Question
What are the points on X-axis whose perpendicular distance from the straight line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ is a?
✓
Answer
Any point on x-axis is $(\pm\text{a},0)\$\text{x}_1,\text{y}_1)$
Perpendicular distance from a line bx + ay = ab is
$\Big|\frac{\text{ax}_1+\text{by}_1+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|=\text{a}$
where,
$\text{a}=\text{b}, \ \text{b}=\text{a}, \ \text{c}=\text{-ab}, \ \text{x}_1=\pm\text{a}, \ \text{y}_1=0$
$=\Big|\frac{\text{b}(\text{x})+\text{a}(0)-\text{ab}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|=\text{a}$
$\text{a}=0$ or
$=\frac{\text{b}(\text{x})+\text{a}(0)-\text{ab}}{\sqrt{\text{a}^2+\text{b}^2}}=\text{a}$
$\frac{\text{b}}{\text{a}}\text{x}=\pm\sqrt{\text{a}^2+\text{b}^2}+\text{b}$
$\text{x}=\frac{\text{a}}{\text{b}}\Big(\text{b}\pm\sqrt{\text{a}^2+\text{b}^2}\Big)$
$\text{x}=0$
Perpendicular distance from a line bx + ay = ab is
$\Big|\frac{\text{ax}_1+\text{by}_1+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|=\text{a}$
where,
$\text{a}=\text{b}, \ \text{b}=\text{a}, \ \text{c}=\text{-ab}, \ \text{x}_1=\pm\text{a}, \ \text{y}_1=0$
$=\Big|\frac{\text{b}(\text{x})+\text{a}(0)-\text{ab}}{\sqrt{\text{a}^2+\text{b}^2}}\Big|=\text{a}$
$\text{a}=0$ or
$=\frac{\text{b}(\text{x})+\text{a}(0)-\text{ab}}{\sqrt{\text{a}^2+\text{b}^2}}=\text{a}$
$\frac{\text{b}}{\text{a}}\text{x}=\pm\sqrt{\text{a}^2+\text{b}^2}+\text{b}$
$\text{x}=\frac{\text{a}}{\text{b}}\Big(\text{b}\pm\sqrt{\text{a}^2+\text{b}^2}\Big)$
$\text{x}=0$
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