Albert1
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P=(3,7),Q=(1,5) ,A=(x,0),B=(0,y)
if k=minimum(PA+AB+BQ)
find: x+y+k
if k=minimum(PA+AB+BQ)
find: x+y+k
your answer is correct:)Opalg said:[sp]
https://www.physicsforums.com/attachments/2267
Let $Q'$ be the reflection of $Q$ in the $y$-axis and let $Q''$ be the reflection of $Q'$ in the $x$-axis. The shortest distance between $P$ and $Q''$ is the straight line between them. The shortest path between $P$ and $Q$ (via $A$ and $B$) is obtained by reflecting parts of that line in the axes, as appropriate (the red path in the diagram). The length $k$ of the path is the same as the distance from $P$ to $Q''$, which is $4\sqrt{10}.$ The point $A$ is at $(2/3,0)$ and $B$ is at $(0,2)$. So $x+y+k = 4\sqrt{10} + \frac83$.[/sp]