Find the equation of the locus of points equidistant from the lines $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$.

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

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?