Question

Let a variable line of slope $m>0$ passing through the point $(4, -9)$ intersect the coordinate axes at the points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is:

• A. 25
• B. 30
• C. 15
• D. 10

Answer 1

Answer: (A)

25

Answer 2

The equation of the line is:

$y + 9 = m(x - 4).$

Find $A$ and $B$ (intersection points with the axes):

At $y = 0$, $x = \frac{9}{m} + 4 \implies A\left(\frac{9}{m} + 4, 0\right)$.

At $x = 0$, $y = -9 - 4m \implies B(0, -9 - 4m)$.

The sum of distances:

$OA + OB = \sqrt{\left(\frac{9}{m} + 4\right)^2} + \sqrt{(-9 - 4m)^2}.$

Using AM-GM inequality, the minimum value occurs when:

$m = \frac{3}{2}.$

Substitute $m$ to get:

$OA + OB = 25.$