Question

The equation of the internal bisector of $\angle BAC$ of $\Delta ABC$ with vertices $A(5,2)$, $B(2,3)$ and $C(6,5)$ is

• A. $2x+y+12=0$
• B. $x+2y-12=0$
• C. $2x+y-12=0$
• D. $2x-y-12=0$

Answer 1

The correct option is C $2x+y-12=0$

Now, slope of $AB$ is $m_1=\frac{3-2}{2-5}=-\frac{1}{3}$ and that of $AC$ is $m_2=\frac{5-2}{6-5}$ $=3$.

We, have $m_1.m_2=-1$, so $\angle$A $={90}^{\circ }$.

Let, $p(x,y)$ be any point on the internal bisector of $\angle$ BAC. Then

${\tan }^{-1}\frac{m_1-m_3}{1+m_3{m}_1}=$ ${45}^{\circ }$, where $m_3$ is slope of $AP$ and $m_3=\frac{(y-2)}{(x-5)}$.

So,

$1=\frac{m_1-m_3}{1+m_1{m}_3}$

or, $1=\frac{-\frac{y-2}{x-5}-\frac{1}{3}}{1-\frac{(y-2)}{3(x-5)}}$

or, $3(x-5)-(y-2)=-3(y-2)-(x-5)$

or, $2(y-2)+4(x-5)=0$

or, $2x+y-12=0$