Question

The plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$meets the co-ordinate axes in $A,B,C$. The centroid of the triangle ABC is

• A. $\left( \frac{a}{3},\frac{b}{3},\frac{c}{3} \right)$
• B. $\left( \frac{3}{a},\frac{3}{b},\frac{3}{c} \right)$
• C. $\left( \frac{1}{a},\frac{1}{b},\frac{1}{c} \right)$
• D. $(a,b,c)$

Answer 1

Answer: (D)

$(a,b,c)$

Answer 2

Obviously, co-ordinates of A, B, C are respectively $( 3 a , 0,0 )$ $( 0,3 b , 0 )$ and $( 0,0,3 c )$

Hence centroid is $( a , b , c )$