Proof of Formula For Least Distance of Two Dimensional Line From Origin

Find the least distance of the line  
\[ax+by=c\]
  from the origin.
Rearrange the line as  
\[y=- \frac{a}{b}x+ \frac{c}{b}\]
.
The gradient of this line is  
\[- \frac{a}{b}\]
.
The gradient of the perpendicular line is  
\[\frac{b}{a}\]
.
The  
\[y\]
  intercept of this line is  
\[0\]
  so the equation of the line is  
\[y=\frac{b}{a} x\]
.
Now find the point of intersection of these two lines by solving the simultaneous equations  
\[ax+by=c, \; y=\frac{b}{a}x\]
, equivalent to  
\[ax+by=c\]
  (1)
\[bx-ay=0\]
  (2)
Add  
\[a\]
  times (1) to  
\[b\]
  times (2) to give
\[a^2x+b^2x=ac \rightarrow x= \frac{ac}{a^2+b^2}\]
.
Then from (2),  
\[y=\frac{b}{a}x= \frac{b}{a} \times \frac{ac}{a^2+b^2} = \frac{bc}{a^2+b^2}\]
.
The distance  
\[d\]
  is then the distance between the points  
\[(0,0)\]
  and  
\[(\frac{ac}{a^2+b^2}, \frac{ab}{a^2+b^2})\]

\[d=\sqrt{(\frac{ac}{a^2+b^2}-0)^2+(\frac{bc}{a^2+b^2}-0)^2}=\frac{c}{\sqrt{a^2+b^2}}\]
.