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NCERT Exemplar · Q16

Q.If the origin is the centriod of a triangle ABC having vertices A(a,1,3)A(a,1,3), B(−2,b,−5)B(-2,b,-5) and C(4,7,c)C(4,7,c), find the values of aa, bb, cc.

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The centroid of a triangle is the average of its vertices’ coordinates. Setting the origin (0,0,0)(0,0,0) equal to the centroid gives three simple equations, yielding a=−2a = -2, b=−8b = -8, c=2c = 2.

We are told that the origin (0,0,0)(0,0,0) is the centroid of triangle ABCABC. The centroid of a triangle in 3D (or any dimension) is simply the arithmetic mean of the coordinates of its three vertices. That is the core idea — no fancy geometry, just averaging.

If you ever forget the centroid formula, think of it as the “balance point”: if equal masses are placed at the vertices, the centre of mass is the average position. That’s exactly what we use here.

Let’s write the vertices:

A(a,1,3),B(−2,b,−5),C(4,7,c)A(a,1,3),\quad B(-2,b,-5),\quad C(4,7,c)

The centroid GG has coordinates:

G=(xA+xB+xC3,  yA+yB+yC3,  zA+zB+zC3)G = \left( \frac{x_A + x_B + x_C}{3},\; \frac{y_A + y_B + y_C}{3},\; \frac{z_A + z_B + z_C}{3} \right)

We are given that G=(0,0,0)G = (0,0,0). So each coordinate average must equal zero.

  1. For the xx-coordinate:

a+(−2)+43=0\frac{a + (-2) + 4}{3} = 0

Multiply by 3:

a−2+4=0⇒a+2=0a - 2 + 4 = 0 \quad\Rightarrow\quad a + 2 = 0

Hence:

a=−2a = -2

  1. For the yy-coordinate:

1+b+73=0\frac{1 + b + 7}{3} = 0

Multiply by 3:

1+b+7=0⇒b+8=01 + b + 7 = 0 \quad\Rightarrow\quad b + 8 = 0

Hence:

b=−8b = -8

  1. For the zz-coordinate:

3+(−5)+c3=0\frac{3 + (-5) + c}{3} = 0

Multiply by 3:

3−5+c=0⇒c−2=0…3 - 5 + c = 0 \quad\Rightarrow\quad c - 2 = 0 …

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