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 $\Delta BCD,\Delta CD\Lambda$, $\Delta DAB$ and $\Delta ABC.$ 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 $(4,2,2)$.