Given: The mid-points of the sides of the triangle are P(1, 2, -3), Q(3, 0, 1) and R(-1, 1, -4).
To find: the coordinates of the centroid
Formula used:
Centroid of triangle ABC whose vertices are A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) is given by,
\(\Big(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_2}{3}\Big)\)
Section Formula:
A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).
The coordinates of C is given by,
\(\Big(\frac{nx+ma}{m+n},\frac{ny+mb}{m+n},\frac{nz+mc}{m+n}\Big)\)
We know the mid-point divides side in the ratio of 1:1.
Therefore,
The coordinates of C is given by,
\(\Big(\frac{x+a}{2},\frac{y+a}{2},\frac{z+c}{2}\Big)\)
P(1, 2, -3) is mid-point of A(x1, y1, z1) and B(x2, y2, z2) Therefore,
Centroid of the triangle
\(\Big(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_2}{3}\Big)\)
= \(\Big(\frac{3}{3},\frac{3}{3},\frac{-6}{3}\Big)\)
= (1, 1, -2)
Hence, the centroid of the triangle is (1, 1, -2)
