Question

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

• A. $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$
• B. $\frac{x}{2p}+\frac{2y}{q}+\frac{z}{2r}=1$
• 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 equation of the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Then coordinates of $A,B,C$ are $(a,0,0),(0,b,0),(0,0,c).$
So, the centroid of the triangle ABC is $(\frac{a}{3},\frac{b}{3},\frac{c}{3})$
The coordinates of centroid is $(p,q,r)$
$\Rightarrow a=3p,b=3q,c=3r$
So, the equation of the required plane is $\frac{x}{3p}+\frac{y}{3q}+\frac{z}{3r}=1$