Question

find the centroid of the triangle whose vertices are a (6,1) b (8,3) and (10,-5)​

Answer 1

Answer:-

Given:

Vertices of a triangle are (6 , 1) , (8 , 3) & (10 , - 5)

We know that,

Centroid of a triangle with vertices (x₁ , y₁) , (x₂ , y₂) and (x₃ , y₃) is :

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

Let,

• x₁ = 6

• x₂ = 8

• x₃ = 10

• y₁ = 1

• y₂ = 3

• y₃ = - 5

According to the question,

$\implies \sf \: g(x \: \: , \: \: y) = \bigg( \frac{6 + 8 + 10}{3} \: \: , \: \: \frac{1 + 3 - 5}{3} \bigg) \\ \\ \\ \implies \sf \: g(x \: \: ,\: \: y) = \bigg( \frac{24}{3} \: \: , \: \: \frac{ - 1}{3} \bigg) \\ \\ \\ \implies \boxed{\sf \: g(x \: \:, \: \: y) = \bigg(8 \: \: , \: \: \frac{ - 1}{3} \bigg) }$

∴ The centroid of the given triangle is ( 8 , - 1/3).

Answer 2

Answer:

$\huge \bf \: solution$

As we are knowing that centroid of triangle whose vertices are a (6,1) (8,3) and (10,-5).

Now,

We will apply the Formula

$\sf \: g \: (x \: and \: y) = (\frac{x1 + x2 + x3}{3} \frac{y1 + y2 + y3} {3})$

Now,

x₁ = 6

x₂ = 8

x₃ = 10

y₁ = 1

y₂ = 3

y₃ = -5

Putting values

$\frac{6 + 8 + 10}{3}$

$\frac{1 + 3 - 5}{3}$

$(\frac{24}{3} \: , \frac{ - 1}{3} )$

$\huge \bf \: centroid \: = 8 \: and \frac{ - 1}{3}$