Discussion Forum

If  a triangle ABC with two vertices A(5,4,6) and B(1,-1,3) has its centroid  at ($\frac{10}{3},2,\frac{11}{3})$ then the third vertex C is 

Answer:

Explanation:

 We have $\triangle$ ABC

 A=(5,4,6). B=(1,-1,3), C=(x,y,z)

 Centroid of $\triangle$ ABC is    $\left(\frac{10}{3},2, \frac{11}{3}\right)$

$\therefore$      $\frac{10}{3} = \frac{5+1+x}{3} \Rightarrow   x=4$

                   $2= \frac{4-1+y}{3} \Rightarrow   y=3$

                 $\frac{11}{3}=  \frac{6+3+z}{3} \Rightarrow  z=2$

 $\therefore$   Vertices  C= (4,3,2)