Given vertices $A(1, 1)$,$B(4, -2)$,and $C(5, 5)$ of a triangle,find the equation of the line perpendicular to the internal angle bisector of $\angle A$ and passing through vertex $C$.

The equation of the line which bisects the obtuse angle between the lines $x - 2y + 4 = 0$ and $4x - 3y + 2 = 0$ is:

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

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

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.

The equation of the angle bisectors between the lines $3x + 4y - 7 = 0$ and $12x + 5y + 17 = 0$ is: