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 $\frac{1}{4}$. 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 ?

• A. A only
• B. B only
• C. C only
• D. All the three

Hint:

Any condition of the form $\frac{x_1}{a} + \frac{y_1}{b} = 1$ implies the line always passes through the fixed point $(x_1, y_1)$.

Answer 1

The Correct Option is B

Step 1: Let the line be $\frac{x}{a} + \frac{y}{b} = 1$.
Step 2: Given $\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{1}{4} \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{2}$.
Step 3: Rearranging the condition: $\frac{2}{a} + \frac{2}{b} = 1$.
Step 4: Comparing this with $\frac{x}{a} + \frac{y}{b} = 1$, we see that the point $(2, 2)$ must lie on the line.
Step 5: Therefore, stone B is on the path.