Let the triangle have vertices $A(1,1)$, $B(x_1,y_1)$ and $C(x_2,y_2)$.

The midpoints of the two sides through vertex $A$ are given.

• Midpoint of $AB$ is $(-1,2)$
• Midpoint of $AC$ is $(3,2)$

Use the midpoint formula to find $B$ and $C$.

For $AB$: $\left(\frac{1+x_1}{2},\frac{1+y_1}{2}\right)=(-1,2)$ So, $\frac{1+x_1}{2}=-1 \Rightarrow 1+x_1=-2 \Rightarrow x_1=-3$ $\frac{1+y_1}{2}=2 \Rightarrow 1+y_1=4 \Rightarrow y_1=3$ Hence, $B=(-3,3)$

For $AC$: $\left(\frac{1+x_2}{2},\frac{1+y_2}{2}\right)=(3,2)$ So, $\frac{1+x_2}{2}=3 \Rightarrow 1+x_2=6 \Rightarrow x_2=5$ $\frac{1+y_2}{2}=2 \Rightarrow 1+y_2=4 \Rightarrow y_2=3$ Hence, $C=(5,3)$

Now find the centroid $G$ of triangle $ABC$.

The centroid is $G=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right)$

Substituting $A(1,1)$, $B(-3,3)$, $C(5,3)$: $G=\left(\frac{1+(-3)+5}{3},\frac{1+3+3}{3}\right)$ $G=\left(\frac{3}{3},\frac{7}{3}\right)$ $G=\left(1,\frac{7}{3}\right)$

Compare with the options:

• A: $\left(-1,\frac{7}{3}\right)$
• B: $\left(-\frac{1}{3},\frac{7}{3}\right)$
• C: $\left(1,\frac{7}{3}\right)$
• D: $\left(\frac{1}{3},\frac{7}{3}\right)$

Therefore, the correct option is C.