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My problem is that I can't create a line that passes through both points AND is the shortest path AND touches both axis lines.

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3 Answers 3

Hint: A path can be a sequence of segments. You can go from $(8,2)$ to $(0,y)$ to $(x,0)$ to $(3,5)$ Calculate the length of each segment, add them to get the total path length. Then you can differentiate to find the minimum in each variable-you will get two equations in two unknowns. Or, you can remember angle of incidence=angle of reflection and use that. I don't warrant that this you should go to the $y$ axis first, but it looks likely.

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Hint: fix one point and look at paths from this point to a reflection of the other in both axes (you need to cross both) - eg from (3,5) to (-8,-2)

Think about how the proof by reflection works for one mirror, and apply it with two.

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You follow path of light.

First point is (3,5), then comes (0,a) and (b,0) and finally (8,2).
Line equation for the first segment is $y=\frac{5-a}{3}x+a$.
Line equation for the second segment is $y=-\frac{a}{b}x+a$.
Line equation for the third (last) segment is $y=\frac{2}{8-b}x-\frac{2b}{8-b}$.

Because it's reflection in both cases, $k_1=-k_2$ and $k_2=-k_3$:
1) $\frac{5-a}{3}=-\left( -\frac{a}{b}\right)$
2) $-\frac{a}{b}=-\frac{2}{8-b}$

Solve it!

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Better see answer of Mark Bennet. –  zaarcis Feb 10 '13 at 19:02
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