Angle Bisectors

Given vertices $A(1, 1), B(4,-2) \& C(5, 5)$of a triangle, find the equation of the perpendicular dropped from $C$ to the interior bisector of the angle $A.$

Equation of angle bisector between the lines $3x+4y-7=0$ and $12x+5y+17=0$ are

If $x \cos \alpha +y \sin \alpha -p=0, x \cos {\alpha }^{\text{'}}+y \sin {\alpha }^{\text{'}}-p^{\text{'}}=0$ are two intersecting straight lines, then show that the equations of the bisectors of the angles between the lines are:

$x(\cos \alpha +\cos \alpha )+y(\sin \alpha +\sin \alpha )-(p+p)=0$ and $x\left(\cos \alpha -\cos {\alpha }^{\text{'}}\right)+y\left(\sin \alpha -\sin {\alpha }^{\text{'}}\right)-\left(p-p^{\text{'}}\right)=0$.

The internal and external bisectors of the angles between two lines $y=\sqrt{3}x+3$ and $\sqrt{3}y=x+3\sqrt{3}$ meet $x$-axis and $y$-axis at $A$ and $B$ respectively. Find the length of the line segment $AB$.

Find the equations of the internal bisectors of the angles of the triangle formed by the straight line $3x-4y+12=0$ and the axes of co-ordinates. Then verify that the internal bisectors of a triangle are concurrent

The vertices of a triangle are $A(1,7), B(-5,-1)$ and $C(-1,2) .$ Then, the equation of the bisector of the $\angle ABC$ is

If the pairs of straight lines $x^2-2pxy-y^2=0$ and $x^2-2qxy-y^2=0$ bisect the angles between each other, then which of the following is correct?

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

A ray of light coming along the line $3x+4y-5=0$ gets reflected from the line $ax+by-1=0$ and goes along the line $5x-12y-10=0,$ then

If $ax+y-16=0$ and $\mathrm{x}+ay\mathit{-}6=0$ represent two adjacent sides of a rhombus whose diagonals intersect at $\left(3, 5\right)$ then sum of possible value of a is

If $A\left(x_1,y_1\right), B\left(x_2,y_2\right), C\left(x_3,y_3\right)$ are the vertices of the triangle then find the equation of angle bisector through $A,$ where $a,b,c$ units are the lengths of the sides opposite to the vertices $A,B,C$ respectively.
 

The incident ray is along the line $3x-4y-3=0$ and the reflected ray is along the line $24x+7y+5=0$. Find the equation of mirrors.

Find the equation of the bisector of the obtuse angle between the lines $3x-4y+7=0$ and $12x+5y-2=0$.

Find the area of the triangle formed by the straight line $\mathrm{x}+\mathrm{y}=4$ and angle bisectors of pair of lines $x^2+2y=1+y^2$:

The equation of the angle bisector of the acute angle between the lines $3x+7=4y$ and $12x+5y=2$ is given by

 In a $∆\mathrm{ABC},$if $A=(1, 2)$ and internal angle bisectors through $C$ and $B$ are $\mathrm{y} =-2\mathrm{x}$ and$\mathrm{y} =x$, Then the inradius $r$ of $∆\mathrm{ABC}$ is :

If $r\left(1-m^2\right)+m\left(p-q\right)=0,$then find the bisector of angle between the lines represented by the equation $p{x}^2-2rxy+q{y}^2=0,$ (where $m$ is slope of line $y=\mathit{mx}$)

The point $Q$ is symmetric to $P(5, -2)$ with respect to the bisector of the first quadrant. The distance between the points $P$ and $Q$ is  _____

The straight lines $3x-4y+7=0$ and $12x+5y-2=0$ has equation of the acute angle bisector as:

The coordinates of the points on the line $3x+2y=5$ which are equidistant from the lines $4x+3y-7=0$ and $2y-5=0$ can be