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 with the obtuse angle at \( A \), we can follow these steps:
### Step 1: Find the equation of the circle with diameter \( BC \)
The endpoints of the diameter are \( B(-2, 0) \) and \( C(1, 1) \). The equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[
(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0
\]
Substituting the coordinates of points \( B \) and \( C \):
\[
(x + 2)(x - 1) + (y - 0)(y - 1) = 0
\]
Expanding this:
\[
(x + 2)(x - 1) + y(y - 1) = 0
\]
\[
x^2 + x - 2 + y^2 - y = 0
\]
Rearranging gives us:
\[
x^2 + y^2 + x - y - 2 = 0
\]
### Step 2: Determine the condition for point \( A(0, a) \) to be inside the circle
For point \( A(0, a) \) to be inside the circle, we substitute \( x = 0 \) and \( y = a \) into the circle's equation:
\[
0^2 + a^2 + 0 - a - 2 < 0
\]
This simplifies to:
\[
a^2 - a - 2 < 0
\]
### Step 3: Factor the quadratic inequality
We can factor the quadratic:
\[
a^2 - a - 2 = (a - 2)(a + 1)
\]
Now we need to find the intervals where this product is less than zero. The critical points are \( a = -1 \) and \( a = 2 \).
### Step 4: Test intervals
We test the intervals \( (-\infty, -1) \), \( (-1, 2) \), and \( (2, \infty) \):
1. For \( a < -1 \) (e.g., \( a = -2 \)):
\[
(-2 - 2)(-2 + 1) = (-4)(-1) > 0
\]
2. For \( -1 < a < 2 \) (e.g., \( a = 0 \)):
\[
(0 - 2)(0 + 1) = (-2)(1) < 0
\]
3. For \( a > 2 \) (e.g., \( a = 3 \)):
\[
(3 - 2)(3 + 1) = (1)(4) > 0
\]
Thus, the solution to \( a^2 - a - 2 < 0 \) is:
\[
-1 < a < 2
\]
### Step 5: Exclude the line \( BC \)
Next, we need to ensure that point \( A \) does not lie on the line segment \( BC \). We find the equation of line \( BC \) using the two points \( B(-2, 0) \) and \( C(1, 1) \).
The slope \( m \) of line \( BC \) is:
\[
m = \frac{1 - 0}{1 - (-2)} = \frac{1}{3}
\]
Using point-slope form, the equation of the line is:
\[
y - 1 = \frac{1}{3}(x - 1)
\]
Rearranging gives:
\[
y = \frac{1}{3}x + \frac{2}{3}
\]
Substituting \( x = 0 \) to find \( a \):
\[
a = \frac{1}{3}(0) + \frac{2}{3} = \frac{2}{3}
\]
Since \( A(0, a) \) must not lie on this line, we exclude \( a = \frac{2}{3} \) from our solution.
### Final Answer
Thus, the complete set of values of \( a \) such that the triangle is obtuse at \( A \) is:
\[
a \in (-1, 2) \quad \text{and} \quad a \neq \frac{2}{3}
\]
