Question

Find the equation of line which is perpendicular on straight line $\frac{x}{a}-\frac{y}{b}=1$ at the point where it meets the $x-$ axis

Answer 1

Let equation of line $AB$ is $\frac{x}{a}-\frac{y}{b}=1$, the co-ordinates of that point where it meets with $x-$ axis will be $(a,0)$. We get equation of line $\perp$ to this line.
Equation of $AB\frac{x}{a}-\frac{y}{b}=1$
$\frac{y}{b}=\frac{x}{a}-1$
$\Rightarrow$ Comparing $y=\frac{b}{a}x-b$ with $y=mx+c$
$m=\frac{b}{a}$
$\therefore$ Equation of line $CD$ perpendicular to line $AB$
$y=-\frac{1}{m}x+c$
$\Rightarrow y=-\dfrac {1}{\left( \dfrac ba \right)}
x+c$
$\Rightarrow y=-\frac{a}{b}x+c$
$\because$ This line passes through $(a,0)$
$\therefore 0=\frac{-a}{b}a+c$
$\to c=\frac{a^2}{b}$
$\therefore$ Thus, equation $y=-\frac{a}{b}x+\frac{a^2}{b}$
$\Rightarrow ax+by=a^2$