Lets have a straight talk!

Similar solution with @Dheeman Kuaner 's

Suppose that line $L$ cuts the $x$ - and $y$ -axis at $P$ and $Q$ respectively. Let $OP=a$ , $OQ=b$ and $\angle QPO = \theta$ . Then we have:

$\begin{cases} \tan \theta = \dfrac 2{a-8} & \implies a = \dfrac 2{\tan \theta} + 8 \\ \tan \theta = \dfrac ba & \implies b = a \tan \theta = 2 + 8 \tan \theta \end{cases}$

Then we have: $OP+OQ = a+b = \dfrac 2{\tan \theta} + 10 + 8 \tan \theta$ . Using AM-GM inequality $\dfrac 2{\tan \theta} + 8 \tan \theta \ge 2 \sqrt {16} = 8$ . Therefore $OP+OQ \ge \boxed{18}$

Let $m \in \mathbb{R^{+}}$ be the slope of line L in question, which is represented as $y - 2 = -m(x - 8) \Rightarrow y = -mx + (8m+2).$ The corresponding y and x-intercepts, $P$ and $Q$ , are given as:

$P(0, 8m+2)$ ; $Q(8 + \frac{2}{m}, 0)$

and $OP + OQ = L(m) = (8m+2) + (8 + \frac{2}{m}) = 10 + 8m + \frac{2}{m}.$ as a function of the slope $m.$ Differentiating $L(m)$ produces:

$L'(m) = 8 - \frac{2}{m^{2}} = 0 \Rightarrow 8m^2 = 2 \Rightarrow m = \frac{1}{2}$ (since we require $m > 0).$

and $L''(m) = \frac{4}{m^{3}} \Rightarrow L''(\frac{1}{2}) > 0$ (hence, a minimum).

Thus, the minimum value of $OP + OQ$ is $L(\frac{1}{2}) = 10 + 8(\frac{1}{2}) + \frac{2}{\frac{1}{2}} = \boxed{18}.$