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?

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 $12x+5y-2=0$.

If  $A(–1,–7), B(5,1)$ and $C(1,4)$ are vertices of $∆ABC$, then the equation of the internal angle bisector of $\angle ABC$

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

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 $y=\mathit{mx}$)

Comprehension Question: 5 to 6

In a $ABC$ given $\angle A={60}^o$ and AD is angle bisector of angle $A$ so that $\frac{\mathrm{\text{BD}}}{\mathrm{\text{DC}}}=\frac{2}{3}$ and $A{B}^2-B{D}^2=6$ then

tan C equals

The locus of the point $P$ which is equidistant from $3x+4y+5=0$ and $9x+12y+7=0$ is

The equation of  bisector of the acute angle formed between the lines $4x-3y+7=0$ and $3x-4y+14=0$ has the equation is 

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

The equation of the bisector of the angle between the lines $x+2y-11=0, 3x-6y-5=0$ which contains the point $\left(1,3\right)$ is

Equation of angle bisectors between $x$ and $y$ - axes are

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