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 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.

Lines $L_1: y - x = 0$ and $L_2: 2x + y = 0$ intersect the line $L_3: y + 2 = 0$ at points $P$ and $Q$ respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$.
Statement-$1$: The ratio $PR:RQ$ is equal to $2\sqrt{2} : \sqrt{5}$.
Statement-$2$: In any triangle,the angle bisector divides the opposite side in the ratio of the sides containing the angle.

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:

If the straight line $2x + 3y + 1 = 0$ bisects the angle between two other straight lines,one of which is $3x + 2y + 4 = 0$,then the equation of the other straight line is

The equation of the perpendicular bisector of the line segment joining the points $(1, 2)$ and $(-2, 0)$ is: