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

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:

The lines $L_1: y-x=0$ and $L_2: 2x+y=0$ intersect the line $L_3: y+2=0$ at $P$ and $Q$ respectively. The bisector of the angle between $L_1$ and $L_2$ divides the line segment $PQ$ internally at $R$.
Statement-$I$: $PR:RQ = 2\sqrt{2}:\sqrt{5}$
Statement-$II$: In any triangle,the bisector of an angle divides the opposite side in the ratio of the sides containing the angle.

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

If $P(-1, 0)$,$Q(0, 0)$,and $R(3, 3\sqrt{3})$ are three points,then the equation of the bisector of the $\angle PQR$ is