Question

A plane meets the co-ordinates axes $A,B,C.$ The centroid of triangle $ABC$ is $(2,r,r^2)$ . If the plane passes through $(2,-5,7).$ Then $r=$

• A. $1$
• B. $-\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $-\frac{7}{2}$

Answer 1

Answer: (A)

$1$

Given centroid $\equiv (2,r,r^2)$

Let equation of plane $\frac{x}{A}+\frac{y}{B}+\frac{z}{C}=1$

$A\equiv (x,0,0),B\equiv (0,y,0),C\equiv (0,0,z)$

$\frac{A}{3}=2,\frac{B}{3}=r,\frac{C}{3}=r^2\to A=6,B=3r,C=3r^2$

Now

$\frac{x}{6}+\frac{y}{3r}+\frac{z}{3r^2}=1$

$=\frac{x}{2}+\frac{y}{r}+\frac{z}{r^2}=3$,

Substitute

$(x,y,z)\equiv (2,-5,7)$

We get

$=1-\frac{5}{r}+\frac{7}{r^2}=3$

$=2r^2+5r-7=0=(r-1)(2r+7)=0$

$r=1,r=\frac{-7}{2}$