The coordinates of the centroid of the triang | vedclass

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

Question

(A) The coordinates of the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are given by the formula:$G = \left(

Vedclass · Accepted Answer

$(1, 4)$

Vedclass · Answer

(A) The coordinates of the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are given by the formula:$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$Given vertices are $(x_1, y_1) = (2, 3)$,$(x_2, y_2) = (4, 7)$,and $(x_3, y_3) = (-3, 2)$.Substituting these values into the formula:$x = \frac{2 + 4 + (-3)}{3} = \frac{3}{3} = 1$$y = \frac{3 + 7 + 2}{3} = \frac{12}{3} = 4$Therefore,the coordinates of the centroid are $(1, 4)$.

The coordinates of the centroid of the triangle with vertices $(2, 3)$,$(4, 7)$,and $(-3, 2)$ are:

Similar Questions

Prove that $(2a, 4a)$, $(2a, 6a)$, and $(2a + \sqrt{3}a, 5a)$ are the vertices of an equilateral triangle.

If the vertices of $\square ABCD$ are $A(1, 3), B(4, 3), C(4, 5)$ and $D(1, 5)$,then $\square ABCD$ is a $\ldots$

The coordinates of the midpoint of the line segment joining the points $(20, 10)$ and $(6, 8)$ are:

Prove that the point $P(7, 5)$ is equidistant from the points $A(2, 4)$ and $B(6, 10)$.

The segment joining $A (-2, 1)$ and $B (7, 8)$ is divided into five congruent segments. Find the coordinates of the third point from $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module