What is the equation of the bisector of the acute angle between the lines $3x - 4y + 7 = 0$ and $12x + 5y - 2 = 0$?

The family of lines,forming an isosceles triangle with the lines $3x - 4y - 2 = 0$ and $12x - 5y + 6 = 0$,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:

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

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.