JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given, position of A=(1,1)
Position of B=(2,2)
Position of C=(4,4)Let x-intercept be a and y-intercept be b.
Equation of line traced is
‌

x

a

+‌

y

b

=1
This is the equation of path, let a point (h,k) lie on this path. Then
‌

h

a

+‌

k

b

=1
Also, AM of reciprocal of a and b=‌

1

4

∴ ‌

‌ 1 a + 1 b

2

=‌

1

4

‌

1

a

+‌

1

b

=‌

1

2

On comparing Eqs. (i) and (ii), we get (h,k)=(2,2)
Hence, the required stone is B(2,2).