Question

A triangle has a vertex at (1,2) and the midpoints of the two sides through it are
(-1,1) and (2,3). Then the centroid of the triangle is
$\begin{align}
  & \left( A \right)\left( \dfrac{1}{3},1 \right) \\
 & \left( B \right)\left( \dfrac{1}{3},2 \right) \\
 & \left( C \right)\left( 1,\dfrac{7}{3} \right) \\
 & \left( D \right)\left( \dfrac{1}{3},\dfrac{5}{3} \right) \\
\end{align}$

Answer 1

Hint: We solve this question by considering the definition of centroid of a triangle. Then we assume that the other two vertices are $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$. Then we apply the formula for the midpoint of two points to the two sides and equate them to (-1,1) and (2,3) respectively. After getting the all the vertices of the triangle we apply the formula for centroid of triangle, $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$ to find the coordinates of centroid of the triangle.

Complete step by step answer:
First let us go through the definition of the centroid of a triangle.
The point of intersection of the medians of a triangle is known as the Centroid of that triangle. For any triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ the centroid of that triangle is given by $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.

We are given that $\left( 1,2 \right)$ is a vertex of the triangle and $\left( -1,1 \right)$ and $\left( 2,3 \right)$ are midpoints of the sides through it.
Let us consider the vertex $\left( 1,2 \right)$ as A and the other two vertices as B and C whose coordinates are $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$.

Now, by looking at the figure we can see that $\left( -1,1 \right)$ is midpoint of $\text{A}\left( 1,2 \right)$ and $\text{B}\left( {{x}_{1}},{{y}_{1}} \right)$.
Now let us consider the formula for midpoint of two points $\left( {{a}_{1}},{{b}_{1}} \right)$ and $\left( {{a}_{2}},{{b}_{2}} \right)$.
$\Rightarrow \left( \dfrac{{{a}_{1}}+{{a}_{2}}}{2},\dfrac{{{b}_{1}}+{{b}_{2}}}{2} \right)$
Using this formula, we can write the midpoint of $\text{A}\left( 1,2 \right)$ and $\text{B}\left( {{x}_{1}},{{y}_{1}} \right)$ as,
$\begin{align}
  & \Rightarrow \left( \dfrac{1+{{x}_{1}}}{2},\dfrac{2+{{y}_{1}}}{2} \right)=\left( -1,1 \right) \\
 & \Rightarrow \left( 1+{{x}_{1}},2+{{y}_{1}} \right)=\left( -2,2 \right) \\
 & \Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( -2-1,2-2 \right) \\
 & \Rightarrow \left( {{x}_{1}},{{y}_{1}} \right)=\left( -3,0 \right) \\
\end{align}$
So, we get the coordinates of B as (-3,0).
Now, let us apply the above formula for midpoint of $\text{A}\left( 1,2 \right)$ and $\text{C}\left( {{x}_{2}},{{y}_{2}} \right)$ and equate it to $\left( 2,3 \right)$, as $\left( 2,3 \right)$ is midpoint of side AC.
$\begin{align}
  & \Rightarrow \left( \dfrac{1+{{x}_{2}}}{2},\dfrac{2+{{y}_{2}}}{2} \right)=\left( 2,3 \right) \\
 & \Rightarrow \left( 1+{{x}_{2}},2+{{y}_{2}} \right)=\left( 4,6 \right) \\
 & \Rightarrow \left( {{x}_{2}},{{y}_{2}} \right)=\left( 4-1,6-2 \right) \\
 & \Rightarrow \left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,4 \right) \\
\end{align}$
So, we get the co-ordinates of C as (3,4).
So, we can say that the co-ordinates of the vertices of a given triangle are (1,2), (-3,0) and (3,4).
Now let us apply the formula of finding the centroid of the triangle to the given triangle.
$\begin{align}
  & \Rightarrow \left( \dfrac{1+\left( -3 \right)+3}{3},\dfrac{2+0+4}{3} \right) \\
 & \Rightarrow \left( \dfrac{1}{3},\dfrac{6}{3} \right) \\
 & \Rightarrow \left( \dfrac{1}{3},2 \right) \\
\end{align}$
So, we get the coordinates of the centroid of a given triangle as $\left( \dfrac{1}{3},2 \right)$.

So, the correct answer is “Option A”.

Note: The possibility of making a mistake while using the formula for the centroid one might confuse it with the circumcentre of the triangle or the orthocentre. One needs to remember that circumcentre is the point equidistant from the vertices of a triangle while centroid is the point of intersection of medians drawn from the vertices of the triangle. So, one needs to remember the difference between them.