Question

If three vertices of a triangle are(2,0), (6,3) , (-4,5), find the coordinates of the centroid of the triangle​

Answer 1

GENERALLY:-)

We know that centroid(x,y)of the Triangle formed by joining three points

A(x1, y2), B(x2, y2) and C(x3, y3)

is given by

$x = \dfrac{x_1 + x_2 + x_3}{3} \\ and \\ y = \dfrac{y_1 + y_2 + y_3}{3}$

So,

ln the given question coordinates of vertex

of the triangle are (2, 0) , (6, 3) , (-4,5)

coordinates of centroid will be given by

$x = \frac{2 + 6 - 4}{3} \\ \\ x = \frac{4}{3} \\ and \\ \\ y = \frac{0 + 3 + 5}{3} \\ \\ y = \frac{8}{3}$

So,

centroid of the triangle is given by

$centroid = ( \dfrac{4}{3} , \dfrac{8}{3} )$

Answer 2

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

Coordinates of the centroid :

$\sf{(x,y)\:=\:{\dfrac{4}{3}}},{\dfrac{8}{3}}$

$\bold{\underline{\underline{\large{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• Three Vertices Of Triangle :

1. A ( 2,0 )
2. B ( 6,3 )
3. C ( -4,5 )

To FiNd :

• The coordinates of the centroid of the triangle

SoLuTiOn :

Let's consider Δ ABC.

Let the coordinates of the centroid of the Δ ABC be (x, y)

Where,

• Point of vertice A (2,0)

1. Let $\sf{x_1}$ = 2
2. Let $\sf{y_1}$ = 0

• Point of vertice B (6,3)

1. Let $\sf{x_2}$ = 6
2. Let $\sf{y_2}$ = 3

• Point of vertice C (-4,5)

1. Let $\sf{x_3}$ = -4
2. Let $\sf{y_3}$ = 5

FoRmUlA :

$\sf{\large{\boxed{\red{\sf{(x,y)\:=\:{\dfrac{x_1\:+\:x_2\:+\:x_3\:}{3}},{\dfrac{y_1\:+\:y_2\:+\:y_3}{3}}}}}}}$

Block in the values,

$\longrightarrow$$\sf{(x,y)\:=\:{\dfrac{2\:+\:6\:+\:(-4)}{3}}},{\dfrac{0\:+\:3\:+\:5}{3}}$

$\longrightarrow$$\sf{(x,y)\:=\:{\dfrac{8\:+(-4)}{3}}},{\dfrac{3\:+\:5}{3}}$

$\longrightarrow$$\sf{(x,y)\:=\:{\dfrac{4}{3}}},{\dfrac{8}{3}}$

•°• Coordinates of centroid of Δ ABC,

• x = $\sf{\dfrac{4}{3}}$
• y = $\sf{\dfrac{8}{3}}$