If a vertex of a triangle is $$(1, 1)$$ and the mid-points of two sides through this vertex are $$(-1, 2)$$ and $$(3, 2)$$, then the centroid of the triangle is

Let the three vertices of the triangle be $$A(1,1)$$ (given), $$B(x_B,y_B)$$ and $$C(x_C,y_C)$$.

The mid-point formula states that the mid-point $$M$$ of the segment joining $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ is $$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$$.

Given mid-point $$M_1(-1,2)$$ lies on $$AB$$, so

$$\left(\dfrac{x_B+1}{2},\dfrac{y_B+1}{2}\right)=(-1,2).$$

Equating coordinates: $$\dfrac{x_B+1}{2}=-1 \;\Rightarrow\; x_B+1=-2 \;\Rightarrow\; x_B=-3,$$ $$\dfrac{y_B+1}{2}=2 \;\Rightarrow\; y_B+1=4 \;\Rightarrow\; y_B=3.$$

Thus $$B(-3,3).$$

Given mid-point $$M_2(3,2)$$ lies on $$AC$$, so

$$\left(\dfrac{x_C+1}{2},\dfrac{y_C+1}{2}\right)=(3,2).$$

Equating coordinates: $$\dfrac{x_C+1}{2}=3 \;\Rightarrow\; x_C+1=6 \;\Rightarrow\; x_C=5,$$ $$\dfrac{y_C+1}{2}=2 \;\Rightarrow\; y_C+1=4 \;\Rightarrow\; y_C=3.$$

Thus $$C(5,3).$$

The centroid $$G(x_G,y_G)$$ of a triangle with vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ is

$$\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right).$$

Substituting $$A(1,1), B(-3,3), C(5,3):$$

$$x_G=\dfrac{1+(-3)+5}{3}=\dfrac{3}{3}=1,$$ $$y_G=\dfrac{1+3+3}{3}=\dfrac{7}{3}.$$

Therefore, the centroid is $$\left(1,\dfrac{7}{3}\right).$$

Option C which is: $$\left(1, \dfrac{7}{3}\right).$$