Minimising #1

Coordinates of a point on the line are of the form of: $\left(x,\dfrac{2x-9}{3}\right)$

We have to find the minimum of :

$f(x)= \sqrt{(x+2)^2+\left(\dfrac{2x-9}{3}\right)^2} + \sqrt{x^2+\left(\dfrac{2x-9}{3}-4\right)^2}$

After simplifying ,

$f(x)=\sqrt{\dfrac{13}{9}}\left[\sqrt{x^2+9}+\sqrt{\left(x-\dfrac{42}{13}\right)^2+\left(\dfrac{63}{13}\right)^2}\right]$

Finding the minimum of the above function is equivalent to finding the point $(x,0)$ such that the sum of distances between $(0,3)$ ; $(x,0)$ and $\left(\dfrac{42}{13},\dfrac{-63}{13}\right)$ ; $(x,0)$ is minimised.

This will happen at the point where the line joining $\left(\dfrac{42}{13},\dfrac{-63}{13}\right)$ and $(0,3)$ meets the $x$ -axis

Equation of the line is :

$y-3=\left(\dfrac{-51}{21}\right)x$

Setting $y=0$ , we get :

$x=\dfrac{21}{17}$

Using the equation , $2x-3y=9$ to get the $y$ - coordinate:

$y=\dfrac{-37}{17}$

So, the coordinates of the required point are:

$\boxed{\left(\dfrac{21}{17}\right),\left(\dfrac{-37}{17}\right)}$