Question

Find the centroid of a triangle, mid-points of whose sides are (1,2,-3), (3,0,1) and (-1,1,-4).

Answer 1

Hint: Start by letting the coordinates of the vertices of the triangle be $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ , $B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ . We know that the midpoint of the line segment joining two points has coordinates equal to half of the sum of the corresponding coordinates of the two points while the coordinate of the centroid is one third of the sum of the corresponding coordinates of the vertices of a triangle. So, use the formulas to and the data given in the question to reach the answer.

Complete step-by-step answer:
Before starting the solution to the above question, let us know about the centroid of a triangle. The centroid of a triangle is defined as the meet point of the medians of the triangles drawn from the three of its vertices. The centroid of a triangle is generally denoted by G.
Now let us start the solution to the above question by letting the coordinates of the vertices of the triangle be $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ , $B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ and drawing a diagram of the situation given in the question for better visualisation. Also, let the centroid of the triangle be G(a,b,c).

We know that the coordinate of the centroid is one third of the sum of the corresponding coordinates of the vertices of a triangle.
$a=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}.........(i)$
$b=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}........(ii)$
$c=\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3}...........(iii)$
Let us first solve the equation (i).
 $a=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}=\dfrac{\dfrac{{{x}_{1}}+{{x}_{2}}}{2}+\dfrac{{{x}_{2}}+{{x}_{3}}}{2}+\dfrac{{{x}_{3}}+{{x}_{1}}}{2}}{3}$
Now, we know that the mid-point of the line segment joining two points has coordinates equal to half of the sum of the corresponding coordinates of the two points. So, $\dfrac{{{x}_{1}}+{{x}_{2}}}{2}$ is the x-coordinate of D, i.e., 1, $\dfrac{{{x}_{2}}+{{x}_{3}}}{2}$ is the x-coordinate of F, i.e., -1 and so on.
 $a=\dfrac{\dfrac{{{x}_{1}}+{{x}_{2}}}{2}+\dfrac{{{x}_{2}}+{{x}_{3}}}{2}+\dfrac{{{x}_{3}}+{{x}_{1}}}{2}}{3}=\dfrac{1-1+3}{3}=1$
Similarly, b and c will also be one third of the coordinates of the midpoints of each side.
$b=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}=\dfrac{\dfrac{{{y}_{1}}+{{y}_{2}}}{2}+\dfrac{{{y}_{2}}+{{y}_{3}}}{2}+\dfrac{{{y}_{3}}+{{y}_{1}}}{2}}{3}=\dfrac{2+0+1}{3}=1$
$c=\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3}=\dfrac{\dfrac{{{z}_{1}}+{{z}_{2}}}{2}+\dfrac{{{z}_{2}}+{{z}_{3}}}{2}+\dfrac{{{z}_{3}}+{{z}_{1}}}{2}}{3}=\dfrac{-3+1-4}{3}=-2$
So, the coordinates of the centroid is $G\left( 1,1,-2 \right)$ .


Note:You could have also used the property that the centroid of a triangle and the centroid of the triangle formed by joining the mid-points of each side is the same. Don’t get confused between the properties of different special points related to triangles like centroid, incentre, circumcentre etc. It is also important that you learn all the properties related to different special points as they are used very often.