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