Question

In $△ABC,G$ is the centroid, $D$ is the midpoint of $BC$. If $A=\left(2,3\right)$ and $G=\left(7,5\right)$, then the point $D$ is

• A. $\left(\frac{9}{2},4\right)$
• B. $\left(\frac{19}{2},6\right)$
• C. $\left(\frac{11}{2},\frac{11}{2}\right)$
• D. $\left(8,\frac{13}{2}\right)$

Answer 1

The correct option is B

$\left(\frac{19}{2},6\right)$

Explanation for the correct option.

Find the coordinates of the point $D$.

For the $△ABC$, it is given that the coordinate of $A$ is $\left(2,3\right)$.

Let us assume the coordinate of $B$ and $C$ be $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

Now coordinates of the centroid of a triangle whose coordinates of vertices are $\left(2,3\right),\left(x_1,y_1\right),\left(x_2,y_2\right)$ is given as: $\left(\frac{2+x_1+x_2}{3},\frac{3+y_1+y_2}{3}\right)$.

But it is given the coordinates of the centroid of the triangle is $G\left(7,5\right)$. So comparing the $x$ coordinates:

$$\begin{gathered} \frac{2+x_1+x_2}{3}=7 \\ \Rightarrow 2+x_1+x_2=21 \\ \Rightarrow x_1+x_2=19 \end{gathered}$$

Now compare the $y$ coordinates:

$$\begin{gathered} \frac{3+y_1+y_2}{3}=5 \\ \Rightarrow 3+y_1+y_2=15 \\ \Rightarrow y_1+y_2=12 \end{gathered}$$

Now, it is given that $D$ is the midpoint of $BC$. So midpoint of $B\left(x_1,y_1\right)$ and $C\left(x_2,y_2\right)$ is given as:

$\begin{array}{rcl}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)& =& \left(\frac{19}{2},\frac{12}{2}\right)\left[x_1+x_2=19;y_1+y_2=12\right]\\ & =& \left(\frac{19}{2},6\right)\end{array}$

So the coordinates of point $D$ is $\left(\frac{19}{2},6\right)$.

Hence, the correct option is B.