The equations of the sides A B ,   B C   &   C A of a triangle A B C are 2 x + y = 0 , x…

The equations of the sides AB, BC & CA of a triangle ABC are 2x+y=0x+py=21a a0 and x-y=3 respectively. Let P2,a be the centroid of the triangle ABC, then BC2 is equal to

Solution

We have,

Since, G2,a, so

1+α+β+33=2 and -2-2α+β3=a

α+β=2 and -2α+β=3a+2

So, α=-a, β=2+a

So, B-a,2a, C5+a,2+a

Now, B and C lies on the line x+py=21a, so

-a+2pa=21a

p=11

Also,

5+a+112+a=21a

27=9aa=3

So,

B-3,6, C8,5

So,

BC2=121+1=122

Asked in: JEE Main 2023 (24 Jan Shift 2)

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