Question

Given A((1, 1), B(4, -2) and C(5, 5) are vertices 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

The internal bisector of the angle A will divide the opposite side BC at D in the ratio of arms of the angle i.e., $AB=3\sqrt{2}$ and $AC=4\sqrt{2}$. Hence by ratio formula the point D is $(\frac{31}{7},1)$. Slope of AD by $\frac{y_2-y_1}{x_2-x_1}=0$.
$\therefore$ slope of a line perpendicular to AD is $\infty$.
Any line through C perpendicular to this bisector is $\frac{y-5}{x-5}=m=\infty$; $\therefore$ x-5=0