Q96 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Mathematics
Let the centroid of a triangle formed by the points $A(4, x, 1), B(y,-5,2)$ and $C(7,8,3)$ be $G(3,5,2)$ and $C G$ meet $A B$ in $F$. Then, $F=$
- A$\left(\frac{5}{2}, \frac{3}{2}, \frac{5}{2}\right)$
- B$\left(\frac{11}{2}, 10,2\right)$
- Correct answerC$\left(1, \frac{7}{2}, \frac{3}{2}\right)$
- D$(10,12,5)$
Solution
Coordinate of
$$
G=\left(\frac{4+y+7}{3}, \frac{x-5+8}{3}, \frac{1+2+3}{3}\right)
$$

$$ \begin{aligned} & \Rightarrow(3,5,2)=\left(\frac{4+y+7}{3}, \frac{x+3}{3}, 2\right) \\ & \Rightarrow \frac{x+3}{3}=5 \Rightarrow x=12 \\ & \Rightarrow \frac{11+y}{3}=3 \Rightarrow y=-2 \end{aligned} $$ $F$ is mid-point of $A B$ $$ \begin{aligned} & =F\left(\frac{4+y}{2}, \frac{x-5}{2}, \frac{1+2}{2}\right) \\ & =F\left(\frac{4-2}{2}, \frac{12-5}{2}, \frac{3}{2}\right)=F\left(1, \frac{7}{2}, \frac{3}{2}\right) . \end{aligned} $$

$$ \begin{aligned} & \Rightarrow(3,5,2)=\left(\frac{4+y+7}{3}, \frac{x+3}{3}, 2\right) \\ & \Rightarrow \frac{x+3}{3}=5 \Rightarrow x=12 \\ & \Rightarrow \frac{11+y}{3}=3 \Rightarrow y=-2 \end{aligned} $$ $F$ is mid-point of $A B$ $$ \begin{aligned} & =F\left(\frac{4+y}{2}, \frac{x-5}{2}, \frac{1+2}{2}\right) \\ & =F\left(\frac{4-2}{2}, \frac{12-5}{2}, \frac{3}{2}\right)=F\left(1, \frac{7}{2}, \frac{3}{2}\right) . \end{aligned} $$
Hence, the correct answer is option: 'C'.