The vertices of a triangle are $A(1, 7)$,$B(-5, -1)$,and $C(-1, 2)$. Then,the equation of a bisector of the $\angle ABC$ is

The equal sides of an isosceles triangle are given by the equations $7x-y+3=0$ and $x+y-3=0$. If the slope $m$ of the third side is an integer,then $m=$

Let $P \equiv (-5, 0)$,$Q \equiv (0, 0)$,and $R \equiv (2, 2\sqrt{3})$ be three points. Then the equation of the bisector of the angle $\angle PQR$ is

The sides of a rhombus $ABCD$ are parallel to the lines $x - y + 2 = 0$ and $7x - y + 3 = 0$. If the diagonals of the rhombus intersect at $P(1, 2)$ and the vertex $A$ (different from the origin) is on the $y$-axis,then the ordinate of $A$ is

Let $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$ be three points. The equation of the bisector of the angle $\angle PQR$ is:

The lines $L_1: y-x=0$ and $L_2: 2x+y=0$ intersect the line $L_3: y+2=0$ at $P$ and $Q$ respectively. The bisector of the angle between $L_1$ and $L_2$ divides the line segment $PQ$ internally at $R$.
Statement-$I$: $PR:RQ = 2\sqrt{2}:\sqrt{5}$
Statement-$II$: In any triangle,the bisector of an angle divides the opposite side in the ratio of the sides containing the angle.