Q83 / 180 · KCET 2007 · Mathematics
$A B C$ is triangle, $G$ is the centroid, $D$ is the mid-point of $B C$. If $A=(2,3)$ and $G=(7,5)$, then the point $D$ is
- A$\left(\frac{9}{2}, 4\right)$
- Correct answerB$\left(\frac{19}{2}, 6\right)$
- C$\left(\frac{11}{2}, \frac{11}{2}\right)$
- D$\left(8, \frac{13}{2}\right)$
Solution
Since, $D$ is the mid-point of $B C$. So, coordinate of $B C$ are $\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$
Given, $G(7,5)$ is the centroid of $\triangle A B C$.
$\therefore \quad 7=\frac{2+x_{2}+x_{3}}{3}$ and $5=\frac{3+y_{2}+y_{3}}{3}$
$\Rightarrow \quad x_{2}+x_{3}=21-2$
and $\quad y_{2}+y_{3}=15-3$
$\begin{array}{ll}\Rightarrow & x_{2}+x_{3}=19 \\ \text { and } & y_{2}+y_{3}=12\end{array}$

$\Rightarrow \quad \frac{x_{2}+x_{3}}{2}=\frac{19}{2}$ and $\quad \frac{y_{2}+y_{3}}{2}=6$ $\therefore$ Coordinate of $D$ are $\left(\frac{19}{2}, 6\right)$.

$\Rightarrow \quad \frac{x_{2}+x_{3}}{2}=\frac{19}{2}$ and $\quad \frac{y_{2}+y_{3}}{2}=6$ $\therefore$ Coordinate of $D$ are $\left(\frac{19}{2}, 6\right)$.
Hence, the correct answer is option: 'B'.