Question

find the centroid of the triangle whose vertices are A(-1,0) B(5,-2) C(8,2) ​

Answer 1

Centroid Of A Triangle -

For a triangle ABC with the coordinates being A ( x¹ , y¹ ) , B ( x², y² ) and C ( x³, y³ )

Centroid -

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

Here , the vertices are -

A = ( -1 , 0 )

B = ( 5, -2 )

C = ( 8, 2 )

X¹ = -1

x² = 5

x³ = 8

y¹ = 0

y² = -2

y³ = 2

[ x¹ + x² + x³ ]

=> 12

[ x¹ + x² + x³ ] / 3

=> 4

[ y¹ + y² + y³ ]

=> 0

[ y¹ + y² + y³ ] / 3

=> 0

Thus , the required coordinates of the centroid becomes -

=> ( 4, 0 ) .

This is the answer .

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Answer 2

$\huge{\underline{\pink{QuesTion:}}}$

find the centroid of the triangle whose vertices are A(-1,0) B(5,-2) C(8,2).

$\huge{\underline{\pink{AnswEr:}}}$

GIVEN:

centroid of the triangle whose vertices areA(-1,0) B(5,-2) C(8,2) ..

here,

$x_1$=-1

$y_1$=0

$x_2$=5

$y_2$=-2

$x_3$=8

$y_3$=2

Formula used.

$G =( \frac{(x1 + x2 + x3)}{3}. \frac{(y1 + y2 + y3)}{3} )$

substituting,

x=1/3($x_1$+$x_2$+$x_3$)

=1/3(-1+5+8)

=1/3(12)

=4

y=1/3($y_1$+$y_2$+$y_3$)

=1/3(0-2+2)

=1/3(0)

=0

therefore⭐

centroid of the triangle whose vertices are(4, 0)....