Finding the shortest line segment joining $2$ line segments in $\Bbb R^2$ On an online class, a lecturer gave the following question.


Let $A$ be any point on a line segment joining $(4,0)$ and $(0,2)$, and $B$ is any point on a line segment joining $(8,0)$ and $(0,6)$. What is the length of the shortest line segment $AB$?


What I tried to do is write A and B in parametric form using trigonometry as $A(4\cos^2\theta,2\sin^2\theta)$ and $B(8\cos^2\theta,6\sin^2\theta)$
Finding $AB$ comes out to be a $2$ variable expression with which I am not able relate to with any inequality. How to solve this?
 A: Disclaimer: This answer is more of an intuitive and geometric answer than a  linear algebra or optimization answer.
Let $y_1=6-\frac{3}{4}x$ and $ y_2=2-\frac{1}{2}x$. These represent line segments $B$ and $A$ respectively. The point $(4,0)$ is the point where the shortest line segment connecting $B$ and $A$ will pass through. Intuitively, this is because the lines $y_1$ and $y_2$ will "converge" to a single point when moving rightwards along the x-axis and "diverge" away from each other when moving leftwards along the x-axis. That means that the closest point on $A$ from $B$ is the point $(4,0)$.
Now, the shortest distance to a point on $B$ from $A$ is a line. We will try to construct a right triangle in order to find this distance.
Link to image of triangle. The line perpendicular to $y_1$ passing through point $(4,0)$ is line $y_3=\frac{4}{3}x-\frac{16}{3}$. This creates our right triangle. We create another right triangle by dropping a perpendicular from the intersection of lines $y_1$ and $y_3$. The triangle created by lines $y=0, y_3,$ and the perpendicular will be the right triangle we refer to through out the rest of the answer.
Setting $y_1=y_3$ to solve for $x$ yields $x=\frac{136}{25}$. This distance, however, is not the distance of the horizontal leg of the triangle, but $\frac{136}{25}-4=\frac{36}{25}$ is. To solve for the height of the triangle, we plug in $x=\frac{136}{25}$ into equation $y_1$ to yield $y=\frac{192}{100}$. By the Pythagorean theorem, the shortest distance, $d$, is  $d=\sqrt{\big(\frac{36}{25}\big)^2+\big(\frac{192}{100}\big)^2}=\frac{12}{5}$.
A: By Lagrange multipliers:
Let $(p_1,p_2)$ be the point on the first segment, and $(q_1,q_2)$ be the point on the second segment that minimizes the distance.
$$L=(p_1-q_1)^2+(p_2-q_2)^2 - \lambda \left(\frac{p_1}{4}+\frac{p_2}{2}-1\right) - \mu \left(\frac{q_1}{8}+\frac{q_2}{6}-1\right)$$
$$\begin{aligned}
0= \frac{1}{2} \frac{\partial L}{\partial p_1} &= p_1 - q_1-\frac{\lambda}{8}\\
0= \frac{1}{2} \frac{\partial L}{\partial p_2} &= p_2 -q_2-\frac{\lambda}{4}\\
0= \frac{1}{2} \frac{\partial L}{\partial q_1} &= q_1 -p_1-\frac{\mu}{16}\\
0= \frac{1}{2} \frac{\partial L}{\partial p_2} &= q_2 -p_2-\frac{\mu}{12}
\end{aligned}$$
Solving this system of equations:
$$\bbox[5px, border: 1pt solid green]{\lambda=\mu=0} $$
$$\bbox[5px, border: 1pt solid green]{(p_1,p_2)=(q_1,q_2)}$$
Thus, the Lagrange multiplier solution requires that the two points are actually the same point, and lie at the intersection of the two lines.
Since the problem specifies that only the two line segments are allowable, this method fails to produce the solution.
@TonyK pointed out in the comments, if the line segments were permitted to continue (they are not), the solution would have been the point of intersection.  Since this is not in the feasible region, the solution will be at the boundary (an endpoint of one of the segments).  The distance is the perpendicular distance from $(0,4)$ to the other segment.  @CSquared worked it out yesterday.
A: When two segments in the plane are given this is a complicated geometric situation. You cannot expect that there is a simple once-for-all method that solves a particular problem connected with two arbitrary such segments. Draw a figure!
The triangle $T$ over the segment $\sigma_2=[(8,0), (0,6)]$ with right angle at$(0,0)$ contains the other segment $\sigma_1=[(4,0), (0,2)]$.   All points in $T$ having the same distance $d$ to $\sigma_2$ are lying on a parallel to $\sigma_2$. Draw the parallel through $A:=(4,0)\in \sigma_1$. Since parallels nearer to $\sigma_2$ contain no points of $\sigma_1$ the point $A$ is nearest to $\sigma_2$ among all points of $\sigma_1$. You find the point $B\in\sigma_2$ by drawing the perpendicular $y={4\over3}(x-4)$ from $A$ to $\sigma_2$. A little computation then gives $$d=|AB|={12\over5}\ .$$
