1. Equation of the variable line

A line with slope $m>0$ passing through $(4,-9)$ is

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

Rewriting:

$y=mx-4m-9.$

2. Intercepts on the axes

Let the line meet the $x$-axis at $A$ and the $y$-axis at $B$.

• For the $x$-axis, $y=0$: $0=mx-4m-9$ $x=4+\frac{9}{m}.$ So, $A\left(4+\frac{9}{m},0\right).$

• For the $y$-axis, $x=0$: $y=-4m-9.$ So, $B(0,-4m-9).$

3. Distances of $A$ and $B$ from the origin

Since $A$ lies on the $x$-axis,

$OA=\left|4+\frac{9}{m}\right|=4+\frac{9}{m}$ because $m>0$.

Since $B$ lies on the $y$-axis,

$OB=|-4m-9|=4m+9.$

Hence the required sum is

$S=OA+OB=\left(4+\frac{9}{m}\right)+(4m+9)=13+4m+\frac{9}{m}.$

4. Minimize the expression

We need the minimum of

$S=13+4m+\frac{9}{m}, \qquad m>0.$

Using AM-GM:

$4m+\frac{9}{m} \ge 2\sqrt{4m\cdot \frac{9}{m}}=2\sqrt{36}=12.$

So,

$S \ge 13+12=25.$

Equality holds when

$4m=\frac{9}{m} \Rightarrow 4m^2=9 \Rightarrow m=\frac{3}{2}$ (since $m>0$).

Thus the minimum value is

$\boxed{25}.$

5. Option check

• A: $30$ ✗
• B: $15$ ✗
• C: $10$ ✗
• D: $25$ ✓

So the correct option is D.