Question

What are the points on the $x$ axis whose perpendicular distance from the straight line $\frac{x}{a}+\frac{y}{b}=1$ is $a$?

• A. $\{\frac{a}{b}(b\pm \sqrt{a^2-b^2}),0\}$
• B. $\{\frac{a}{b}(b\pm \sqrt{a^2+b^2}),0\}$
• C. $(a\pm \sqrt{a^2-b^2}),0$
• D. $(b\pm \sqrt{a^2-b^2}),0$

Answer 1

The correct option is B $\{\frac{a}{b}(b\pm \sqrt{a^2+b^2}),0\}$

Soln $\Rightarrow$ Let perpendicular distance be $\left(d\right)$ is given as

$d=\left|\frac{Ah+BK-C}{A^2+B^2}\right|$

So from given question eqn of line is $bx+ay=ab$ & let the point be $(p,0)$ or x-axis then distance is given a which is

$\Rightarrow \left|\frac{bp+a\times 0-ab}{\sqrt{b^2+a^2}}\right|=a\Rightarrow \left|\frac{bp-ab}{\sqrt{b^2+a^2}}\right|=a$

$\Rightarrow \frac{bp-ab}{\sqrt{b^2+a^2}}=\pm a$

So,

$\frac{b(p-a)}{\sqrt{b^2+a^2}}=a$ or $\frac{b(p-a)}{\sqrt{b^2+a^2}}=-a$

$\Rightarrow (p-a)=\frac{a}{b}\left(\sqrt{1^2+a^2}\right)$ or $(p-a)=-\frac{a}{b}\sqrt{b^2+a^2}$

$\Rightarrow p=\frac{a\sqrt{b^2+a^2}+ab}{b}$ or $p=\frac{ab-a\sqrt{b^2+a^{@}}}{b}$

Points are

$\left(\frac{a\sqrt{b^2+a^2}+ab}{b}\right)$ & $(\frac{ab-a\sqrt{b^2+a^2}}{b},0)$