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 $\angle ABC$ is

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

Answer 1

Answer: (A)

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