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 then use differentiation to minimize the sum.

Step 2: Key Formula or Approach:

Equation of line passing through (4, 9):
y - 9 = m(x - 4)

Sum of intercepts:
S = x-intercept + y-intercept

Step 3: Detailed Explanation:

Let the slope be m, where m < 0.

Equation of the line:
y - 9 = m(x - 4)

Find x-intercept (OA):

Set y = 0:
-9 = m(x - 4)
x - 4 = -9/m
x = 4 - 9/m

So, the x-intercept is:
OA = 4 - 9/m

Find y-intercept (OB):

Set x = 0:
y - 9 = m(0 - 4)
y - 9 = -4m
y = 9 - 4m

So, the y-intercept is:
OB = 9 - 4m

Therefore, the sum of intercepts is:
S = OA + OB
S = (4 - 9/m) + (9 - 4m)
S = 13 - 9/m - 4m

To minimize S, differentiate with respect to m:
dS/dm = 9/m2 - 4

Set dS/dm = 0:
9/m2 - 4 = 0
9/m2 = 4
m2 = 9/4
m = ±3/2

Since m < 0, we take:
m = -3/2

Now substitute into S:
S = 13 - 9/(-3/2) - 4(-3/2)
S = 13 + 6 + 6
S = 25

Step 4: Final Answer:

The minimum value is 25.