If the vertices of a triangle are $A(-1, -7)$,$B(5, 1)$,and $C(1, 4)$,then the equation of the angle bisector of $\angle ABC$ is:

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 acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$.
$STATEMENT-1$ : The ratio $PR:RQ$ equals $2\sqrt{2}:\sqrt{5}$.
$STATEMENT-2$ : In any triangle,the angle bisector divides the opposite side in the ratio of the sides containing the angle.

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

The equation of the bisector of the acute angle between the lines $2x - y + 4 = 0$ and $x - 2y - 1 = 0$ is

Statement $-I$: Two lines which pass through a given fixed point and are equally inclined to two other lines passing through the same point,are always perpendicular to each other.
Statement $-II$: Angle bisectors of two intersecting lines are always perpendicular to each other.

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 $m=$