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NCERT Exemplar · Q11

Q.Find the third vertex of triangle whose centroid is origin and two vertices are (2,4,6)(2,4,6) and (0,−2,−5)(0,-2,-5).

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The centroid of a triangle is the average of its vertices' coordinates. Given the centroid is the origin and two vertices, we find the third vertex by setting up equations for each coordinate, resulting in the third vertex being (−2,−2,−1)\boxed{(-2, -2, -1)}.

In coordinate geometry, the centroid of a triangle is a fundamental point. It's the point where the three medians of the triangle intersect. A median connects a vertex to the midpoint of the opposite side. The beauty of the centroid is that its coordinates are simply the arithmetic mean (average) of the coordinates of the three vertices. This holds true whether the triangle is in 2D (with xx and yy coordinates) or in 3D (with x,y,x, y, and zz coordinates).

The intuition behind this is that the centroid represents the "center of mass" of the triangle if its mass were uniformly distributed along its perimeter or area. Averaging the coordinates effectively finds this central balancing point.

Let's apply this concept to find the missing vertex.

  1. Identify the given information:

    We are given:

    • The centroid of the triangle, G=(0,0,0)G = (0,0,0) (the origin).
    • Two vertices, A=(2,4,6)A = (2,4,6) and B=(0,−2,−5)B = (0,-2,-5).
    • We need to find the third vertex, let's call it C=(x3,y3,z3)C = (x_3, y_3, z_3).
  2. Recall the Centroid Formula for 3D:

    If the three vertices of a triangle 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 coordinates of its centroid G=(xG,yG,zG)G=(x_G, y_G, z_G) are given by:

    xG=x1+x2+x33x_G = \frac{x_1+x_2+x_3}{3}

    yG=y1+y2+y33y_G = \frac{y_1+y_2+y_3}{3}

    zG=z1+z2+z33z_G = \frac{z_1+z_2+z_3}{3}

  3. Substitute the known values into the formula:

    We have x1=2,y1=4,z1=6x_1=2, y_1=4, z_1=6 for vertex AA.

    We have x2=0,y2=−2,z2=−5x_2=0, y_2=-2, z_2=-5 for vertex BB.

    We have xG=0,yG=0,zG=0x_G=0, y_G=0, z_G=0 for the centroid GG.

    We need to find x3,y3,z3x_3, y_3, z_3.

    For the xx-coordinate:

    0=2+0+x330 = \frac{2+0+x_3}{3} …

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