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

The bisector of the acute angle formed between the lines $4x - 3y + 7 = 0$ and $3x - 4y + 14 = 0$ has the equation

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.

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

Given vertices $A(1, 1)$,$B(4, -2)$,and $C(5, 5)$ of a triangle,find the equation of the line perpendicular to the internal angle bisector of $\angle A$ and passing through vertex $C$.

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