Question

A plane meets the coordinate axes at points $A,B,C$ such that the centroid of the $△ABC$ is $(\alpha ,\beta ,\gamma )$. Show that the equation of the plane is $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$ .

Answer 1

Let the eqation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Then the coordinates of the points $A,B$ and $C$ are $(a,0,0),(0,b,0)$ and $(0,0,c)$ respectively.

Now, the centriod of the $△ABC$ is
$(\frac{x_1+y_1+z_1}{3},\frac{x_2+y_2+z_2}{3},\frac{x_3+y_3+z_3}{3})=(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3})$
$=(\frac{a}{3},\frac{b}{3},\frac{c}{3})$

But, the centriod of the $△ABC$ is $(\alpha ,\beta ,\gamma )$
On comparing both the coordinates we have
$a=3\alpha ,b=3\beta ,c=3\gamma$

Putting the values of $a,b$ and $c$ in the eqation of plane, we get
$\frac{x}{3\alpha }+\frac{y}{3\beta }+\frac{z}{3\gamma }=1$
$\Rightarrow \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3$ (proved)