The midpoints of the sides of triangle $\mathrm{ABC}$ are the points $\mathrm{D}, \mathrm{E},$ and $\mathrm{F}.$ Given that the coordinates of $\mathrm{D}, \mathrm{E},$ and $\mathrm{F}$ are $(5, 1), (6, 5),$ and $(0, 3),$ respectively, find the coordinates of $\mathrm{A}, \mathrm{B},$ and $\mathrm{C}.$

Let $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$.

Given, mid point of BC, CA and AB are D(5, 1), E(6, 5) and F(0, 3) respectively.

$$\text{Now, } BC = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (5, 1)$$

$$E(6, 5) \text{ of } CA \Rightarrow \frac{(x_3 + x_1)}{2} = 6,$$

$$\frac{(y_3 + y_1)}{2} = 5$$

$$F(0, 3) \text{ of } AB \Rightarrow \frac{(x_1 + x_2)}{2} = 0,$$

$$\frac{(y_1 + y_2)}{2} = 3$$

Solve for x-coordinate

$$x_2 + x_3 = 10, x_3 + x_1 = 12, x_1 + x_2 = 0$$

Add first two: $x_1 + x_2 + 2x_3 = 22$

Using $x_1 + x_2 = 0$: $2x_3 = 22 \Rightarrow x_3 = 11$

Then:

$$x_2 = 10 - 11 = -1$$

$$x_1 = 0 - (-1) = 1$$

Solve for y-coordinate

$$y_2 + y_3 = 2, y_3 + y_1 = 10, y_1 + y_2 = 6$$

Add first two: $y_1 + y_2 + 2y_3 = 12$

Using $y_1 + y_2 = 6$: $2y_3 = 6 \Rightarrow y_3 = 3$

Then:

$$y_2 = 2 - 3 = -1$$

$$y_1 = 6 - (-1) = 7$$

Hence, coordinates of A, B and C are (1, 7), (-1, -1) and (11, 3) respectively.

Use the standard convention: D = midpoint of BC, E = midpoint of CA, F = midpoint of AB. The three midpoint relations give:

$$B + C = 2D = (10, 2) \quad \text{—(i)}$$

$$\begin{aligned} C + A &= 2E = (12, 10) & -(ii) \\ A + B &= 2F = (0, 6) & -(iii) \end{aligned}$$

Add all three: $2(A + B + C) = (22, 18)$ , so $A + B + C = (11, 9)$ .

Subtract each equation from this total:

$$\begin{aligned} A &= (A + B + C) - (B + C) = (11, 9) - (10, 2) \\ &= (1, 7). \end{aligned}$$

$$\begin{aligned} B &= (A + B + C) - (C + A) = (11, 9) - (12, 10) \\ &= (-1, -1). \end{aligned}$$

$$\begin{aligned} C &= (A + B + C) - (A + B) = (11, 9) - (0, 6) \\ &= (11, 3). \end{aligned}$$