Question

If the lenght of the perpendicular from the point ($1,1$) to the line ax-by+c=$0$ be unity show that $-\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=\frac{c}{2ab}$

Answer 1

We know that distance of a point $(x_1,y_1)$ from a line $ax+by+c=0$ is,

distance $=\frac{|{ax}_1+{by}_1+c|}{\sqrt{a^2+b^2}}$

Distance of $(1,1)$ from ax $+by-c=0$ is 1$so$

$\frac{|a+b-c|}{\sqrt{a^2+b^2}}=1$

$|a+b-c|=\sqrt{a^2+b^2}$

Squaring we get,

$a^2+b^2+c^2+2ab-2bc-2ca=a^2+b^2$

$c^2+2ab-2bc-2ca=0$

$ab-bc-ca=-\frac{c^2}{2}$

Dividing by - abc we get,

$\frac{-ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=\frac{c^2}{2abc}$

$\frac{-1}{c}+\frac{1}{a}+\frac{1}{b}=\frac{c}{2ab}$