Question

Given vertices $A(1,1),B(4,-2)$ and $C(5,5)$ of a triangle, then the equation of the perpendicular dropped from $C$ to the interior bisector of the angle $A$ is

• A. $y-5=0$
• B. $x-5=0$
• C. $y+5=0$
• D. $x+5=0$

Answer 1

The correct option is B $x-5=0$

slope of $AC\left(m_1\right)=\frac{5-1}{5-1}=1$

slope of $AB\left(m_2\right)=\frac{-2-1}{4-1}=-1$

Angle $\tan A=\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|=\left|\frac{-1-1}{1-1}\right|=\frac{2}{0}\infty$

$A={90}^o$

Slope of bisector line $AD=m$

Angle $b/wAD$&$AC={45}^o$

$\Rightarrow \tan {45}^o=\left|\frac{m-1}{1+m}\right|$

$\Rightarrow \left|\frac{m-1}{1+m}\right|=1$

$\frac{m-1}{1+m}=+1$

$\Rightarrow m-1=-(1+m)$

$2m=0$

$m=0$

then slope of the perpendicular on AD

$m=\infty$

$\Rightarrow$ eqn of line

$\Rightarrow (y-5)=\frac{1}{0}(x-5)$

$\Rightarrow x-5=0$

So, option (B) is correct.