Question

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

• A. $x-7y+2=0$
• B. $x+y-6=0$
• C. $x+2y-7=0$
• D. $116x-39y=157$

Answer 1

The correct option is A $x-7y+2=0$

Given vertices of $\Delta ABC$ are $A(-1,-7)B(5,1)$ and $C(1,4)$

Then it the bisector of angle $B$ meets side $AC$ at $D$ then,

$\Rightarrow \frac{AD}{DC}=\frac{BA}{BC}$

Where $(BA{)}^2=(5+1{)}^2+(-1-7{)}^2$

$=36+64=100$

$BA=10$

and $B{C}^2=(1-5{)}^2+(4-1{)}^2$

$=16+9$

$BC=5$

So $\frac{AD}{DC}=\frac{10}{5}=2$

The coordinate for point $D$B are

$\left(\frac{2}{3}\right)C+\left(\frac{1}{3}\right)A=\left(\frac{2}{3}\right)(1,4)+\left(\frac{1}{3}\right)(-1,-7)$

$=\left(\frac{1}{3}\right)(1,1)$

$=(\frac{1}{3},\frac{1}{3})$

$Slope=(1-\frac{1}{3})/(5-\frac{1}{3})$

$=\frac{2}{3}/\frac{14}{3}=\frac{1}{7}$

Equation of times $BD$ passes through $B$ is given as

$\Rightarrow y-1=\left(\frac{1}{7}\right)(x-5)$

$\Rightarrow 7y=x+2$

$\Rightarrow x-7y+2=0$