Shortest line segment that is cut off by the first quadrant and passes through a given point

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Edit: this was posted before the question was edited to reflect the clarification. The following is the answer to a different problem:

If $b<a$, it's the line segment along $x=a$ with length $b$. Otherwise, it's the line segment along $y=b$ with length $a$ (if $a=b$, then they're the same length). You can see this by noting that if you fix the point at (a,b) but move the line, the distance will increase according to the Pythagorean theorem. So the answer is $\min(a,b)$.

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Chris
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Chris

Updated on July 31, 2022

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  • Chris
    Chris 3 months

    Let $a$ and $b$ be two positive numbers. Find the length of the shortest line segment that is cut off by the first quadrant and passes through the point $(a,b)$.

    I attempted to let the $y$-intercept be $(0, Y_c)$ where the slope from there to $(a, b)$ is $(b - Y_c ) / a$ and $x$-intercept be $(X_c, 0)$ and slope from there to $(a, b)$ is $(b / a - X_c)$. Since the two slopes are equal, they can be used to solve for $X_c$ in terms of $Y_c$ and $a, b$. Then use the distance formula to find distance $D$ between $x$-intercept and $y$-intercept points. Then differentiate $ D$ to find its max/min value.

    • Rory Daulton
      Rory Daulton over 7 years
      Welcome to MathSE! However, this is not a homework-answering site. You need to give us more details, such as the attempts you have made and where you are stuck, as well as the level of math you want in the solution. You should also clarify what you are asking. Do you mean the shortest segment through $(a,b)$ with endpoints on the positive $x$ and $y$ axes?
    • Chris
      Chris over 7 years
      College level calculus 1
    • Rory Daulton
      Rory Daulton over 7 years
      Good: now the other questions I asked?
    • Chris
      Chris over 7 years
      I attempted to let the y-intercept be (0, Yc) where the slope from there to (a, b) is (b - Yc ) / a) and x-intercept be (Xc, 0) and slope from there to (a, b) is (b / a - Xc). Since the two slopes are equal, they can be used to solve for Xc in terms of Yc and a, b. Then use the distance formula to find distance D between x-intercept and y-intercept points. Then differentiate D to find it's max/min value.