# Find the minimum length of a line segment with endpoints on the coordinate axes that passes through the point $(1, 1)$

A line passes through the point $P=(1,1)$ and through the $x$ and $y$-axes at points $A$ and $B$ respectively. Find the minimum length of the line segment $AB$.

• Intuitively, the line is length $2*sqrt(2)$. – CAGT Mar 19 '16 at 4:48
• Intuitively you are right. The problem would be nicer if the $x$ and $y$ coordinates would be different. – imranfat Mar 19 '16 at 4:50
• How about non-intuitively? – Calculas Mar 19 '16 at 4:50

$\frac{y-1}{x-1}=m$ defines the line.

$A$ is the $x$ intercept which is $(1-\frac{1}{m}, 0)$ and $B$ is the $y$ intercept which is $(0,1-m)$.

By minimizing the distance between these two points we can calculate $m$ which will then give us the distance. To make the algebra simple we will actually minimize the distance squared, call it d.

Then $d=(\frac{1}{m}-1)^2+(1-m)^2$ and taking the derivative with respect to $m$ gives $\frac{dd}{dm}=-\frac{2}{m^2}(\frac{1}{m}-1)-2(1-m)=\frac{-2+2m}{m^3}-\frac{2m^3(1-m)}{m^3}=\frac{-2+2m-2m^3+2m^4}{m^3}$

Setting this equal to $0$ gives $-2+2m-2m^3+2m^4=0\implies -1+m-m^3+m^4\implies m = 1$ or $m=-1$.

If $m=1$ then both the $x$ and $y$ intercepts will be $0$ so $A=B$ and the distance is $0$.

If $m=-1$ then $A=(2,0)$ and $B=(0, 2)$ so the distance is $\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$

Therefore, the minimum distance is $0$ when $A=B$. If we change the problem to say that $A$ and $B$ are distinct then the minimum distance is $2\sqrt{2}$.

HINT: For a non-intuitive answer, I'm thinking on two possibilities:

1) An algorithm (computer program) to calculate the distance of y-intercept to x-intercept of all line equations that cross (1,1), with slope dependent on line angle to x axis and make a table of angle vs length AB.

2) I don't know an equation which by differentiation to zero, you can find the minimum, but by using the 45º as the intuitive minimum length, demonstrate that two other lines crossing (1,1) with a +/-delta angle $(\Delta a)$, the AB length increases length to both change in angle $({45º+\Delta a})$ and $({45º-\Delta a})$.

Let the equation of $AB$

$$\dfrac xa+\dfrac yb=1$$

As it passes through $P(1,1),$

$$\dfrac1a+\dfrac1b=1$$

Clearly, $a,b$ both can not be $<0$

Case$\#1:$ If $a,b>0$ WLOG $a=\sec^2t,b=\csc^2t$

We need to minimize $\sqrt{\sec^4t+\csc^4t}$ i.e., $\sec^4t+\csc^4t$

Now $F(t)=\sec^4t+\csc^4t=(\sec^2t+\csc^2t)^2-2\sec^2t\csc^2t$

$=\sec^2t\csc^2t(\sec^2t\csc^2t-2)$ as $\sec^2x+\csc^2x=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\sec^2x\csc^2x$

Now $u=\sec^2t\csc^2t=\dfrac4{\sin^22t}\ge4$

$\implies F(t)\ge4(4-2)$

Case$\#2:$ If $a\ge0,b\le0$ WLOG $a=\sin^2t,b=-\tan^2t$

We need to minimize $\sqrt{\sin^4t+\tan^4t}$ i.e., $\sin^4t+\tan^4t$ whose minimum value is clearly $0$

Can you take it from here?