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

Let $P \equiv (-5, 0)$,$Q \equiv (0, 0)$,and $R \equiv (2, 2\sqrt{3})$ be three points. Then the equation of the bisector of the angle $\angle PQR$ 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?

The straight line $x+y+1=0$ bisects an angle between the pair of lines of which one is $2x+3y-4=0$. Then,the equation of the other line 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

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