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Given are the points $P (1,0)$ and $Q (3,2)$. The points $P$ and $Q$ have the same distance to a certain line $l$, which intersects the positive x-axis in the point $A$ and the positive y-axis in the point $B$. The area of the triangle $ABO$ is minimal. Get the equation of $l$.

What I did:

First of all, I got the equation of line $PQ$, which is: $y=x-1$.

Then I got the bisector of $PQ$, because every point on that line is equidistant to $P$ and $Q$. The bisector: $y=-x+3$.

However, this turned out to be wrong, and I had a gut feeling that it would be wrong, since I did nothing with the statement 'The area of triangle of $ABO$ is minimal', because I thought (and still think) that there is just 1 line equidistant from both points.. Can anyone help me with this?

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Let the equation of the line $l(AB)$ be $\frac x a+\frac y b=1$

So, $A(a,0)$ and $B(0,b)$ and $a>0$ and $b>0$.

The area of $\triangle ABO=\frac{ab}2$

The distance of $l:b x+ay-ab=0$ from $P(1,0)$ is $\frac{\mid b-ab\mid}{\sqrt{a^2+b^2}}$ and that of from $Q(3,2)$ is $\frac{\mid 3b+2a-ab\mid}{\sqrt{a^2+b^2}}$

So, $(b-ab)^2=(3b+2a-ab)^2$ $\implies 4(a+b)(ab-a-2b)=0\implies ab-a-2b=0$ as $ab>0$

So, $a=\frac{2b}{(b-1)}\implies b>1$

So, the area of $\triangle ABO=\frac{ab}2=\frac{b^2}{b-1}$

Its minimum value can be derived in the following ways:

(1) $\frac{b^2}{b-1}=\frac{b^2-1}{b-1}+\frac1{b-1}=b+1+\frac1{b-1}$ as $b\ne 1$

So, $\frac{b^2}{b-1}=b-1+\frac1{b-1}+2$ $=\left(\sqrt{b-1}-\frac1{\sqrt{b-1}}\right)^2+2+2\ge 4$ as $b-1>0$

So, the minimum are will be $4$ Square Unit for $\sqrt{b-1}=\frac1{\sqrt{b-1}}\implies b=2$

(2) $\frac{b^2}{b-1}=c$(say) So,$c>0$ as $b>1$

So, $b^2-bc+c=0$ as $b$ is real, the discriminant $(c^2-4c)$ must be $\ge 0$

$\implies c\ge 4$ or $c\le 0$ which is impossible

So, the minimum area is $4$ Square Unit.

$\implies b^2-4b+4=0\implies b=2$

$\implies a=\frac{2b}{(b-1)}=4$

So, the equation of the line $l$ is $\frac x 4+\frac y 2=1$

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@Zafars, please find the inserted image if it can help you out. – lab bhattacharjee Dec 2 '12 at 14:17
Yes, I finally was able to solve this problem, but in a different way. I found that $B (0,-2m+1)$ and $A (\dfrac{2m-1}{m},0)$ , m being the slope (which we want). We then do $\dfrac{1}{2} . -2m+1 . \dfrac{2m-1}{m} $ , that's the area of the triangle. After that use graphing calculator and calc minimum, you'll get $m=-\dfrac{1}{2}$, and then just fill in to get $y=-\dfrac{1}{2} + 2$. I however don't completely understand your method. – JohnPhteven Dec 2 '12 at 14:23

The thing is that line $l$ from your problem must be at equal distances from $P$ and $Q$, but any single point $T \in l$ need not be equidistant from $P$ and $Q$. You are only interested in the line "as a whole", not in every individual point on that line.

There are infinitely many lines equidistant from points $P$ and $Q$. There are two families of such lines. Family 1: all the lines that are parallel to line $PQ$. Family 2: all the lines that contain the middle point of segment $PQ$.

This is probably enough to get you going in the right direction, but if you have any further questions - please ask!

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Oh, wow, that is a subtle yet huge difference. – JohnPhteven Nov 7 '12 at 15:17

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