Question

If two verticies of a triangle ABC are $(1,2,3)$ and $(3,-2,7)$ and medians intersect each other at $(4,-3,6)$ then find the third vertex of the triangle ABC.

Answer 1

Let $(x,y,z)$ be the third vertex of $△ABC$.

The point of intersection of medians of a triangle is known as its centroid.

Hence, from the given data $(4,-3,6)$ is the centroid of $△ABC$.

$(4,-3,6)=(\frac{x+1+3}{3},\frac{y+2-2}{3},\frac{z+3+7}{3})$ ...(from formula of centroid for a triangle)

On equating and solving each part seperately, we get

$\frac{x+1+3}{3}=4⟹x+4=12⟹x=8$,

$\frac{y+2-2}{3}=-3⟹\frac{y}{3}=-3⟹y=-9$ and

$\frac{z+3+7}{3}=6⟹z+10=18⟹z=8$.

Hence, the third vertex of $△ABC$ is $(8,-9,8)$.