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

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

In a $\triangle ABC$,suppose $y=x$ is the equation of the angle bisector of $\angle B$ and the equation of the side $AC$ is $2x-y=2$. If $2AB=BC$ and the points $A$ and $B$ are $(4,6)$ and $(\alpha, \beta)$ respectively,then $\alpha+2\beta$ is equal to

Let $A, B, C$ be three points in the $xy$-plane,whose position vectors are given by $\sqrt{3} \hat{i} + \hat{j}$,$\hat{i} + \sqrt{3} \hat{j}$,and $a \hat{i} + (1 - a) \hat{j}$ respectively with respect to the origin $O$. If the distance of the point $C$ from the line bisecting the angle between the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is $\frac{9}{\sqrt{2}}$,then the sum of all the possible values of $a$ 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 equations of the angle bisectors between the $x$-axis and $y$-axis are: