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

Let's solve the problem step-by-step:

The equation of the line in terms of intercepts on the axes is given by:

\frac{x}{a} + \frac{y}{b} = 1

where a and b are the x-intercept and y-intercept respectively.

The arithmetic mean of the reciprocals of the intercepts is given as:

\frac{1/a + 1/b}{2} = \frac{1}{4}

Simplifying the above expression:

\frac{1}{a} + \frac{1}{b} = \frac{1}{2}

Or,

\frac{a+b}{ab} = \frac{1}{2}

This implies:

2(a + b) = ab

We can rearrange this to:

ab - 2a - 2b = 0

Adding 4 to both sides, we get:

ab - 2a - 2b + 4 = 4

The above equation can be factored as:

(a - 2)(b - 2) = 4

Therefore, the line can be represented with intercept form:

\frac{x}{a} + \frac{y}{b} = 1 with conditions (a - 2)(b - 2) = 4.

We are to check the points (1, 1), (2, 2), and (4, 4) one-by-one to find if they lie on the line:

• Point (1, 1):
  • Plugging into ax + by = ab gives a + b = ab.
  • However, it does not satisfy (a - 2)(b - 2) = 4.
• Point (2, 2):
  • Substitute into ax + by = ab gives 2a + 2b = ab.
  • Dividing by 2: a + b = \frac{ab}{2}, which satisfies (a - 2)(b - 2) = 4.
• Point (4, 4):
  • Substitute into ax + by = ab gives 4a + 4b = ab.
  • This equation does not satisfy (a - 2)(b - 2) = 4.

Therefore, we conclude that only point B (2, 2) lies on the path of the man.

Hence, the correct answer is: B only