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Step-by-Step Solution
Step 1: Identify the Known Vertex and Midpoints
We are given one vertex of the triangle as
$ (1,\, 1) $. Call this vertex $A(1, 1)$.
The midpoints of the sides that meet at $A$ are
$(-1, 2)$ and $(3, 2)$.
Let the other two vertices of the triangle be
$B(x_B, y_B)$ and $C(x_C, y_C)$.
Step 2: Apply the Midpoint Formula
The midpoint formula states that the midpoint $M$ of a segment between points
$P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by
$ M\bigl(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\bigr). $
Step 3: Use Midpoint Conditions to Find the Coordinates of B and C
Let $M_1(-1, 2)$ be the midpoint of $AB$. Then
$$
\frac{x_A + x_B}{2} = -1, \quad \frac{y_A + y_B}{2} = 2.
$$
Substituting $A(1,1)$:
$$
\frac{1 + x_B}{2} = -1 \quad \Longrightarrow \quad 1 + x_B = -2 \quad \Longrightarrow \quad x_B = -3,
$$
$$
\frac{1 + y_B}{2} = 2 \quad \Longrightarrow \quad 1 + y_B = 4 \quad \Longrightarrow \quad y_B = 3.
$$
Therefore, $ B(-3,\,3). $
Similarly, let $M_2(3, 2)$ be the midpoint of $AC$. Then
$$
\frac{x_A + x_C}{2} = 3, \quad \frac{y_A + y_C}{2} = 2.
$$
Substituting $A(1,1)$:
$$
\frac{1 + x_C}{2} = 3 \quad \Longrightarrow \quad 1 + x_C = 6 \quad \Longrightarrow \quad x_C = 5,
$$
$$
\frac{1 + y_C}{2} = 2 \quad \Longrightarrow \quad 1 + y_C = 4 \quad \Longrightarrow \quad y_C = 3.
$$
Therefore, $ C(5,\,3). $
Step 4: Find the Centroid
The centroid $G$ of a triangle with vertices
$A(x_A, y_A)$, $B(x_B, y_B)$, and $C(x_C, y_C)$
is given by
$$
G\Bigl(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\Bigr).
$$
Substituting $ A(1,1)$, $B(-3,3)$, and $C(5,3)$:
$$
x_G = \frac{1 + (-3) + 5}{3} = \frac{3}{3} = 1,
$$
$$
y_G = \frac{1 + 3 + 3}{3} = \frac{7}{3}.
$$
Thus the centroid is $ (1, \tfrac{7}{3}) $.
Step 5: Final Answer
Therefore, the centroid of the given triangle is
$ \left(1, \frac{7}{3}\right). $
