Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches) 
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.

I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\min(\sqrt{x^2+y^2})=3$.
I'm wondering is it possible to solve this problem differently?


$3x+4y=15$  represents a straight line and $\sqrt{x^2+y^2}$ represents the distance of the point $(x,y)$ from the origin. So the question is basically telling us to find the minimum distance of any point lying on the line $3x+4y=15$, from the origin.
This shortest distance must be the perpendicular distance from the origin to the line.
The perpendicular distance of a point $(h,k)$ from the line $ax+by+c=0$ is $\Bigg|\dfrac{ah+bk+c}{\sqrt{a^2+b^2}}\Bigg|$. Replacing $(h,k)$ with $(0,0)$ and the line with $3x+4y-15=0$ gives us the minimum value as 3.
