Find the Shortest Segment Cut Off by the First Quadrant

Suppose a line passes through a fixed point $(a,b)$ in the first quadrant, with $a>0$ and $b>0$. It meets the positive $x$- and $y$-axes at two points, and the portion between those intercepts is cut off by the first quadrant. The goal is to choose the line so that this segment is as short as possible.

The answer depends on the coordinates of the fixed point. The method is to express the segment length in terms of its two intercepts and minimize subject to the condition that the line passes through $(a,b)$.

Write the line using its intercepts

Let the intercepts be $(u,0)$ and $(0,v)$, where $u,v>0$. The segment length is

$$L=\sqrt{u^2+v^2}.$$

A line with these intercepts has equation

$$\frac{x}{u}+\frac{y}{v}=1.$$

Since it passes through $(a,b)$, the intercepts must satisfy

$$\frac{a}{u}+\frac{b}{v}=1.$$

Thus, the problem is to minimize $u^2+v^2$ subject to this constraint. Minimizing the squared length is equivalent to minimizing the length, and it avoids carrying a square root through the calculus.

Minimize with the constraint

Use a Lagrange multiplier for

$$f(u,v)=u^2+v^2, \qquad g(u,v)=\frac{a}{u}+\frac{b}{v}-1=0.$$

The multiplier equations are

$$\begin{aligned} 2u&=\lambda\left(-\frac{a}{u^2}\right),\\ 2v&=\lambda\left(-\frac{b}{v^2}\right). \end{aligned}$$

Eliminating $\lambda$ gives

$$\frac{u^3}{a}=\frac{v^3}{b},$$

so the minimizing intercepts have the ratio

$$\frac{u}{v}=\left(\frac{a}{b}\right)^{1/3}.$$

This ratio, together with the point constraint, determines both intercepts. Set

$$S=a^{2/3}+b^{2/3}.$$

Then the intercepts are

$$u=a^{1/3}S, \qquad v=b^{1/3}S.$$

Indeed, substituting them into the constraint gives

$$\frac{a}{u}+\frac{b}{v} =\frac{a^{2/3}+b^{2/3}}{S}=1.$$

Minimum length and worked example

Substitute the minimizing intercepts into the length formula:

$$\begin{aligned} L_{\min} &=\sqrt{(a^{1/3}S)^2+(b^{1/3}S)^2}\\ &=S\sqrt{a^{2/3}+b^{2/3}}\\ &=\left(a^{2/3}+b^{2/3}\right)^{3/2}. \end{aligned}$$

For example, take the fixed point $(8,1)$. Then

$$S=8^{2/3}+1^{2/3}=4+1=5.$$

The best intercepts are

$$u=8^{1/3}(5)=10, \qquad v=1^{1/3}(5)=5.$$

So the segment runs from $(10,0)$ to $(0,5)$ and has length

$$\sqrt{10^2+5^2}=\sqrt{125}=5\sqrt{5}.$$

The line is $x/10+y/5=1$, and it passes through $(8,1)$ because $8/10+1/5=1$.

Why this is a minimum

The constraint implies $u>a$ and $v>b$. Solving it for $v$ gives $v=bu/(u-a)$, with $u>a$. As $u\to a^+$, $v$ and hence the segment length tend to infinity; as $u\to\infty$, the segment length also tends to infinity. Therefore, the continuous length function on this feasible curve attains a minimum at an interior point. At any such minimum, the Lagrange equations apply. They yield the ratio above, and that ratio together with the constraint has the unique positive solution $u=a^{1/3}S$, $v=b^{1/3}S$. Thus this pair is the unique global minimizer.

Common mistakes

• Using the slope without enforcing the point condition. A candidate line must satisfy $a/u+b/v=1$; optimizing over arbitrary intercepts does not solve the stated problem.
• Minimizing the sum of intercepts. The segment length is $\sqrt{u^2+v^2}$, not $u+v$. Those objectives generally produce different lines.
• Forgetting the square root at the end. The minimized quantity during the calculation is the squared length. Take its square root to report the segment length.
• Ignoring the point’s location. The formula assumes $a,b>0$. If the point lies on an axis, the positive-intercept setup becomes a boundary case and needs separate interpretation.