- #1
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
How Do You Calculate the Minimal Path Sum in Coordinate Geometry?
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- Thread starter Albert1
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SUMMARY
The minimal path sum in coordinate geometry can be calculated using reflections of points across the axes. Given points P=(3,7) and Q=(1,5), the shortest path involves reflecting Q across the y-axis to obtain Q' and then reflecting Q' across the x-axis to get Q''. The length of the path k is determined to be 4√10, with points A and B located at (2/3,0) and (0,2) respectively. The final calculation yields x+y+k = 4√10 + 8/3.
PREREQUISITES- Understanding of coordinate geometry and reflections
- Familiarity with distance formulas in a Cartesian plane
- Knowledge of basic algebraic manipulation
- Ability to interpret geometric diagrams
- Study the properties of reflections in coordinate geometry
- Learn about distance calculations between points in a Cartesian plane
- Explore optimization techniques in geometric problems
- Investigate the application of geometric reflections in real-world scenarios
Students and professionals in mathematics, particularly those focused on geometry, optimization problems, and anyone interested in advanced coordinate geometry techniques.
- #2
Opalg
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MHB
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[sp]
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]
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]
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Albert1
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your answer is correct:)Opalg said:
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