If the vertices of a triangle are $A(-1, -7)$,$B(5, 1)$,and $C(1, 4)$,then the equation of the angle bisector of $\angle ABC$ is:

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

Two sides of a rhombus are along the lines $x - y + 1 = 0$ and $7x - y - 5 = 0$. If its diagonals intersect at $(-1, -2)$,then which one of the following is a vertex of this rhombus?

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

The equation of the bisector of the acute angle between the lines $2x - y + 4 = 0$ and $x - 2y - 1 = 0$ is

If the line $l_1: 3y - 2x = 3$ is the angular bisector of the lines $l_2: x - y + 1 = 0$ and $l_3: \alpha x + \beta y + 17 = 0$,then $\alpha^2 + \beta^2 - \alpha - \beta$ is equal to