Question

Find the equation of the line at a distance of:
$2$ units form the origin and passing through the point $(4,-2)$

Answer 1

Let the equation of the line be

$ax+by+c=0$

Distance of the line from origin is

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

$\Rightarrow \frac{c}{\sqrt{a^2+b^2}}=2⟶\left(1\right)$

Now the line passes through $(4,-2)$

$\therefore$ $4a-2b+c=0⟶\left(2\right)$

From $\left(1\right)$, Squaring we get

$c^2=4(a^2+b^2)⟶\left(3\right)$

From $\left(2\right)$,

$c=2b-4a⟶\left(4\right)$

Squaring

$c^2={4b}^2+{16a}^2-16ab⟶\left(5\right)$

Putting the value of $c^2$ from equation $\left(3\right)$ we get

${4a}^2+{4b}^2={4b}^2+{16a}^2-16ab$

$\Rightarrow {12a}^2-16ab=0$

$\Rightarrow 4a(3a-4b)=0$

$\therefore$ $3a=4b$ or $a=0$ (does not give the line passing through $(4,-2)$)

Now putting

$3a=4b$

$\Rightarrow a=4b/3$ in equation $\left(2\right)$ we get

$c=-4a+2b=\frac{-16b}{3}+2b=\frac{-16b}{3}+2b=\frac{-10b}{3}$

$\therefore$ equation of line will be

$\frac{4b}{3}x+by-\frac{10b}{3}=0\Rightarrow 4x+3y-10=0$