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. 7x+y-2=0
• C. 2y-7x-5=0
• D. 2x+7y+5=0

Answer 1

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

Let BD be the bisector of $\angle ABC$ then,

$AD:DC=AB:BCandAB=\sqrt{(5+1{)}^2+(1+7{)}^2}=10BC=\sqrt{(5-1{)}^2+(1-4{)}^2}=5\therefore AD:DC=2:1$
Therefore, by section formula, $D=(\frac{1}{3},\frac{1}{3})$ therefore the equation of BD is
$y-1=\frac{\frac{1}{3}-1}{\frac{1}{3}-5}(x-5)ory-1=\frac{-\frac{2}{3}}{-\frac{14}{3}}(x-5)or7y-7=x-5x-7y+2=0$