Math, asked by seemakakanda, 9 months ago

find the centroid of triangle if vertices are (2,1) (3,7)( 8,4 )​

Answers

Answered by ravibharathi22
2

Answer:

(13/3, 4)

Step-by-step explanation:

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Answered by MisterIncredible
28

Given : -

The vertices of the triangle are (2,1) , (3,7) & (8,4)

Required to find : -

  • Centroid of a triangle

Concept used : -

Before ! solving this question we need to know an important concept related to the centroid of a triangle .

How to find the co-ordinates of the centroid of a triangle .

Consider a traingle ABC such that the co-ordinates of the vertices of the triangle A = (x1,y1) , B = (x2,y2) & C = (x3,y3) .

A centroid is a point where all the 3 altitudes bisect each other .

Since, the vertices of a triangle are 3 . so, the no. of altitudes are also 3 .

A centroid divides the altitude of a triangle in the ratio of 1 : 2 .

If we want to find the centroid of a triangle then we need to draw the corresponding altitudes . where all the 3 altitudes bisect that point is termed as the centroid .

According to physics , it will be called as the centre of gravity .

So,

The formula to find the co-ordinates of centroid of a triangle is ;

\sf{ G =  \left( \dfrac{ x_1 + x_2 + x_3}{ 3 } , \dfrac{ y_1 + y_2 + y_3 }{ 3} \right) }

Here, G is the centroid of the traingle .

Using this concept let's crack that question .

Solution : -

The vertices of the triangle are (2,1) , (3,7) & (8,4) .

We need to find the co-ordinates of the centroid .

So,

Let the vertices of the traingle be ;

  • A = (2,1)

  • B = (3,7)

  • C = (8,4)

This implies ;

x1 = 2 , y1 = 1

x2 = 3 , y2 = 7

x3 = 8 , y3 = 4

Now,

Using the formula let's find the co-ordinates of the centroid .

\sf{ G =  \left( \dfrac{ x_1 + x_2 + x_3}{ 3 } , \dfrac{ y_1 + y_2 + y_3 }{ 3} \right) }

Substituting the values in it

\sf \longrightarrow G =\left( \dfrac{ 2 + 3 + 8}{3} , \dfrac{1 + 7 + 4}{3} \right) \\  \\ \sf \longrightarrow G = \left( \dfrac{13}{3} , \dfrac{12}{3}\right)  \\  \\ \sf \longrightarrow G = \left( \dfrac{13}{3} , 4 \right) \\  \\  \\ \therefore \tt Co-ordinates \; of \;  the \; centroid = \left( \dfrac{13}{3} , 4 \right)

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