• question_answer 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                                              [Roorkee 1994] A)            $y-5=0$                                    B)            $x-5=0$ C)            $y+5=0$                                   D)            $x+5=0$

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 $\left( \frac{31}{7},1 \right)$. Slope of $AD$by $\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=0$.                    \ Slope of a line perpendicular to $AD$is $\infty$.                    Any line through C perpendicular to this bisector is $\frac{y-5}{x-5}=m=\infty$; \ $x-5=0$.