Question

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

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

Answer 1

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

Line is $bx+ay=ab$

Let point on $x$ axis be $(l,0)$

then $\left|\frac{bl-ab}{\sqrt{a^2+b^2}}\right|=a$

$bl-ab=\pm a\sqrt{a^2+b^2}$

$bl=ab+a\sqrt{a^2+b^2}$

$l=\frac{a}{b}(b\pm \sqrt{a^2+b^2})$

Point is $\frac{a}{b}(b\pm \sqrt{a^2+b^2},0)$

$B$ is correct.