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

$x–5=0$

Explanation for the correct option:

Step 1.Finding the angle at $A$:

Given, $A(1,1),B(4,-2)$ and $C(5,5)$ are the vertices of triangle

Now, Slope of $AB$, $\begin{array}{rcl}{m}_1& =& \frac{(y_2–y_1)}{(x_2–x_1)}\end{array}$

$\begin{array}{rcl}& =& \frac{(-2-1)}{(4-1)}\\ & =& -\frac{3}{3}\\ & \mathbf{=}& \mathbf{}\mathbf{-}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Slope of $AC$, $\begin{array}{rcl}{m}_2& =& \frac{(5-1)}{(5-1)}\end{array}$

$\begin{array}{rcl}& =& \frac{4}{4}\\ & \mathbf{=}& \mathbf{}\mathbf{1}\mathbf{}\end{array}$

$\therefore m_1{m}_2=-1$

$\Rightarrow$$AB\perp AC$

Step 2. Let slope of bisector line $AD=m$

$\because \angle A=90°$, the angle between $AD$ and $AC$ is $45°$

As we know,

$\tan \alpha =\left|\frac{(m_1–m_2)}{(1+m_1{m}_2)}\right|$

so, $\tan 45°=\left|\frac{(m–1)}{(1+m)}\right|$

$\Rightarrow$ $\pm 1=\left|\frac{(m–1)}{(1+m)}\right|$

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

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

$\Rightarrow$$-1-m-m+1=0$

$\Rightarrow$ $2m=0$

$\Rightarrow$ $m=0$

$\Rightarrow$The bisector is parallel to $x$-axis so the line perpendicular to it will be parallel to $y$-axis, so it will be in the form of $x=a,$as it passes through$\left(5,5\right)$, we have $5=a$

then slope of the perpendicular on $AD,m=\infty$

$\therefore$The equation of line is $y–5=\left(\frac{1}{0}\right)(x–5)$

$\Rightarrow$ $\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{5}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$

$\Rightarrow$ $\mathit{x}\mathbf{=}\mathbf{5}$

Hence, option(B) is correct.