Let the circumcentre of `DeltaABC` is `S(-1, 0)` and the midpoints of the sides AB and AC are `E(1, -2)` and `F(-2, -1)` respectively, then the coordinates of A are

Options

A. `(0, -3)`
B. `(0, 3)`
C. `(-3, 0)`
D. `(3, 0)`

Answer

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: Define the coordinates of points
Let the coordinates of point A be \( A(x_1, y_1) \), point B be \( B(x_2, y_2) \), and point C be \( C(x_3, y_3) \). We know:
- Circumcenter \( S(-1, 0) \)
- Midpoint \( E(1, -2) \) of side AB
- Midpoint \( F(-2, -1) \) of side AC

### Step 2: Use the midpoint formula for E
The coordinates of midpoint E can be expressed as:
\[ E = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (1, -2) \]
From this, we can set up two equations:
1. \( \frac{x_1 + x_2}{2} = 1 \) → \( x_1 + x_2 = 2 \) (Equation 1)
2. \( \frac{y_1 + y_2}{2} = -2 \) → \( y_1 + y_2 = -4 \) (Equation 2)

### Step 3: Use the midpoint formula for F
The coordinates of midpoint F can be expressed as:
\[ F = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (-2, -1) \]
From this, we can set up two more equations:
3. \( \frac{x_1 + x_3}{2} = -2 \) → \( x_1 + x_3 = -4 \) (Equation 3)
4. \( \frac{y_1 + y_3}{2} = -1 \) → \( y_1 + y_3 = -2 \) (Equation 4)

### Step 4: Solve the equations
Now we have four equations:
1. \( x_1 + x_2 = 2 \)
2. \( y_1 + y_2 = -4 \)
3. \( x_1 + x_3 = -4 \)
4. \( y_1 + y_3 = -2 \)

From Equation 1, we can express \( x_2 \) in terms of \( x_1 \):
\[ x_2 = 2 - x_1 \]

From Equation 2, we can express \( y_2 \) in terms of \( y_1 \):
\[ y_2 = -4 - y_1 \]

From Equation 3, we can express \( x_3 \) in terms of \( x_1 \):
\[ x_3 = -4 - x_1 \]

From Equation 4, we can express \( y_3 \) in terms of \( y_1 \):
\[ y_3 = -2 - y_1 \]

### Step 5: Find the slopes
The circumcenter S is perpendicular to the line segments connecting the midpoints E and F. Therefore, we can use the slopes to set up the conditions for perpendicularity.

The slope of line segment \( AB \) (from A to B) is given by:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting \( x_2 \) and \( y_2 \):
\[ m_{AB} = \frac{(-4 - y_1) - y_1}{(2 - x_1) - x_1} = \frac{-4 - 2y_1}{2 - 2x_1} \]

The slope of line segment \( ES \) (from E to S) is:
\[ m_{ES} = \frac{0 - (-2)}{-1 - 1} = \frac{2}{-2} = -1 \]

Since the slopes of perpendicular lines multiply to -1:
\[ m_{AB} \cdot m_{ES} = -1 \]

### Step 6: Set up the equation
Setting up the equation:
\[ \left(\frac{-4 - 2y_1}{2 - 2x_1}\right) \cdot (-1) = -1 \]
This simplifies to:
\[ -4 - 2y_1 = 2 - 2x_1 \]
Rearranging gives:
\[ 2x_1 + 2y_1 = 6 \quad \Rightarrow \quad x_1 + y_1 = 3 \quad (Equation 5) \]

### Step 7: Solve Equations 2 and 5
Now we can solve Equations 2 and 5 together:
1. \( y_1 + 2 = -4 \) → \( y_1 = -3 \)
2. Substitute \( y_1 = -3 \) into Equation 5:
\[ x_1 - 3 = 3 \quad \Rightarrow \quad x_1 = 6 \]

### Conclusion
Thus, the coordinates of point A are:
\[ A(6, -3) \]