Consider two straight lines
$$y=1, \quad y=-1$$
and the point $\mathrm{A}(0,3)$ in the $xy$-plane. Take a point P on the straight line $y=1$ and a point Q on the straight line $y=-1$ such that
$$\angle\mathrm{PAQ}=90^\circ.$$
Let the two points P and Q move preserving the above conditions. We are to find the minimum value of the length of the line segment PQ.
First, denote the coordinates of P by $(\alpha,1)$ and the coordinates of Q by $(\beta,-1)$. Then the condition $\angle\mathrm{PAQ}=90^\circ$ is reduced to the conditions $\alpha\neq 0$, $\beta\neq 0$ and
$$\alpha\beta = \mathbf{AB}.$$
Since we know that $\alpha$ and $\beta$ have opposite signs, let us assume that $\alpha<0<\beta$.
Then we have
$$\begin{aligned} \mathrm{PQ}^2 &= (\beta-\alpha)^2+\mathbf{C} \\ &= \alpha^2+\beta^2+\mathbf{DE} \\ &\geqq 2|\alpha\beta|+\mathbf{DE} = \mathbf{FG}. \end{aligned}$$
So we have
$$\mathrm{PQ}\geqq\mathbf{H}.$$
Hence, when
$$\alpha=\mathbf{IJ}\sqrt{\mathbf{K}} \text{ and } \beta=\mathbf{L}\sqrt{\mathbf{M}},$$
PQ takes the minimum value $\mathbf{H}$.