Point J(−2,5) is 1 unit away from a line L that passes through point K(−3,0). Find the line L.

Question

how do i find the equation of the L line given its distance from point J which is 1 unit away from it. there could be two Lines to this equation but I don't know how to find it

Answer 1

Let the normal to the line be the unit vector $(\cos \theta, \sin \theta)$

Since $K(-3,0)$ is on the line, then the equaiton of the line is

$ \cos \theta (x + 3) + \sin \theta y = 0 $

Since $J(-2, 5)$ is $1 $ unit away from the line then,

$ 1 = |\cos \theta + 5 \sin \theta |$

Define $\phi = \text{ATAN2}(1, 5) = \cos^{-1} \dfrac{1}{\sqrt{26}}$ , then

$1 = \sqrt{26} | \cos(\theta - \phi ) |$

From which,

$ \cos(\theta - \phi) = \pm \dfrac{1}{\sqrt{26}} $

Hence,

$\theta = \phi + \cos^{-1} \pm \dfrac{1}{\sqrt{26}} = 2 \phi , \pi $

If $\theta = \pi$ then the equation of the line is: $-(x + 3) = 0$, i.e. $\boxed{x = -3}$

If $\theta = 2 \phi$, then the equation of the line is $\cos 2 \phi (x + 3) + \sin 2 \phi = 0 $

Now, $\cos 2 \phi = 2 \cos^2 \phi - 1 = \dfrac{2}{26} - 1 = -\dfrac{12}{13}$

And $\sin 2 \phi = \dfrac{5}{13} $

Hence, the equation is: $\boxed{-12 (x + 3) + 5 y = 13}$