Q. The point on the axis of $x$, whose perpendicular distance from the straight line $\frac{x}{a} + \frac{y}{b} = 1\ is\ a,\$are

A) $\frac{b}{a}(a \pm \sqrt{a^{2} + b^{2}},\ 0$
B) $\left( \frac{a}{b}\left( \ b \pm \sqrt{a^{2} + b^{2}} \right),\ 0 \right)$
C) $\frac{b}{a}(a + b,0)$
D) $\frac{a}{b}(a \pm \sqrt{a^{2} + b^{2}},\ 0)$
Explanation

(b)
Let the point on the $x$-axis be $(h,0)$.
The perpendicular distance from $(h,\ 0)$ to the line
$= \frac{\left| \frac{h}{a} - 1 \right|\ \ }{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} = a\ \lbrack given\rbrack$
$\Rightarrow \frac{h}{a} - 1 = \pm a\frac{\sqrt{a^{2} + b^{2}}}{ab}$
$\Rightarrow h - a = \pm \frac{a}{b}\sqrt{a^{2} + b^{2}}$
$\Rightarrow h = a \pm \frac{a}{b}\sqrt{a^{2} + b^{2}}$
$= \frac{a}{b}\left( b \pm \ \sqrt{a^{2} + b^{2}} \right)$
$\therefore Required\ point\ is\ \left( \frac{a}{b}\left( b \pm \sqrt{a^{2} + b^{2}} \right),\ 0 \right)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.