The equations of the angle bisectors between the lines $3x - 4y + 7 = 0$ and $12x - 5y - 8 = 0$ are:

Let the line $L_1 : x + 3 = 0$ intersect the lines $L_2 : x - y = 0$ and $L_3 : 3x + y = 0$ at the points $A$ and $B$, respectively. Let the bisector of the obtuse angle between the lines $L_2$ and $L_3$ intersect the line $L_1$ at the point $C$. Then $BC^2 : AC^2$ is equal to:

Let the point $P(\alpha, \beta)$ be at a unit distance from each of the two lines $L_{1}: 3x - 4y + 12 = 0$ and $L_{2}: 8x + 6y + 11 = 0$. If $P$ lies below $L_{1}$ and above $L_{2}$,then $100(\alpha + \beta)$ is equal to

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:

The vertices of a triangle are $A(1, 7)$,$B(-5, -1)$,and $C(-1, 2)$. Then,the equation of a bisector of the $\angle ABC$ is

Let $u \equiv ax + by + a \sqrt[3]{b} = 0$ and $v \equiv bx - ay + b \sqrt[3]{a} = 0$ where $a, b \in R$ be two straight lines. The equation of the bisectors of the angle formed by $k_1u - k_2v = 0$ and $k_1u + k_2v = 0$ for non-zero real $k_1$ and $k_2$ are: