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Triangles $PBA, ACQ$ are similar $\Rightarrow$ $\frac{y} {8} = \frac{2} {x} \Rightarrow xy=16$ .

$OP+OQ=2+y+8+x=10+x+y$ .

$OP+OQ$ has minimal value for minimal value of $x+y$ . This is equivalent of asking for minimal perimeter of a rectangle with area of 16 which is a square with side 4. This means $x=y=4, OP+OQ = \boxed{18}$ .

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

Let the gradient of $L$ be $-m$ , where $m > 0$ . Then the equation of $L$ is:

$\begin{aligned} \frac {y-2}{x-8} & = -m \\ y - 2 & = -mx + 8m \\ \implies y & = 8m+2-mx \end{aligned}$

Let $P(x_0,0)$ and $Q(0,y_0)$ , then we have:

$\begin{cases} 0 = 8m+2-mx_0 & \implies x_0 = 8 + \dfrac 2m & \implies OP = 8 + \dfrac 2m \\ y_0 = 8m+2-m(0) & \implies y_0 = 8m + 2 & \implies OQ = 8m + 2 \end{cases}$

Therefore,

$\begin{aligned} OP+OQ & = 8 + \dfrac 2m + 8m + 2 \\ & = 10 + \dfrac 2m + 8m & \small {\color{#3D99F6}\text{By AM-GM inequality}} \\ & \ge 10 + 2\sqrt{\dfrac 2m \cdot 8m} \\ & \ge \boxed{18} \end{aligned}$

Let the line in question have slope $-m$ (with $m > 0$ ) and be modeled as:

$y-2 = -m(x-8) \Rightarrow y =-mx + (8m+2)$ (i)

The $x$ and $y-$ intercepts of (i) are respectively: $P(8 + \frac{2}{m}, 0), Q(0, 8m+2)$ . We wish to calculate the least distance of $OP + OQ$ , or:

$L(m) = 8 + \frac{2}{m} + 8m + 2 = 10 + 8m + \frac{2}{m}$ (ii)

Taking $L'(m)= 0$ , we get $8 - \frac{2}{m^2} = 0 \Rightarrow m^2 = \frac{1}{4} \Rightarrow m = \frac{1}{2}.$ Evaluating this critical slope at the second derivative of (ii) yields:

$L''(m) = \frac{4}{m^3} \Rightarrow L''(1/2) = 32 > 0$ (hence a global minimum). Finally, the least distance computes to:

$L(1/2) = 10 + 8(\frac{1}{2}) + 2(2) = \boxed{18}.$