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

Open in App

Solution

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

$∴$ Coordinates of centriod of $Δ P Q R$ is

$(\frac{2 a - 4 + 8}{3}, \frac{2 + 3 b + 14}{3}, \frac{6 - 10 + 2 c}{3})$

= $(\frac{2 a + 4}{3}, \frac{3 b + 16}{3}, \frac{2 c - 4}{3})$

But is it given that coordinates of centriod is $(0, 0, 0)$

$∴ \frac{2 a + 4}{3} = 0 \Rightarrow 2 a + 4 = 0 \Rightarrow a = - 2$

$\frac{3 b + 16}{3} = 0 \Rightarrow 3 b + 16 = 0$

$\Rightarrow b = \frac{- 16}{3}$

$\frac{2 c - 4}{3} = 0 \Rightarrow 2 c - 4 = 0 \Rightarrow c = 2.$

Suggest Corrections

Similar questions

Q. If the origin is the centroid of the triangle

P Q R

with vertices

P (2 a, 2, 6), Q (- 4, 3 b, - 10)

and

R (8, 14, 2 c),

then find the values of

a, b

and

c .

If the origin is the centroid of the triangle with vertices P(2a, 2, 6), Q(-4, 3b, -10) and R(8, 14, 2c), find the values of a,b and c. Also, determine the value of $a^{2} + b^{2} - c^{2}$ .

Q. In

△ P Q R

A

and

B

are the mid point of the sides

P Q

and

P R

respectively, then the ratio of area of

(△ G A Q + △ G B R + △ G Q R)

to the area of

△ P Q R

, where

G

is the centriod is

If the vertices of a triangle are A(k,2k), B(-2,6), C(3,1) and its area is 5 sq.units, then the values of k are

Q. If the centriod of a triangle formed by points

(a, b), (b, c)

and

(c, a)

(0, 0)

, then find

a^{3} + b^{3} + c^{3}

Join BYJU'S Learning Program

If the origin is the centriod 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.

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