Question

A variable straight-line L with negative slope passes through the point (4,9) and cuts the positive coordinate axes in A and B. If O is the origin, then the minimum value of OA + OB is

• A. 25
• B. 12
• C. 13
• D. 5

Hint:

For a line passing through \( (a, b) \), the minimum sum of positive intercepts is \( (\sqrt{a} + \sqrt{b})^2 \). Here, \( (\sqrt{4} + \sqrt{9})^2 = (2+3)^2 = 25 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We model the line passing through a fixed point with variable slope \( m \), express the sum of intercepts in terms of \( m \), and use calculus (differentiation) to minimize the sum.
Step 2: Key Formula or Approach:

Equation of line: \( y - 9 = m(x - 4) \). Sum of intercepts \( S = \text{x-intercept} + \text{y-intercept} \).
Step 3: Detailed Explanation:

Let the slope be \( m \) (given \( m \textless 0 \)). Equation: \( y - 9 = m(x - 4) \). Find x-intercept (OA):
Set \( y=0 \). \( -9 = m(x - 4) \implies x - 4 = -9/m \implies x = 4 - 9/m \). Find y-intercept (OB):
Set \( x=0 \). \( y - 9 = -4m \implies y = 9 - 4m \). Sum \( S = OA + OB = (4 - 9/m) + (9 - 4m) = 13 - \frac{9}{m} - 4m \). To minimize S, differentiate w.r.t \( m \): \[ \frac{dS}{dm} = -9(-m^{-2}) - 4 = \frac{9}{m^2} - 4 \] Set \( \frac{dS}{dm} = 0 \): \[ \frac{9}{m^2} = 4 \implies m^2 = \frac{9}{4} \implies m = \pm \frac{3}{2} \] Since \( m \textless 0 \), we take \( m = -1.5 \). Calculate min S: \[ S = 13 - \frac{9}{-1.5} - 4(-1.5) = 13 + 6 + 6 = 25 \]
Step 4: Final Answer:

The minimum value is 25.