Question

If the plane meets the coordinate axes in $A,B,C$ such that the centroid of $△ABC$ is $(p,q,r)$ then the equation of the plane is:

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

Answer 1

The correct option is C $\frac{x}{3p}+\frac{y}{3q}+\frac{z}{3r}=1$

Let the coordinates points $A,B,C$ are $\left(\right)a,0,0),(0,b,0),(0,0,c)$ respectively.

then,

Coordinates of centroid of $\Delta ABC$

$=(\frac{a}{3},\frac{b}{3},\frac{c}{3})$

$(\frac{a}{3},\frac{b}{3},\frac{c}{3})=(p,q,r)$

$a=3p,b=3q,c=3r$

therefore,

equation of the plane is

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

$\frac{x}{3p}+\frac{y}{3q}+\frac{z}{3r}=1$