To find the coordinates of point A in triangle ABC given the circumcenter S and the midpoints E and F, we can follow these steps:
### Step 1: Understand the given points
We have:
- Circumcenter \( S(-1, 0) \)
- Midpoint of \( AB \) \( E(1, -2) \)
- Midpoint of \( AC \) \( F(-2, -1) \)
Let the coordinates of point A be \( A(x_1, y_1) \), point B be \( B(x_2, y_2) \), and point C be \( C(x_3, y_3) \).
### Step 2: Use the midpoint formula for point E
Since \( E \) is the midpoint of \( AB \):
\[
E = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (1, -2)
\]
From this, we can set up the equations:
1. \( \frac{x_1 + x_2}{2} = 1 \) → \( x_1 + x_2 = 2 \) → \( x_2 = 2 - x_1 \) (Equation 1)
2. \( \frac{y_1 + y_2}{2} = -2 \) → \( y_1 + y_2 = -4 \) → \( y_2 = -4 - y_1 \) (Equation 2)
### Step 3: Use the midpoint formula for point F
Since \( F \) is the midpoint of \( AC \):
\[
F = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (-2, -1)
\]
From this, we can set up the equations:
3. \( \frac{x_1 + x_3}{2} = -2 \) → \( x_1 + x_3 = -4 \) → \( x_3 = -4 - x_1 \) (Equation 3)
4. \( \frac{y_1 + y_3}{2} = -1 \) → \( y_1 + y_3 = -2 \) → \( y_3 = -2 - y_1 \) (Equation 4)
### Step 4: Calculate the slopes
The circumcenter \( S \) is the intersection of the perpendicular bisectors of the sides of the triangle. The slopes of the lines can be used to find relationships between the coordinates.
#### Slope of line ES
The slope of line \( ES \) is given by:
\[
\text{slope of } ES = \frac{-2 - 0}{1 - (-1)} = \frac{-2}{2} = -1
\]
#### Slope of line AB
Since \( ES \) is perpendicular to \( AB \):
\[
\text{slope of } AB \cdot \text{slope of } ES = -1 \implies \text{slope of } AB = 1
\]
Thus,
\[
\frac{y_1 + 2}{x_1 - 1} = 1 \implies y_1 + 2 = x_1 - 1 \implies x_1 - y_1 = 3 \quad (Equation 5)
\]
#### Slope of line SF
The slope of line \( SF \) is given by:
\[
\text{slope of } SF = \frac{-1 - 0}{-2 - (-1)} = \frac{-1}{-1} = 1
\]
Since \( SF \) is perpendicular to \( AC \):
\[
\text{slope of } AC \cdot \text{slope of } SF = -1 \implies \text{slope of } AC = -1
\]
Thus,
\[
\frac{y_1 + 1}{x_1 + 2} = -1 \implies y_1 + 1 = -x_1 - 2 \implies x_1 + y_1 = -3 \quad (Equation 6)
\]
### Step 5: Solve the equations
Now we have two equations:
1. \( x_1 - y_1 = 3 \) (Equation 5)
2. \( x_1 + y_1 = -3 \) (Equation 6)
Adding these two equations:
\[
(x_1 - y_1) + (x_1 + y_1) = 3 - 3 \implies 2x_1 = 0 \implies x_1 = 0
\]
Substituting \( x_1 = 0 \) into Equation 6:
\[
0 + y_1 = -3 \implies y_1 = -3
\]
### Final Result
Thus, the coordinates of point A are:
\[
A(0, -3)
\]
