perpendicular distance from $ax-by = 1$ to origin I have a problem from a basic number theory book that asks for the perpendicular distance from the line $ax - by = 1$ to the origin. 
My approach was to find the area of the triangle formed by the line and the axes, find the base of the triangle situated at the diagonal, then use those to find the height of the triangle, which would be the perpendicular distance.
I found that the $x$ intercept is $\frac1a$, and the y intercept is $-\frac1b$. So if you consider the base to be the diagonal, it is $\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}$. Then the area of the triangle is $\frac{1}{2ab}$. Thus the height is $\frac{1}{2ab\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$ = $\frac{1}{2\sqrt{(a^2 + b^2}}$, which I thought would be the answer.
However, the book states that the answer is $\frac{1}{\sqrt{a^2 + b^2}}$, so I'm off by $\frac12$. Did I make a mistake with the area of the triangle? If not, where did I go wrong?
Thank you!

Minimum Distance
Since
$$
\begin{align}
1
&=\left|\,ax-by\,\right|\\
&=\left|\,(a,-b)\cdot(x,y)\,\right|\\
&\le\left|\,(a,-b)\,\right|\left|\,(x,y)\,\right|
\end{align}
$$
we have
$$
\begin{align}
\left|\,(x,y)\,\right|
&\ge\frac1{\left|\,(a,-b)\,\right|}\\
&=\frac1{\sqrt{a^2+b^2}}
\end{align}
$$
If $(x,y)=\frac{(a,-b)}{a^2+b^2}$, then $ax-by=1$ and $\left|\,(x,y)\,\right|=\frac1{\sqrt{a^2+b^2}}$
Thus, the minimum of $\left|\,(x,y)\,\right|$ is $\frac1{\sqrt{a^2+b^2}}$.

Perpendicular Distance
If $ax_1-by_1=1$ and $ax_2-by_2=1$, then
$$
(a,-b)\cdot(x_1-x_2,y_1-y_2)=1-1=0
$$
Thus, $(a,-b)$ is perpendicular to the line containing $(x_1,y_1)$ and $(x_2,y_2)$; i.e. the line $ax-by=1$.
This means the vector from the origin to $\frac{(a,-b)}{a^2+b^2}$ is perpendicular to the line $ax-by=1$ and the point $\frac{(a,-b)}{a^2+b^2}$ is on the line $ax-by=1$.
Thus, the perpendicular distance from the origin to the line is
$$
\left|\frac{(a,-b)}{a^2+b^2}\right|=\frac1{\sqrt{a^2+b^2}}
$$
