Question

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

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

Answer 1

Answer: (C)

Centroid of the $△PQR$ 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., $\left(\frac{2a-4+8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3}\right)$
Origin is the centroid i.e., the coordinates of the centroid are $(0,0,0)$. Then,
$\frac{2a-4+8}{3}=0$
$\Rightarrow 2a+4=0\Rightarrow a=-2$

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