Equation of a line that intersects two lines in a segment of length $2$

Question

How can I find the line that passes through the point $(1,1)$ and intersects the lines $x+y=0$ and $x-y-1=0$ in a segment of length $2$?

I've tried assuming that we know the points of the intersections and then find the conditions, but an equation of the fourth degree appeared.
So can anybody suggest any other idea?

Answer 1

Let the direction vector of the line passing through $(1,1)$ be the unit vector $(\cos(\theta), \sin(\theta) ) $, then the parametric equation of the line is

$ P = (1, 1) + t (\cos(\theta) , \sin(\theta) ) = (1 + t \cos(\theta) , 1 + t \sin(\theta) ) $

where $t $ is the parameter, $t \in \mathbb{R} $

Plug in the point $P$ into the two lines given. For the first line we have

$ 2 + t (\cos(\theta) + \sin(\theta) ) = 0 $

So, $ t_1 = \dfrac{-2}{ \cos(\theta) + \sin(\theta) } $

For the second line, we get

$ t_2( \cos(\theta) - \sin(\theta) ) - 1 = 0 $

So

$ t_2 = \dfrac{1}{ \cos(\theta) - \sin(\theta) } $

The distance between the two points of intersection is $| t_1 - t_2 |$.

We have

$ t_1 - t_2 = \dfrac{-2}{\cos(\theta) + \sin(\theta) } - \dfrac{1}{\cos(\theta) - \sin(\theta) } = \dfrac{ -3 \cos(\theta) + \sin(\theta) }{\cos(2 \theta)} $

Therefore, we know have two cases:

Case I:

$ -3 \cos(\theta) + \sin(\theta) = 2 \cos(2 \theta) $

Case II:

$ -3 \cos(\theta) + \sin(\theta) = -2 \cos(\theta) $

The solutions of Case I are

$\theta_1 = 0.992125 $ and $ \theta_2 = 4.971958 $

The solutions of Case II are

$\theta_3 = 1.830366 $ and $ \theta_4 = 4.133717 $

These four solutions are actually only two solutions, because the angles in the second case are separted by $\pi$ from the solutions of the first case.

Hence there are two possible lines whose equations are

$ -\sin(\phi_i) (x - 1) + \cos(\phi_i) (y - 1) = 0 $

with $ \phi_1 = 0.992125 $ and $\phi_2 = 1.830366 $