Question

$A(3,2,-1),\, B(4,1,1),\, C(6,2,5)$ and $D(3,3,3)$ are four points. $G_{1},\, G_{2},\, G_{3}$ and $G_{4}$ respectively are the centroids of the triangles $\Delta B C D, \Delta C D \Lambda$, $\Delta D A B$ and $\Delta A B C .$ The point of concurrence of the lines $A G_{1},\, B G_{2},\, C G_{3}$ and $D G_{4}$ is

• A. (4,2,2)
• B. (2,4,2)
• C. (2,2,4)
• D. (2,2,2)

Answer 1

Answer: (A)

Given points,
$A(3,2,-1), B(4,1,1), C(6,2,5)$ and $D(3,3,3)$,
So, $G_{1}$ is centroid of triangle $BCD ,$
$G_{1} \equiv\left(\frac{13}{3}, \frac{6}{3}, \frac{9}{3}\right)$
$G_{2}$ is centroid of triangle $CDA$
$G_{2} \equiv\left(\frac{12}{3}, \frac{7}{3}, \frac{7}{3}\right)$
$\because$ The line $A G_{1}, B G_{2}, C G_{3}$ and $D G_{4}$ are concurrent, so point of concurrence of these four lines is point of intersection of lines $A G_{1}$ and $B G_{2}$.
Equation of line $A G_{1}$ is
$\frac{x-3}{4 / 3}=\frac{y-2}{0}=\frac{z+1}{\frac{12}{3}}=r_{1}$ (let)
So, point on line this $A G_{1}$ is
$\left(3+\frac{4}{3} r_{1}, 2,-1+\frac{12}{3} r_{1}\right)$
and equation of line $B G_{2}$ is
$\frac{x-4}{0}=\frac{y-1}{4 / 3}=\frac{z-1}{4 / 3}=r_{2} ($ let $) $
So, point on line $B G_{2}$ is
$\left(4,1+\frac{4}{3} r_{2}, 1+\frac{4}{3} r_{2}\right)$
Let the above point is the point of intersection,
so $3+\frac{4}{3} r_{1}=4$
$\Rightarrow 2=1+\frac{4}{3} r_{2}$
and $-1+\frac{12}{3} r_{1}=1+\frac{4}{3} r_{2}$,
from these we are getting
$r_{1}=\frac{3}{4}$ and $r_{2}=\frac{3}{4}$
So, required point of concurrence is $(4,2,2)$.