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Miscellaneous Exercise · Q3

Q.If the origin is the centroid of the triangle PQRPQR with vertices P(2a,2,6)P(2a, 2, 6), Q(−4,3b,−10)Q(-4, 3b, -10) and R(8,14,2c)R(8, 14, 2c), then find the values of aa, bb and cc.

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The centroid of a triangle is the average of its vertices' coordinates. Setting that average equal to the origin (0,0,0)(0,0,0) gives three simple equations, yielding a=−2a = -2, b=−163b = -\frac{16}{3}, and c=2c = 2.

The centroid of a triangle in 3D works exactly the same way as in 2D — it's the arithmetic mean of the coordinates of the three vertices. If you have points P(x1,y1,z1)P(x_1, y_1, z_1), Q(x2,y2,z2)Q(x_2, y_2, z_2), and R(x3,y3,z3)R(x_3, y_3, z_3), the centroid GG is:

G=(x1+x2+x33,y1+y2+y33,z1+z2+z33)G = \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)

Why does this work? The centroid is the point where the three medians intersect, and it divides each median in the ratio 2:12:1. That ratio leads directly to the average formula. No matter how the triangle is oriented in space, this holds.

Here, we're told the centroid is the origin (0,0,0)(0,0,0). That means each coordinate average must be zero. So we just set up three equations and solve.

  1. x-coordinate equation The x-coordinates are 2a2a, −4-4, and 88. Their average is zero:

2a+(−4)+83=0\frac{2a + (-4) + 8}{3} = 0

Multiply through by 3:

2a+4=02a + 4 = 0

2a=−42a = -4

a=−2a = -2

  1. y-coordinate equation The y-coordinates are 22, 3b3b, and 1414. Their average is zero:

2+3b+143=0\frac{2 + 3b + 14}{3} = 0

Multiply by 3:

3b+16=03b + 16 = 0

3b=−163b = -16

b=−163b = -\frac{16}{3}

  1. z-coordinate equation …

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