Question

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

• A. $\left\lbrack \frac{a}{b}(b \pm \sqrt{a^{2} + b^{2}}),0 \right\rbrack$
• B. $\left\lbrack \frac{b}{a}(b \pm \sqrt{a^{2} + b^{2}}),0 \right\rbrack$
• C. $\left\lbrack \frac{a}{b}(a \pm \sqrt{a^{2} + b^{2}}),0 \right\rbrack$
• D. None of these

Answer 1

Answer: (A)

$\left\lbrack \frac{a}{b}(b \pm \sqrt{a^{2} + b^{2}}),0 \right\rbrack$

Answer 2

Let the point be $(h,0)$ then $a = \pm \frac{bh + 0 - ab}{\sqrt{a^{2} + b^{2}}}$

⇒ $bh = \pm a\sqrt{a^{2} + b^{2}} + ab$ ⇒ $h = \frac{a}{b}(b \pm \sqrt{a^{2} + b^{2}})$

Hence the point is $\left\lbrack \frac{a}{b}(b \pm \sqrt{a^{2} + b^{2}}),0 \right\rbrack$