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

• A

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

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

  $\left(\frac{1}{3}, \frac{7}{3}\right)$
• Correct answer
  D

  $\left(1, \frac{7}{3}\right)$

Solution

$\begin{aligned} & \text { Midpoint of } \mathrm{AB}=(-1,2) \\ & \Rightarrow\left(\frac{\mathrm{x}_2+1}{2}, \frac{\mathrm{y}_2+1}{2}\right)=(-1,2)\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{x_2+1}{2}=-1 ; \frac{y_2+1}{2}=2 \\ & \Rightarrow x_2+1=-2 ; y_2+1=4 \\ & \Rightarrow x_2=-3, y_2=3 \end{aligned}$ Midpoint of $\hat{A} \mathrm{C}=(3,2)$ $\begin{aligned} & \Rightarrow\left(\frac{x_1+1}{2}, \frac{y_1+1}{2}\right)=(3,2) \\ & \Rightarrow x_1+1=6 ; y_1+1=4 \\ & \Rightarrow x_1=5, y_1=3 \end{aligned}$ So, vertices of triangle ABC are $(1,1),(-3,3),(5,3)$ So, centroid $=\left(\frac{1-3+5}{3}, \frac{1+3+3}{3}\right)=\left(\frac{3}{3}, \frac{7}{3}\right)=\left(1, \frac{7}{3}\right)$

Hence, the correct answer is option: 'D'.