Question

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

• A. $3x-7y-8=0$
• B. $x-7y+2=0$
• C. $3x-3y-7=0$
• D. None of these

Answer 1

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

If the bisector of angle $B$ meets side $AC$ at $D$ then

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

$B{A}^2=(5-(-1){)}^2+(1-(-7){)}^2=36+64=100$

$BA=10$

$BC=(5-1{)}^2+(1-4{)}^2=16+9=25$

$BC=5$

so

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

The co-ordinates of $D$ are

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

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

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

The slope of $DB$ is $m=\frac{(1-\frac{1}{3})}{(5-\frac{1}{3})}=\frac{\left(\frac{2}{3}\right)}{\left(14/3\right)}=\frac{1}{7}.$

The line $BD$ passes through $B$ so it has equation

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

$⟹7y-7=x-5$

$⟹7y=x+2.$