Inequality on a plane

Consider the combination of translations:

$x \rightarrow x-11, \begin{pmatrix} x \\ y \end{pmatrix} \rightarrow \begin{pmatrix} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\\ -\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

Note all lengths are preserved by this translation (the matrix comes from rotating the line which gives a $1-2-\sqrt{5}$ triangle).

The line is carried to $y=0$ , the points are carried to $A'=(-5 \sqrt{5},4 \sqrt{5})$ and $B'=(-3 \sqrt{5},2 \sqrt{5})$ .

Reflect B' across the line $y=0$ to $C=(-3 \sqrt{5},-2 \sqrt{5})$ . We have $B'M=CM$ for any point $M'$ on the line $y=0$ .

Therefore $AM+BM=A'M'+B'M=A'M'+CM'$ is minimised when $A'MC$ is a straight line so $A'M+CM'=A'C$

$A'C=\sqrt{(-5 \sqrt{5}- (-3 \sqrt{5}))^2+(4 \sqrt{5}- (-2 \sqrt{5}))^2}=\sqrt{20+180}=\sqrt{200}=\boxed{10 \sqrt{2}}$

$\text{Let }M\left(x,\frac{1}{2}x-\frac{11}{2}\right)$ $\text{distance}=\sqrt{(x+3)^2+\left(\frac{1}{2}x-\frac{17}{2}\right)^2}+\sqrt{(x-3)^2+\left(\frac{1}{2}x-\frac{13}{2}\right)^2}$ $\frac{\text{d}(\sqrt{(x+3)^2+\left(\frac{1}{2}x-\frac{17}{2}\right)^2}+\sqrt{(x-3)^2+\left(\frac{1}{2}x-\frac{13}{2}\right)^2}) }{\text{d}x}=\frac{\sqrt{5}}{2}\left(\frac{x-5}{\sqrt{x^2-10x+41}}+\frac{x-1}{\sqrt{x^2-2x+65}}\right)$ $\text{Setting that to zero, we get: }x=\frac{11}{3}$ $\text{Substituting back into the original equation, we get the minimum distance to be 14.14}$