Question

If the vertices of a triangle $ABC$ are $A (1,7), B (-5,-1)$ and $C (7,4)$, then the equation of a bisector of $\angle A B C$ is

• A. $7 x-9 y+26=0$
• B. $9 x-7 y+38=0$
• C. $7 x+9 y+44=0$
• D. $9 x+7 y+52=0$

Answer 1

Answer: (A)

$ \because \frac{A D}{D C}=\frac{B A}{B C} $
$ A B=\sqrt{(1+5)^2+(7+1)^2} $
$ =\sqrt{36+64}$
$ =\sqrt{100}=10 $
$ B C=\sqrt{(7+5)^2+(4+1)^2} $
$ =\sqrt{12^2+5^2}=13 $
$ \text { So, } \frac{A D}{D C}=\frac{10}{13} $
Coordinate of $D=\left(\frac{10 \times 7+13 \times 1}{10+13}, \frac{10 \times 4+13 \times 7}{10+13}\right)$
$=\left(\frac{83}{23}, \frac{131}{23}\right)$
Equation of $B D$ in straight line joining points
$B(-5,-1)$ and $D\left(\frac{83}{23}, \frac{131}{23}\right)$ is given by
$ \frac{y+1}{131}+1=\frac{x+5}{\frac{83}{23}+5} $
$\Rightarrow \frac{y+1}{154}=\frac{x+5}{198} $
$ \Rightarrow \frac{y+1}{22 \times 7}=\frac{x+5}{22 \times 9} $
$ \Rightarrow 9 y+9=7 x+35 $
$ 7 y-9 y+26=0$