Question

If $A(0,0),B(0,3)$ and $C(4,0)$ are the vertices of a triangle then equation of internal bisector of $\angle B$ is given by:-

• A. $2x+y+3=0$
• B. $2x+y-3=0$
• C. $2x-y+3=0$
• D. None of these

Answer 1

The correct option is B $2x+y-3=0$

The given sides are,$AB=3unitsBC=5unitsAC=4units$
Using angle bisector theorem
$\frac{AD}{DC}=\frac{AB}{BC}\frac{AB}{BC}=\frac{3}{5}AD:DC=3:5$
$\therefore$ Co-ordinates of $D=(\frac{12+0}{8},\frac{0+0}{8})$
Co-ordinates of $D(\frac{3}{2},0),B(0,3)$
Slope of line BD= -2
The equation of AD
$(y-3)=-2(x-0)\Rightarrow 2x+y-3=0$