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