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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
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The correct answer is:
25
equation of line is
$\begin{aligned}
& y+9=m(x-4) \\
& \therefore \quad A=\left(\frac{9+4 m}{m}, 0\right) \\
& \quad B=(0,-9-4 m) \\
& \therefore \quad O A+O B=\frac{9+4 m}{m}+9+4 m
\end{aligned}$
$\begin{aligned} & \because \mathrm{m}>0 \\ & =13+\frac{9}{\mathrm{~m}}+4 \mathrm{~m} \\ & \because \frac{4 \mathrm{~m}+\frac{9}{\mathrm{~m}}}{2} \geq \sqrt{36} \Rightarrow 4 \mathrm{~m}+\frac{9}{\mathrm{~m}} \geq 12 \\ & \therefore \mathrm{OA}+\mathrm{OB} \geq 25\end{aligned}$
$\begin{aligned}
& y+9=m(x-4) \\
& \therefore \quad A=\left(\frac{9+4 m}{m}, 0\right) \\
& \quad B=(0,-9-4 m) \\
& \therefore \quad O A+O B=\frac{9+4 m}{m}+9+4 m
\end{aligned}$
$\begin{aligned} & \because \mathrm{m}>0 \\ & =13+\frac{9}{\mathrm{~m}}+4 \mathrm{~m} \\ & \because \frac{4 \mathrm{~m}+\frac{9}{\mathrm{~m}}}{2} \geq \sqrt{36} \Rightarrow 4 \mathrm{~m}+\frac{9}{\mathrm{~m}} \geq 12 \\ & \therefore \mathrm{OA}+\mathrm{OB} \geq 25\end{aligned}$
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