Find point of intersection and the bisectors of the angles made by line $k$ passing through point $(3, -1)$ with direction vector $-\vec i + 2\vec j$ and line $l$ passing through point $(-2, 5)$ with direction vector $3\vec i - 2j$.
Assuming the lines intersect I went about finding the point of intersection like so,
Direction vector of line $k$ and $l$,
$m_k = (-1, 2), m_l = (3, -2)$
Vector equation of line $k$ is,
$r_k = (3, -1) + \lambda (-1, 2) = (-\lambda + 3, 2 \lambda - 1)$
And Vector equation of line $l$ is,
$r_l = (-2, 5) + \mu (3, -2) = (-2 + 3\mu, 5 - 2\mu)$
At intersection,
$(-\lambda + 3, 2 \lambda - 1) = (-2 + 3\mu, 5 - 2\mu)$
Which gives the 2 equations,
$\lambda + 3\mu = 5$ $\lambda + \mu = 3$
Solving for $\lambda$ and $\mu$ gives,
$\lambda = 2, \mu = 1$
And thus the point of intersection is $(1, 1)$
The next part of the problem is eluding me. I figured that if the quadrilateral formed by the direction vectors was a rhombus/kite, the angle bisectors would be the diagonals ie:- $m_k + m_l$ and $m_k - m_l$. But the magnitude of $m_k$ and $m_l$ are $\sqrt{5}$ and $\sqrt{13}$ respectively. So that idea doesn't work.
How do i find the equation of the bisectors?
As always, thanks for all your help, Much appreciated!