Show that the centroid of the triangle with vertices A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) is
((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3).

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Solution

Let D be the midpoint of BC. Join A and D.
Then, the coordinates of D are
x2+x3/2, y2+y3/2, z2+z3/2.
Let G be the centroid of triangle ABC. Then, G lies on AD and divides it in the ratio 2:1
∴ The coordinates of G are (Using section formula)
(2.(x2+x3/2)+1.x1)/(2+1), (2.(y2+y3/2)+1.y1)/(2+1), (2.(z2+z3/2)+1.z1)/(2+1)
i.e., (x1+x2+x3/3, y1+y2+y3/3, z1+z2+z3/3)

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Similar questions

Show that the coordinates of the centroid of the triangle with vertices $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ and $C (x_{3}, y_{3}, z_{3})$ is $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$ .

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Q. If

x_{1} = 3 y_{1} + 2 y_{2} - y_{3}, y_{1} = z_{1} - z_{2} + z_{3} x_{2} = - y_{1} + 4 y_{2} + 5 y_{3}, y_{2} = z_{2} + 3 z_{3} x_{3} = y_{1} - y_{2} + 3 y_{3}, y_{3} = 2 z_{1} + z_{2}

express

x_{1},

in terms of

z_{1}, z_{2}, z_{3}

.
we get

x_{1} = z_{1} - 2 z_{2} + k z_{3}

, what is the value of k?

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Show that the centroid of the triangle with vertices A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) is ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3).

Show that the centroid of the triangle with vertices A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) is
((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3).