Question

(2,0)is centroid of triangle ABC,if (1,3) is midpoint of BC, then A=​

Answer 1

Answer:  A (4,-6)

Step-by-step explanation:

GIVEN THAT [2,0] is the centroid of the triangle.

CASE 1 :

Centroid of the triangle ABC =  G{(x1+x2+x3)/3, (y1+y2+y3)/3}

So, (2,0)={(x1+x2+x3/3),(y1+y2+y3/3)}

      2=x1+x2+x3/3  ,  0= y1+y2+y3/3

        6=x1+x2+x3     ,   0=y1+y2+y3  

        6=x1+x2+x3 ⇒ (i)

        0=y1+y2+y3 ⇒ (ii)

 CASE 2 :

so,Mid point of BC = { x1+x2/2 , y1+y2/2 }

                        (1,3) = { x1+x2/2 , y1+y2/2 }

                          1=x1+x2/2  , 3=y1+y2/2

                           2=x1+x2    , 6= y1+y2  

 by substituting x1+x2 =2 and y1+y2 =6  in equation (i) and (ii)

    you  get  6=2+x3  then x3=4

                     0=6+y3  then y3=-6

Answer 2

Answer:-

[X3,Y3]=[4,-6]

Step-by-step explanation:-

In The Question,[2,0] is the centroid of the triangle

CASE 1 :

Centroid of the triangle ABC = G{(x1+x2+x3)/3, (y1+y2+y3)/3}

So, (2,0)={(x1+x2+x3/3),(y1+y2+y3/3)}

2=x1+x2+x3/3 , 0= y1+y2+y3/3

6=x1+x2+x3 , 0=y1+y2+y3

6=x1+x2+x3 ⇒ (i)

0=y1+y2+y3 ⇒ (ii)

CASE 2 :

so,Mid point of BC = { x1+x2/2 , y1+y2/2 }

(1,3) = { x1+x2/2 , y1+y2/2 }

1=x1+x2/2 , 3=y1+y2/2

2=x1+x2 , 6= y1+y2

by substituting x1+x2 =2 and y1+y2 =6 in equation (i) and (ii)

you get 6=2+x3 then x3=4

0=6+y3 then y3=-6

Then the final answer is [4,-6].