If the origin is the centroid of the triangle | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The centroid $(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by: $x = \frac{x_1+x_2+x_3

Vedclass · Accepted Answer

$-2, -\frac{16}{3}, 2$

Vedclass · Answer

(C) The centroid $(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by: $x = \frac{x_1+x_2+x_3}{3}$,$y = \frac{y_1+y_2+y_3}{3}$,$z = \frac{z_1+z_2+z_3}{3}$ Since the origin $(0, 0, 0)$ is the centroid,we have: $0 = \frac{2a - 4 + 8}{3} \Rightarrow 2a + 4 = 0 \Rightarrow a = -2$ $0 = \frac{2 + 3b + 14}{3} \Rightarrow 3b + 16 = 0 \Rightarrow b = -\frac{16}{3}$ $0 = \frac{6 - 10 + 2c}{3} \Rightarrow 2c - 4 = 0 \Rightarrow c = 2$ Thus,the values are $a = -2, b = -\frac{16}{3}, c = 2$.

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 the values of $a, b, c$ respectively are:

Similar Questions

In a triangle $ABC$,if the midpoints of sides $AB, BC, CA$ are $(3,0,0), (0,4,0), (0,0,5)$ respectively,then $AB^2+BC^2+CA^2=$

The perimeter of the triangle with vertices at $(1,0,0), (0,1,0)$ and $(0,0,1)$ is

If $(1,0,3), (2,1,5), (-2,3,6)$ are the mid-points of the sides of a triangle,then the centroid of the triangle is

If $A \equiv (x, 4, -1)$,$B \equiv (3, x, -5)$,and $C \equiv (2, -2, 3)$ are the vertices and $G \equiv (2, 1, -1)$ is the centroid of the triangle $ABC$,then the value of $x$ is

If $D(2, 1, 0)$,$E(2, 0, 0)$,and $F(0, 1, 0)$ are mid-points of the sides $BC$,$CA$,and $AB$ of $\triangle ABC$,respectively,then the centroid of $\triangle ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module