Question

If the origin is the centroid of the $\triangle P Q R$ with vertices $P(2 a, 2,6), Q(-4,3 b,-10)$ and $R(8,14,2 c)$, then

• A. $a=2$
• B. $b=\frac{16}{3}$
• C. $C=2$
• D. $a+b=-4$

Answer 1

Answer: (C)

Centroid of the $\triangle P Q R$ is
$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$
i.e., $\left(\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right)$
Origin is the centroid i.e., the coordinates of the centroid are $(0,0,0)$. Then,
$\frac{2 a-4+8}{3}=0$
$\Rightarrow 2 a+4=0 \Rightarrow a=-2$

Also, $ \frac{2+3 b+14}{3}=0 \Rightarrow 3 b+16=0$
$\Rightarrow b=-\frac{16}{3}$
and $\frac{6-10 +2c}{3} = 0$
$\Rightarrow 2 c-4=0 \Rightarrow c=2 $
$ \therefore a=-2, b=\frac{-16}{3}, c=2$