Math, asked by sunitayarabati02, 11 months ago

A(5, 8) and D(2, -1) is mid point of BC. Find centroid of triangle ABC

Answers

Answered by Anonymous
7

Solution :-

First, Draw a rough figure

[ Refer to attachment ]

D(2, - 1) is the mid point of BC

Let the coordinates be B(x1, y1), C(x2,y2)

Using mid point formula

M(x,y) =  \bigg( \dfrac{x_1 + x_2 }{2} ,\dfrac{y_1 + y_2 }{2} \bigg)

Substituting the Coordinates of D

 \implies D(2, - 1) =  \bigg( \dfrac{x_1 + x_2 }{2} ,\dfrac{y_1 + y_2 }{2} \bigg)

Comparing x coordinates

 \implies 2=  \dfrac{x_1 + x_2 }{2}

 \implies x_1 + x_2  = 4

Comparing y coordinates

 \implies  - 1=  \dfrac{y_1 + y_2 }{2}

 \implies y_1 + y_2  =  - 2

Now, by using centroid formula

G(x,y) =  \bigg( \dfrac{x_1 + x_2  + x_3}{3} ,\dfrac{y_1 + y_2 + y_3 }{3} \bigg)

A(5,8) B(x1,y1) C(x2,y2)

Here,

  • x1 + x2 = 4
  • y1 + y2 = - 2
  • x3 = 5
  • y3 = 8

Substituting the values

 \implies G(x,y) =  \bigg( \dfrac{4  + 5}{3} ,\dfrac{ - 2+ 8}{3} \bigg)

 \implies G(x,y) =  \bigg( \dfrac{9}{3} ,\dfrac{6}{3} \bigg)

 \implies G(x,y) =  (3,2)

Therefore the coordinates of centroid are (3,2)

Attachments:
Answered by Equestriadash
7

Given: A(5, 8) and D(2, -1) are the mid - points through BC.

To find: The centroid of the triangle.

Answer:

(Diagram for reference attached below.)

Mid - point formula:

\tt M(x,\ y)\ =\ \bigg(\dfrac{x_1\ +\ x_2}{2},\ \dfrac{y_1\ +\ y_2}{2}\bigg)

Using this for D(2, -1),

\tt D(2,\ -1)\ =\ \bigg(\dfrac{x_1\ +\ x_2}{2},\ \dfrac{y_1\ +\ y_2}{2}\bigg)\\

Equating the x - coordinates,

\tt 2\ =\ \dfrac{x_1\ +\ x_2}{2}\\\\\\4\ =\ x_1\ +\ x_2

Equating the y - coordinates,

\tt -1\ =\ \dfrac{y_1\ +\ y_2}{2}\\ \\\\-2\ =\ y_1\ +\ y_2

--------------------------------------------------------------------------------------------------

Formula to find the centroid,

\tt P(x,\ y)\ =\ \bigg(\dfrac{x_1\ +\ x_2\ +\ x_3}{3},\ \dfrac{y_1\ +\ y_2\ +\ y_3}{3}\bigg)

We have:

→  \tt x_1\ +\ x_2\ (solved\ above)\ =\ 4

→  \tt y_1\ +\ y_2\ (solved\ above) =\ -2.

→  A(5, 8) - [ \tt x_3\ =\ 5\ and\ y_3\ =\ 8 ]

Using these values,

\tt P(x, y)\ =\ \bigg(\dfrac{4\ +\ 5}{3},\ \dfrac{-2\ +\ 8}{3}\bigg)\\  \\\\\\P(x,\ y)\ =\ \bigg(\dfrac{9}{3},\ \dfrac{6}{3}\bigg)\\\\\\\\P(x,\ y)\ =\ \bigg(3,\ 2\bigg)

Therefore, the centroid of the triangle is (3, 2).

Attachments:
Similar questions