( \lambda > \frac{3}{2} \)
Since the conditions are contradictory, we conclude that there is no value of `lambda` that satisfies all conditions simultaneously. Thus, the point `(lambda, lambda + 1)` cannot be an interior point of triangle ABC.
