If the plane $56 x+4 y+9 z=2016$ meets the coordinate axes in $A, B$ and $C$, then the centroid of the…
If the plane $56 x+4 y+9 z=2016$ meets the coordinate axes in $A, B$ and $C$, then the centroid of the $\triangle A B C$ is
$(12,168,224)$
$(12,168,112)$
$\left(12,168, \frac{224}{3}\right)$
$\left(12,168, \frac{224}{}\right)$
Solution
The given equation of plane is :
$56 x+4 y+9 z=2016$
$\begin{aligned} & \Rightarrow \frac{56 x}{2016}+\frac{y}{504}+\frac{z}{224}=1 \\ & \Rightarrow \frac{x}{36}+\frac{y}{504}+\frac{z}{224}=1\end{aligned}$
Then co-ordinate of $A, B$ and $C$ are $(36,0,0),(0,504,0)$ and $(0,0,224)$.
Now, the centroid is
$\left(\frac{36+0+0}{3}, \frac{0+504+0}{3}, \frac{0+0+224}{3}\right)$
Centroid $=\left(12,168, \frac{224}{3}\right)$.