Let the points`A:(0, a), B:(-2, 0) and C:(1, 1)` form an obtuse angled triangle (obtuse angled at angle A), then the complete set of values of a is

Options

A. `(-2, 1)`
B. `(-2, 1)-{(2)/(3)}`
C. `(-1, 2)`
D. `(-1, 2)={(2)/(3)}`

Answer

To determine the complete set of values of \( a \) such that the points \( A(0, a) \), \( B(-2, 0) \), and \( C(1, 1) \) form an obtuse-angled triangle at angle \( A \), we can follow these steps:

### Step 1: Understand the Condition for Obtuse Angled Triangle
For triangle \( ABC \) to be obtuse at \( A \), the square of the length of side \( BC \) must be greater than the sum of the squares of the lengths of sides \( AB \) and \( AC \):
\[ BC^2 > AB^2 + AC^2 \]

### Step 2: Calculate the Lengths of the Sides
1. **Length of \( AB \)**:
\[ AB = \sqrt{((-2) - 0)^2 + (0 - a)^2} = \sqrt{4 + a^2} \]
Thus,
\[ AB^2 = 4 + a^2 \]

2. **Length of \( AC \)**:
\[ AC = \sqrt{(1 - 0)^2 + (1 - a)^2} = \sqrt{1 + (1 - a)^2} = \sqrt{1 + 1 - 2a + a^2} = \sqrt{2 - 2a + a^2} \]
Thus,
\[ AC^2 = 2 - 2a + a^2 \]

3. **Length of \( BC \)**:
\[ BC = \sqrt{(1 - (-2))^2 + (1 - 0)^2} = \sqrt{(1 + 2)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]
Thus,
\[ BC^2 = 10 \]

### Step 3: Set Up the Inequality
Now we can set up the inequality for the obtuse triangle condition:
\[ 10 > (4 + a^2) + (2 - 2a + a^2) \]
This simplifies to:
\[ 10 > 6 + 2a^2 - 2a \]
\[ 10 - 6 > 2a^2 - 2a \]
\[ 4 > 2a^2 - 2a \]
Dividing through by 2:
\[ 2 > a^2 - a \]
Rearranging gives:
\[ a^2 - a - 2 < 0 \]

### Step 4: Factor the Quadratic
Factoring the quadratic:
\[ (a - 2)(a + 1) < 0 \]

### Step 5: Determine the Intervals
To find the intervals where this inequality holds, we can analyze the sign changes:
- The roots are \( a = -1 \) and \( a = 2 \).
- The intervals to test are \( (-\infty, -1) \), \( (-1, 2) \), and \( (2, \infty) \).

Testing these intervals:
1. For \( a < -1 \) (e.g., \( a = -2 \)): \( (-)(-) > 0 \)
2. For \( -1 < a < 2 \) (e.g., \( a = 0 \)): \( (+)(-) < 0 \)
3. For \( a > 2 \) (e.g., \( a = 3 \)): \( (+)(+) > 0 \)

Thus, the solution to the inequality is:
\[ -1 < a < 2 \]

### Step 6: Exclude Collinear Points
However, we need to ensure that the points are not collinear. The points \( A(0, a) \), \( B(-2, 0) \), and \( C(1, 1) \) are collinear if the area of triangle \( ABC \) is zero. The area can be calculated using the determinant:
\[ \text{Area} = \frac{1}{2} \left| 0(0 - 1) + (-2)(1 - a) + 1(a - 0) \right| = 0 \]
This simplifies to:
\[ \frac{1}{2} \left| -2 + 2a + a \right| = 0 \implies -2 + 3a = 0 \implies a = \frac{2}{3} \]
Thus, we need to exclude \( a = \frac{2}{3} \) from our interval.

### Final Solution
The complete set of values of \( a \) such that triangle \( ABC \) is obtuse at \( A \) is:
\[ a \in (-1, 2) \setminus \left\{ \frac{2}{3} \right\} \]