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NCERT Exemplar · Q49

Q.If the mid-points of the sides of a triangle AB; BC; CA are D(1,2,−3)D(1,2,-3), E(3,0,1)E(3,0,1) and F(−1,1,−4)F(-1,1,-4), then the centriod of the triangle ABC is ________.

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The centroid of a triangle is the average of its vertices, and also the average of the midpoints of its sides. Using the given midpoints, the centroid of triangle ABC is (1,1,−2)\left(1, 1, -2\right).

The key insight here is a beautiful property of centroids in coordinate geometry: the centroid of a triangle coincides with the centroid of the triangle formed by the midpoints of its sides. This means we don't need to find the original vertices at all — we can directly average the given midpoints.

Let me explain why this works. If the vertices of triangle ABC are A(x1,y1,z1)A(x_1, y_1, z_1), B(x2,y2,z2)B(x_2, y_2, z_2), and C(x3,y3,z3)C(x_3, y_3, z_3), then the midpoints are:

  • DD (midpoint of AB): (x1+x22,y1+y22,z1+z22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)
  • EE (midpoint of BC): (x2+x32,y2+y32,z2+z32)\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}, \frac{z_2+z_3}{2}\right)
  • FF (midpoint of CA): (x3+x12,y3+y12,z3+z12)\left(\frac{x_3+x_1}{2}, \frac{y_3+y_1}{2}, \frac{z_3+z_1}{2}\right)

Now, the centroid GG of triangle ABC is (x1+x2+x33,y1+y2+y33,z1+z2+z33)\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right).

If we average the three midpoints, we get:

D+E+F3=(x1+x22+x2+x32+x3+x123,…)=(x1+x2+x33,…)\frac{D+E+F}{3} = \left(\frac{\frac{x_1+x_2}{2} + \frac{x_2+x_3}{2} + \frac{x_3+x_1}{2}}{3}, \ldots\right) = \left(\frac{x_1+x_2+x_3}{3}, \ldots\right)

That's exactly the centroid! So the centroid of the original triangle equals the centroid of the midpoint triangle.

Tip

This is a time-saver: never solve for vertices when midpoints are given. The centroid of the midpoint triangle is the same as the centroid of the original triangle. …

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