contestada

Find the coordinates of the centroid of ARST.
10
9
y
10
7
9
R
5
4
-10 9 8 7 6 5 4 3 21
2
3
-3
T
-7
72274p
S
-10
$ 6
7 8 9 10

Find the coordinates of the centroid of ARST 10 9 y 10 7 9 R 5 4 10 9 8 7 6 5 4 3 21 2 3 3 T 7 72274p S 10 6 7 8 9 10 class=

Respuesta :

roohafzah

Answer :

  • (1,1)

Solution :

the centroid of a triangle is given by,

  • C = ((x1+x2+x3)/3,(y1+y2+y3)/3)

here,

  • (x1,y1) = (-4,5)
  • (x2,y2) = (5,1)
  • ((x3,y3) = (2,-3)

plug in,

  • C = ((-4+5+2)/3,(5+1-3)/3)
  • C = (1,1)
semsee45

Answer:

(1, 1)

Step-by-step explanation:

The formula for the coordinates of the centroid of a triangle is:

[tex]\boxed{\begin{array}{l}\underline{\textsf{Centroid of a triangle}}\\\\\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)\\\\\textsf{where $A(x_1, y_1), B(x_2, y_2)$, and $C(x_3, y_3)$ are the vertices of the triangle.}\end{array}}[/tex]

In this case, the coordinates of the vertices of triangle RST are:

  • R(-4, 5)
  • S(5, 1)
  • T(2, -3)

Substitute the coordinates of vertices R, S and T into the centroid formula:

[tex]\textsf{Centroid}=\left(\dfrac{x_R + x_S + x_T}{3}, \dfrac{y_R + y_S + y_T}{3}\right)\\\\\\\textsf{Centroid}=\left(\dfrac{-4 + 5 + 2}{3}, \dfrac{5+1+(-3)}{3}\right)\\\\\\\textsf{Centroid}=\left(\dfrac{3}{3}, \dfrac{3}{3}\right)\\\\\\\textsf{Centroid}=\left(1,1\right)[/tex]

Therefore, the coordinates of the centroid of ΔRST are:

[tex]\LARGE\boxed{\boxed{(1,1)}}[/tex]