Question

$A(5,6)$ and $C(4,2)$ are the two vertices of a triangle $ABC$ and the equation of internal angle bisector of the $\angle ABC$ is $x+y-8=0.$ The equation of $AB$ is

• A. $x-y+1=0$
• B. $2x-y-4=0$
• C. $2x+y-16=0$
• D. $y=6$

Answer 1

The correct option is C $2x+y-16=0$


Image of $C$ about the angle bisector $BD$ lies on the line $AB.$ Let the image of $C$ be $C^{\prime }(x_1,y_1).$

$\frac{x_1-4}{1}=\frac{y_1-2}{1}=\frac{-2(-2)}{2}\Rightarrow x_1-4=y_1-2=2\Rightarrow (x_1,y_1)=(6,4)$

Slope of $AB=$ Slope of $A{C}^{\prime }$
$\Rightarrow \frac{y-6}{x-5}=\frac{4-6}{6-5}\Rightarrow 2x+y-16=0$

$\therefore$ Equation of $AB$ is $2x+y-16=0$