The equation of the angle bisectors between the lines $3x + 4y - 7 = 0$ and $12x + 5y + 17 = 0$ is:

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

The set of values of $\alpha$ for which the angle bisector of the lines $(\alpha + 1)x + 2y + 5 = 0$ and $4x + \alpha y - 3 = 0$ containing the origin is also the obtuse angle bisector,is:

Let $A(1,0)$, $B(2,-1)$, and $C(\frac{7}{3},\frac{4}{3})$ be three points. If the equation of the internal angle bisector of $\angle ABC$ is $\alpha x+\beta y=5$, then the value of $\alpha^2+\beta^2$ 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

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 :