Question

Find the centroid of the triangle whose vertices are (3, -5) (4, 3) (11, -4) Solution: Let A (3, -5); B (4, 3); C (11, -4) Suppose (x₁, y₁); (x₂, y₂); (x₃, y₃) Let G (x, y) be the centroid ∴ x= (x₁+x₂+x₃)/3 ∴ y= (y₁+y₂+y₃)/3 ∴x=____ ∴y= _____*​

Answer 1

Answer:-

Given:

Vertices of the triangle are (3 , - 5) , (4 , 3) & (11 , - 4).

We know that,

Centroid of a triangle G(x , y) = [ (x₁ + x₂ + x₃) / 3 , (y₁ + y₂ + y₃) / 3 ]

Let,

• x₁ = 3
• x₂ = 4
• x₃ = 11
• y₁ = - 5
• y₂ = 3
• y₃ = - 4.

So,

⟶ G(x , y) = [ (3 + 4 + 11) / 3 , (- 5 + 3 - 4) / 3 ]

⟶ G(x , y) = [ 18/3 , - 6/3 ]

⟶ G(x , y) = ( 6 , - 2)

∴ The centroid of the triangle is (6 , - 2).

Answer 2

$\huge\underline\sf{Solution:-}$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf (3,-5) }\put(0.5,-0.3){ \bf (11,-4) }\put(5.2,-0.3){ \bf (4,3) }\end{picture}$

$\sf{Vertices \: of \: the \: triangle \: are \: (3,-5), \: (4,3) \: (11, - 4).} \\ \\ </p><p></p><p> \sf{We \: know \: that}$

$\\ \sf{G(x, y) [ \frac{x, + x2 + x3}{3}, \frac{ y + y2 + y3}{3}]}$

$\sf \boxed{X= 3} \\ \\ </p><p></p><p> \sf \boxed{X2 = 4} \\ \\ </p><p></p><p> \sf \boxed{ X3 = 11} \\ \\ </p><p></p><p> \sf \boxed{Y= - 5} \\ \\ </p><p></p><p> \sf \boxed{Y2 = 3} \\ \\ </p><p></p><p> \sf \boxed{Y3 = - 4}$

$\\ \sf{G(x, y) = \frac{(3 + 4 + 11)}{3}, \frac{( - 5 + 3 - 4)}{3}}$

$\\ \sf{G(x , y) = \frac{18}{3} , \frac{ - 6}{3}} \\ \\ </p><p></p><p> \sf{G(x , y) = (6,-2)}$

$\boxed{The \: centroid \: of \: the \: triangle \: is \: (6,-2)}$