Question

Find the points on the x-axis such that their perpendicular distance from the line $\frac{x}{a}+\frac{y}{b}=1$ is $a$ $b>0$

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

Answer 1

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

Given

a is the perpendicular distance from point $(x_1,0)$ to line $\frac{x}{a}+\frac{y}{b}=1$

line become $bx+ay-ab=0$

$a=\left|\frac{b{x}_1-ab}{\sqrt{a^2+b^2}}\right|$

On opening mode with both $\pm$ sign simultaneously

$a=\frac{b(x_1-a)}{\pm \sqrt{a^2+b^2}}$

$\frac{a}{b}=\frac{(x_1-a)}{\pm \sqrt{a^2+b^2}}$

$\frac{a}{b}(\pm \sqrt{a^2+b^2})=x_1-a$

$\frac{a}{b}(\pm \sqrt{a^2+b^2})+a=x_1$

$a\left(\frac{1}{b}\right(\pm \sqrt{a^2+b^2})+1)=x_1$

$\frac{a}{b}(b\pm \sqrt{a^2+b^2})=x_1$

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