Question

If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.

Answer 1

add all these vertices as
(x1 + x2 + x3/3), (y1 + y2 = y3/3). (z1 + z2 + z3/3)
by putting the values you will get
(2a + 4/3) (3b + 16/3) (2c-4/3)

getting it equal to the origin that is 0
we get a= -2
b=-16/3
c= 2

Answer 2

Answer:

Centroid Formula:

• X = [ x₁ + x₂ + x₃ ] / 3

• Y = [ y₁ + y₂ + y₃ ] / 3

• Z = [ z₁ + z₂ + z₃ ] / 3

Here, the Points are:

P = ( 2a, 2, 6 ) ; Q = ( -4, 3b, -10 ) ; R = ( 8, 14, 2c )

Now we know that the centroid is ( 0,0,0 ) as it is given in the question.

So adding all the x coordinates of P, Q and R we get,

⇒ X = [ 2a - 4 + 8 ] / 3

⇒ X = [ 2a + 4 ] / 3

⇒ 0 = [ 2a + 4 ] / 3   [ Since X = 0 ]

⇒ 0 = 2a + 4

⇒ 2a = -4 ⇒ a = -4/2 = -2

Now adding all the y coordinates we get,

⇒ Y = [ 2 + 3b + 14 ] / 3

⇒ Y = [ 3b + 16 ] / 3

⇒ 0 = 3b + 16   [ Since Y = 0 ]

⇒ 3b = -16 ⇒ b = -16 / 3

Now adding all the z coordinates we get,

⇒ Z = [ 6 - 10 + 2c ] / 3

⇒ Z = [ -4 + 2c ] / 3

⇒ 0 = 2c - 4   [ Since Z = 0 ]

⇒ 2c = 4 ⇒ c = 4/2 = 2

Hence a = -2, b = -16/3, c = 2

Hope it helped !!