Question

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

Answer 1

Let (t, 0) be a point on the x-axis.

It is given that the perpendicular distance of the line $\frac{x}{a}+\frac{y}{b}=1$ from a point is a.

$$\begin{gathered} \therefore \left|\frac{{\displaystyle \frac{t}{a}}+0-1}{\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}}\right|=a \\ \Rightarrow a^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=\frac{t^2}{a^2}+1-\frac{2t}{a} \\ \Rightarrow 1+\frac{a^2}{b^2}=\frac{t^2}{a^2}+1-\frac{2t}{a} \\ \Rightarrow \frac{a^2}{b^2}=\frac{t^2}{a^2}-\frac{2t}{a} \end{gathered}$$

$$\begin{gathered} \Rightarrow b^2{t}^2-2a{b}^{2}t-a^4=0 \\ \Rightarrow t=\frac{2a{b}^2\pm 2\sqrt{a^2{b}^4+b^2{a}^4}}{2b^2} \\ \Rightarrow t=\frac{a}{b}\left(b\pm \sqrt{a^2+b^2}\right) \end{gathered}$$

Hence, the required points on the x-axis are $\left(\frac{a}{b}\left(b-\sqrt{a^2+b^2}\right),0\right)\mathrm{and}\left(\frac{a}{b}\left(b+\sqrt{a^2+b^2}\right),0\right)$.