A man is walking on a straight line.

Question

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 $\dfrac14$. Three stones $A, B$ and $C$ are placed at the points $(1,1), (2,2)$ and $(4,4)$ respectively. Then which of these stones is/are on the path of the man?

Accepted Answer

Let the line be $y=mx+c$, $x$-intercept $=-\dfrac cm$, $y$-intercept $=c$.
$$\frac12\left(-\frac mc+\frac1c\right)=\frac14 \implies 2(1-m)=c$$

So $y=mx+2(1-m) \implies (y-2)=m(x-2)$: the line always passes through the fixed point $(2,2)$.

Since $(1,1)$ and $(4,4)$ do not necessarily lie on every such line, but $(2,2)$ always does: only stone B is guaranteed to be on the man's path.

Solutions · 1