Question

Find the equation of the straight line which passes through the point (5,0) and is at a perpendicular distance of 4 units from origin.

Answer 1

Let the equation of the line through points $(5,0)$ be $y-0=m(x-5)$

or $mx-y-5m=0⟶\left(1\right)$

The length of perpendicular from origin $(0,0)$ on line is

$\left|\frac{m.0-0-5m}{\sqrt{m^2+1}}\right|=\left|\frac{-5m}{\sqrt{m^2+1}}\right|$

it is given that $\left|\frac{-5m}{\sqrt{m^2+1}}\right|=4$

$\Rightarrow {25m}^2=16(m^2+1)$

$\Rightarrow {9m}^2=16$

$m^2=\frac{16}{9}$ $\therefore$ $m=\pm \frac{4}{3}$

So, when $m=+\frac{4}{3}$ from $\left(1\right)$ equation of straight line

$\frac{4}{3}x-y-\frac{5\times 4}{3}=0\Rightarrow 4x-3y-20=0$

where $m=\frac{-4}{3}$ from $\left(1\right)$ $\frac{-4}{3}x-y-5\left(\frac{-4}{3}\right)=0\Rightarrow 4x+3y-20=0$