Question

what are the point on the axis of x whose perpendicular distance from the straight line x/a+y/b=1 is a​

Answer 1

Answer:

The points are ($a[1 + \sqrt{1 + \frac{a^2}{b^2}}]$, 0) and ($a[1 - \sqrt{1 + \frac{a^2}{b^2}}]$, 0)

Step-by-step explanation:

Perpendicular distance from any point (x1, y1) to a straight line ax + by + c = 0

is $\frac{absolute(ax1 + by1 + c)}{\sqrt{a^2 + b^2} }$

Now points on the X axis have co-ordinates (x1,0)

according to the problem:

$\frac{absolute(\frac{x1}{a} + \frac{0}{b} - 1)}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} } } = a$

$=> \frac{(\frac{x1}{a} + \frac{0}{b} - 1)^2}{\frac{1}{a^2} + \frac{1}{b^2} } = a^2$  [both sides squared]

$=> (\frac{x1}{a} - 1)^2} = a^2 . ({\frac{1}{a^2} + \frac{1}{b^2}}) = 1 + \frac{a^2}{b^2}$

$=> \frac{x1}{a} - 1 =$ ± $\sqrt{1 + \frac{a^2}{b^2}}$

$=> \frac{x1}{a} = 1$ ± $\sqrt{1 + \frac{a^2}{b^2}}$

$=> x1 = a(1$  ± $\sqrt{1 + \frac{a^2}{b^2}})$

$=> x1 = a(1 + \sqrt{1 + \frac{a^2}{b^2}})$

$or, x1 = a(1 - \sqrt{1 + \frac{a^2}{b^2}})$  

∴ The points are ($a[1 + \sqrt{1 + \frac{a^2}{b^2}}]$, 0) and ($a[1 - \sqrt{1 + \frac{a^2}{b^2}}]$, 0)