 Solution

Minimize the squared distance $D^2=(x-x_0{)}^2+(y-y_0{)}^2$ subject to $g=y-mx-c=0$. The Lagrange multiplier condition $\nabla D^2=\lambda \nabla g$ gives $x=x_0-m\lambda /2$ and $y=y_0+\lambda /2$. Enforcing the line constraint gives $\lambda =2(m{x}_0+c-y_0)/(1+m^2)$, henceThe displacement to this point is parallel to the line's normal vector $(-m,1)$, so it is the orthogonal projection. The squared distance restricted to the line has positive quadratic coefficient $1+m^2$, proving this stationary point is the unique global minimum.