To find the coordinates of points A, B, and C given the midpoints D(3, -2), E(-3, 1), and F(4, -3) of triangle ABC, we can use the midpoint formula. The midpoint M of a line segment joining points (x1, y1) and (x2, y2) is given by:
\[ M = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right) \]
Let’s denote the coordinates of points A, B, and C as A(x1, y1), B(x2, y2), and C(x3, y3).
### Step 1: Set up equations using the midpoint formula
1. For midpoint D(3, -2) which is the midpoint of BC:
\[
D = \left( \frac{x2 + x3}{2}, \frac{y2 + y3}{2} \right)
\]
This gives us two equations:
\[
\frac{x2 + x3}{2} = 3 \quad \text{(1)}
\]
\[
\frac{y2 + y3}{2} = -2 \quad \text{(2)}
\]
2. For midpoint E(-3, 1) which is the midpoint of AC:
\[
E = \left( \frac{x1 + x3}{2}, \frac{y1 + y3}{2} \right)
\]
This gives us two more equations:
\[
\frac{x1 + x3}{2} = -3 \quad \text{(3)}
\]
\[
\frac{y1 + y3}{2} = 1 \quad \text{(4)}
\]
3. For midpoint F(4, -3) which is the midpoint of AB:
\[
F = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right)
\]
This gives us the last two equations:
\[
\frac{x1 + x2}{2} = 4 \quad \text{(5)}
\]
\[
\frac{y1 + y2}{2} = -3 \quad \text{(6)}
\]
### Step 2: Solve the equations
Now, we will solve these equations step by step.
**From equation (1):**
\[
x2 + x3 = 6 \quad \text{(7)}
\]
**From equation (2):**
\[
y2 + y3 = -4 \quad \text{(8)}
\]
**From equation (3):**
\[
x1 + x3 = -6 \quad \text{(9)}
\]
**From equation (4):**
\[
y1 + y3 = 2 \quad \text{(10)}
\]
**From equation (5):**
\[
x1 + x2 = 8 \quad \text{(11)}
\]
**From equation (6):**
\[
y1 + y2 = -6 \quad \text{(12)}
\]
### Step 3: Substitute and solve for x-coordinates
From equation (7) \(x3 = 6 - x2\). Substitute \(x3\) into equation (9):
\[
x1 + (6 - x2) = -6 \implies x1 - x2 = -12 \implies x1 = x2 - 12 \quad \text{(13)}
\]
Now substitute \(x1\) from (13) into equation (11):
\[
(x2 - 12) + x2 = 8 \implies 2x2 - 12 = 8 \implies 2x2 = 20 \implies x2 = 10
\]
Now substitute \(x2 = 10\) into (13):
\[
x1 = 10 - 12 = -2
\]
Now substitute \(x2 = 10\) into (7):
\[
x3 = 6 - 10 = -4
\]
### Step 4: Solve for y-coordinates
From equation (8) \(y3 = -4 - y2\). Substitute \(y3\) into equation (10):
\[
y1 + (-4 - y2) = 2 \implies y1 - y2 = 6 \implies y1 = y2 + 6 \quad \text{(14)}
\]
Now substitute \(y1\) from (14) into equation (12):
\[
(y2 + 6) + y2 = -6 \implies 2y2 + 6 = -6 \implies 2y2 = -12 \implies y2 = -6
\]
Now substitute \(y2 = -6\) into (14):
\[
y1 = -6 + 6 = 0
\]
Now substitute \(y2 = -6\) into (8):
\[
y3 = -4 - (-6) = 2
\]
### Final coordinates
Thus, the coordinates of points A, B, and C are:
- \( A(-2, 0) \)
- \( B(10, -6) \)
- \( C(-4, 2) \)
### Summary of Results
- A = (-2, 0)
- B = (10, -6)
- C = (-4, 2)
