Question

Question No. 18

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Question:- Find the Centroid of the triangle whose vertices are x(2,8),y(-2,2)&z(6,-4).

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Answer 1

$\large{\red{\bold{\underline{Given:}}}}$

$\sf \: Coordinates \: of \: the \: vertices \: are: \: x(2,8), \\ \\ \: \sf \: y(-2,2) \: and \: z(6,-4).$

$\large{\green{\bold{\underline{To \: Find:}}}}$

$\sf \: Centroid \: of \: the \: triangle$

$\large{\blue{\bold{\underline{Formula \: Used:}}}} \\ \\ \sf \:Coordinates \: of \:Centroid = \: (\frac{x_{1} + x_{2} + x_{3}}{3} ),( \frac{y_{1} + y_{2} + y_{3}}{3})$

$\large{\red{\bold{\underline{Solution:}}}} \\ \\ \: \sf \: On \: considering \: the \: respective \: coordinates \: as :$

$\sf \: x(2,8) \: \rightarrow \: (x_{1}, y_{1}) \\ \\\sf \: y(-2,2) \: \rightarrow \: (x_{2}, y_{2}) \\ \\ \sf \: z(6,-4) \: \rightarrow \: (x_{3}, y_{3})$

$\large{\pink{\bold{\underline{Now:}}}} \\ \\ \rightarrow \: \sf \: Centroid = ( \frac{2 + ( - 2) + 6}{3}) ,( \frac{8 + 2 + (-4)}{3} ) \\ \\ \rightarrow \sf \: Centroid = ( \frac{2 - 2 + 6}{3}) , (\frac{8 + 2 - 4}{3} ) \\ \\ \rightarrow \sf \: Centroid = ( \frac{6}{3} ),( \frac{10 - 4}{3} ) \\ \\ \rightarrow \sf \: Centroid = ( \frac{\cancel6}{\cancel3} ),( \frac{\cancel6}{\cancel3} ) \\ \\ \rightarrow \sf \: Centroid = (2,2)$

$\large{\orange{\bold{\underline{Therefore:}}}} \\ \\ \sf \: The \: coordinates \: of \: Centroid \: of \: the \: triangle \\ \sf \: is \: (2,2).$

Answer 2

Answer:

(2,2)

Step-by-step explanation:

Given:

X(2,8)

Y(-2,2)

Z(6,-4)

To find:

Centroid of the triangle

In the above question,

A₁ = 2

A₂=-2

A₃=6

B₁ =8

B₂=2

B₃=-4

According to the variables we have considered, the centroid of the triangle will be equal to:

((A₁ +A₂+A₃)/3,(B₁+B₂+B₃)/3)

so the centroid of the triangle= ((2-2+6)/3,(8+2-4)/3)

Centroid = ((0+6)/3,(10-4)/3)

Centroid = ((6)/3,(6)/3)

Centroid = (2,2)

The centroid of the given triangle will have coordinates which are equal to (2,2)