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.

Asked by Pragya Singh | 11 months ago |  185

1 Answer

Solution :-

The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).

Where,

x1 = 2a, y1 = 2, z1 = 6;

x2 = -4, y2 = 3b, z2 = -10;

x3 = 8, y3 = 14, z3 = 2c

We know that the coordinates of the centroid of the triangle, whose vertices are 

(x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), are 

\( \dfrac{(x_1+x_2+x_3)}{3},\dfrac{(y_1+y_2+y_3)}{3},\)

\( \dfrac{(z_1+z_2+z_3)}{3}\)

So, the coordinates of the centroid of the triangle PQR are

For triangle PQR , the coordinates will be,

ΔPQR = \( \dfrac{2a-4+8}{3}=\dfrac{2+3b+14}{3}=\dfrac{6-10+2c}{3}\)

ΔPQR = \( \dfrac{2a+4}{3}=\dfrac{3b+16}{3}=\dfrac{2c-4}{3}\)

Now, we are given that centroid is the origin,

2a + 4 = 0, 3b + 16 = 0, 2c – 4 = 0

a = -2, b = \( \dfrac{-16}{3}\), c = 2

The values of a, b and c are a = -2, 

b = \( -\dfrac{16}{3}\), c = 2

Answered by Pragya Singh | 11 months ago

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