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

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

If the straight line $2x + 3y + 1 = 0$ bisects the angle between a pair of lines,one of which is $3x + 2y + 4 = 0$,then the equation of the other line in that pair is

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

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

If two equal sides of an isosceles triangle are given by the equations $7x-y+3=0$ and $x+y-3=0$,then the equation of its third side passing through the point $(2,-5)$ is