Find the area of a triangle using analytic geometry 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?


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 $\frac x 4+\frac y 2=1$
