Question

The equation of the plane which meets the axes in $A,B$ and $C$ such that the centroid of the $△ABC$ is $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ is given by

• A. $x+y+z=1$
• B. $x+y+z=2$
• C. $\frac{x}{3}+\frac{y}{3}+\frac{z}{3}=3$
• D. $x+y+z=\frac{1}{3}$

Answer 1

The correct option is A $x+y+z=1$

Let the equation of plane be

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ .......... $\left(i\right)$

Since, the plane meets the coordinate axes at $A$, $B$ and $C$.
$\therefore$ The coordinates of these points are $A(a,0,0)$, $B(0,b,0)$ and $C(0,0,c)$, respectively.

Then, centroid of $△ABC$ is $(\frac{a}{3},\frac{b}{3},\frac{c}{3})$, but it is given $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$
$\therefore \frac{a}{3}=\frac{1}{3}$ , $\frac{b}{3}=\frac{1}{3}$, $\frac{c}{3}=\frac{1}{3}$
$\Rightarrow a=1,b=1,c=1$

On putting the values of $a$, $b$ and $c$ in equation $\left(i\right)$, we get
$\frac{x}{1}+\frac{y}{1}+\frac{z}{1}=1$ $\Rightarrow x+y+z=1$