QJEEM21Ma68a8: JEE Main 2021 Mathematics - Class 11 JEEM21Ma68a8 Solution: A man is walking on a straight line. The arithmetic mean of the …

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

${x \over a} + {y \over b} = 1$

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

Then, ${h \over a} + {k \over b} = 1$

Also, AM of reciprocal of a and b = ${1 \over 4}$

$\therefore$ ${{{1 \over a} + {1 \over b}} \over 2} = {1 \over 4}$

${1 \over a} + {1 \over b} = {1 \over 2}$

On comparing Eqs. (i) and (ii), we get (h, k) = (2, 2)

Hence, the required stone is B(2, 2).