A man is walking on a straight line. The arit | vedclass

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(D) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $x$ and $y$ intercepts respectively.The arithmetic mean

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(D) Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $x$ and $y$ intercepts respectively.The arithmetic mean of the reciprocals of the intercepts is given by $\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{1}{4}$.This implies $\frac{1}{a} + \frac{1}{b} = \frac{1}{2}$.Since the line passes through $(x, y)$,we have $\frac{x}{a} + \frac{y}{b} = 1$. If we test the point $(2, 2)$,we get $\frac{2}{a} + \frac{2}{b} = 2(\frac{1}{a} + \frac{1}{b}) = 2(\frac{1}{2}) = 1$.Thus,the line always passes through the point $(2, 2)$.Since stone $B$ is at $(2, 2)$,only stone $B$ is on the path of the man.

$A$ man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\frac{1}{4}$. Three stones $A, B$ and $C$ are placed at the points $(1,1), (2,2)$ and $(4,4)$ respectively. Which of these stones is/are on the path of the man?

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